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J A N L U B I N A

 jaa

n

n

bilu  

A P P L E P I E O R D E R

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2

A P P L E P I E O R D E RIt doesn‘t fit the margin, but it does go in my book. 

„The scientist dose not study Nature because it is useful; hi studies it because hi delights in it,and he delights in it because it is beautiful. If Nature were not beautiful, it would not be worthknowing, and if Nature were not worth knowing, life would not be worth living.‖ 

Jules Henri Poincaré.  

―Mathematicians do not study objects, but relations between objects. Thus, they are free toreplace some objects by others so long as the relations remain unchanged. Content to them is

irrelevant: they are interested in form only‖. 

Jules Henri Poincaré.  

For the memory of 

Katharina LubinaMarch 7, 2009

4

6

7

89

10

12

13

1415

16

17

18

19

2021

22

23

24

2627

28

25

29

30

31

3233

34

36

37

3839

40

35

41

42

43

4445

46

47

48

5051

52

49

53

54

5657

58

55

59

60

61

6263

64

66

67

6869

70

65

71

72

73

7475

76

78

79

8081

82

77

83

84

8687

88

85

89

90

9293

94

9196

97

9899

100

1235

11

95

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3

"Sed omnia in mensura, et numero, et pondere diposuisti.‖ Sapientia 11, 21.

World and all what oneself it on him finds it carries mathematical structures. So God createdhim with mathematical point of sight. Given the man the strength to him from God of reason,plan of his building can the discoveries. It is the mathematics so the key to understanding of 

world. In peer with her development, she went the change of aims what her was placed. Itdoes not serve the mathematics the endeavor to better perception only and the understandingof nature, but it has to permit her to master.

Both points of sight, chief place of mathematics, strength of granted her certainty and theincontestability, which are useful different disciplines scientific character, as also change of this, what oneself it under this notion understands and what it the thanks were wanted was toreach her, they gave the beginning my trials they would solve problem of primes. So beganmy adventure with primes.

Human spirit and human culture they unrolled such formal system of thinking, to can

formulae recognize, to classify and to use. We call him mathematics, because it ismathematician the science of formulae.The only right of existence for mathematician, the desire of discovery of new formulae is andthe inherent in rights of nature regularities, as and announcing this what it will happen.Though looking for formulae and structures it is mathematician's activity, then proper his task is formulation their in irrefutable proofs.Numbers are the simplest mathematical object, and the simplest formulae of nature arenumerical, because perfect relations between numbers reign.

The basic theorem of algebraically theory of numbers sounds: All numbers descend from one..

"O M N I A E X U N O‖ 

Theorem this be leaning on system of certainties, what Italian mathematician Giuseppe Peanoin 1889 r. submitted on unquestionable truth the undemanding proof "parental power‖ of number one, giving the same bearing foundation theorem taking out from one all naturalnumbers.The forcible model of principle "all of one‖/ OMNIA EX UNO / is the draught of naturalnumbers 1, 2, 3, 4,.. in which number one, it is for all numbers the "point of exit‖. One isreally only corner stone the whole draught of numbers on which is bases here. In gathering of natural numbers the number one is the class alone for me, the "Unity‖ is called also from here.  

One is only number, which does not change when oneself it divide her by her, or it increases.Geometric he be introduced as point, by what his elusiveness be expressed. Point's the lack of length, width and height, upper or bottom side, any color, and even the position.It was cannot say even, points are round, because taking at all closely they do not widen. Thisborders on with miracle directly, that attributes number these essential and necessary featureswithout which the whole draught of natural numbers would not can exist. Then she is the"Point‖ of reference, what to which all natural numbers graphic be co-ordinated, introducedon two co-ordinates the a and the b. She is the "source‖ even and odd units also from which itcomes into being whole row of prime and folded numbers. Exists such "Unity‖, from whichthe whole wealth of world results, as one axiom will suffice as foundation the fine edifice of arithmetic.

"It exists such number 1 possessing property, which treats to every number - n‖: 

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4

n · 1 = n = n + 0 1 · p = p

Really comparative size with 2 enters in life, in support about which , all differentmeasurable can pit .She beyond this is with nature the number of "unification‖ from twounit's make one number.

1 =

312  

2 =3

12 +

3

12  

p = a + b b = p - a a =3

1 p 

p = )

3

1(

3

1

p p

 pp = a + ( p - a )

Prime numbers this "building blocks‖, from which be built all different natural numbers. Notwe will find them however in multiplication table, because number first cannot be the result"sensible‖ operation of multiplication, but only addition. Every prime numbers is the sum two components defining her place in draught of naturalnumbers p = a + b.

Component a =3

1 pthen they came into being with divisible numbers even quotient by 3.

Component b = p - a then difference among prime number, and even quotient.

It number 2 is only even prime number and across her principle "larger about one‖ it willbecometransferred on next natural numbers, guaranteeing contact and progress in draught.

1 =3

12  

2 = 1 + 11

3 = 1 + 2 = 1 + ( 3 - 1 )

14 = 2 + 2 1

5 = 2 + 3 =3

15 + ( 5 -

3

15)

16 = 3 + 3

7 = 2 + 5 =3

17 + ( 7 -

3

17 )

1

8 = 3 + 5

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5

All prime numbers precede or they follow after divisible number by 3 eg. 2, 3, 17, 18, 19,23, 24, for except 3 even. Eureka!

p n21 17 + 1 = 18 = 19 – 1 because 3(p n2)1 3(7-1) = 18

p n21 5 + 1 = 6 = 7 – 1 because 3(p n2)1 3(3-1) = 6

5 11 17 23 29 41 47 59 71 83 101 107 113 131

6 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96, 102, 108, 114, 120, 126, 132,

7 13 19 31 37 43 61 67 73 79 97 103 109 127

Prime number is about one larger or smaller from previous or following even divisiblenumber by 3.

p = 2n 1 e.g. 1999 = 1998 + 1

Odd numbers, as this results from every multiplication table, are product of prime numbers,odd and almost prime.

3

n= b n = 3 b (2 b + b)

9 = 2 (3) + 3In odd numbers the relation of even components to odd is always 2 : 1,we can from here write n = 2b + b e.g. 15 = 2(5) + 5 21 = 2(7) + 7

If decomposes the sum of units of number on the components the being in relation expressedin equation n = 2b + b, then it is surely then the odd number.

Triangle of numbers.

"Number is collection of units‖, Euclid defines her in book "Elements‖ so.

"Tria juncta in uno" / Three join in one / In triangle of numbers the Principle "larger aboutone" the links units in integers.

If decomposes the sum of individuals of number on the components the being in relationexpressed in equation p = a + ( p - a) this is surely then prime number.

If every number natural larger from one, can be written in aspect of the sum of unity or the

sum primes, and ―unity ‖ is quotient of the sum of prime and ―unity‖ by next number prime,then the infinite sum of natural numbers is equals infinite sum ―unity‖,  e. g. 4 =  1 + 1 + 1 + 1 N 1  

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6

Since natural numbers is infinitely many, then and primes is infinitely many, because alldifferent with them consist, and what with this goes also pair of twin primes. This is yetcompletely comprehensible! And simultaneously not natural in natural numbers.

All natural numbers which carry in me principle ―larger about one‖, can be written as the sumof ones, or primes 2 and 3.

2k = p +  p…  2k = )2(   1 =312   n = p‘ +  p‘ n = )3(  

1 + 1 = 2 = 1(2)1 + 1 + 1 = 3 = 1(3)

1 + 1 + 1 + 1 = 4 = 2 + 21 + 1 + 1 + 1 + 1 = 5 = 2 + 3

1 + 1 + 1 + 1 + 1 + 1 = 6 = 2 + 2 + 21 + 1 + 1 + 1 + 1 + 1 + 1 = 7 = 2 + 2 + 3

1 + 1 + 1 + 1 + 1 + 1+ 1 + 1 = 8 = 2 + 2 + 2 + 21 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 9 = 3 + 3 + 3

1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 =10= 2 + 2 + 2 + 2 + 2

1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 =11= 2 + 2 + 2 + 2 + 31 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 =12= 2 + 2 + 2 + 2 + 2 + 2

1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 =13= 2 + 2 + 2 + 2 + 2 + 31 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 =14= 2 + 2 + 2 + 2 + 2 + 2 + 2

1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 =15= 3 + 3 + 3 + 3 + 315 + 1 =16= 2 + 2 + 2 + 2 + 2 + 2 + 2 + 216 + 1 =17= 7(2) + 1(3)  p = n(p) + p‘ 17 + 1 =18= 9(2)18 + 1 =19= 8(2) + 1(3)19 + 1 =20= 10(2)24 + 1 =25=5(2) + 5(3) „p‖= n(p) + n(p‘) 34 + 1 =35=7(2) + 7(3)+ 1 =  

k + k = n 1 + 1 = 2 + 1 = 3, 4 = 2 + 2, 5 = 2 + 3, 6 = 4 + 2, 7 = 4 + 3, 8 = 6 + 2, 9 = 6 + 3,...

12

34

56

78 9 10

1112

1314

1516

1718

1920

2122

2324

2526

2728

2930

3132

3334

3536

3738

3940

4142

4344

4546

4748 49

5051

5253

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 223 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3

24

6 68 8 1010

12121414

16161818

20202222

24242626

28283030

32323434

36363838

40404242

44444646

484850

3

50

2

4

0

10

20

30

40

50

60

1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53

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7

Theorem: Every number prime 3 consists from ternary and number even diminish aboutthree.

N = 2k  – 1 p = 3 + [(2k  – 1) – 3] 2k  – 1 – 3 = 2k  – 4 = 2k  – 2 k > 2

1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 +

  \ 3 / + \ 2 / + \ 2 / + \ 2 / + \ 2 / + \ 2 / + \ 2 / + \ 2 / + \ 2 / + \ 2 / + \ 2 / 1 2 3 4 5 6 7 8  9  10 11 12 13 14  15  16 17 18 19 20  21  22 23 24

p = )2(3 n p = 3 + 2k  – 323 = 10(2) + 1(3) 37 = 17(2) + 1(3)29 = 13(2) + 1(3) 41 = 19(2) + 1(3)31 = 14(2) + 1(3) 43 = 20(2) + 1(3)47 = 22(2) + 1(3) 67 = 32(2) + 1(3)53 = 25(2) + 1(3) 71 = 34(2) + 1(3)59 = 28(2) + 1(3) 73 = 35(2) + 1(3)

61 = 34(2) + 1(3) 79 = 38(2) + 1(3)83 = 40(2) + 1(3) 89 = 43(2) + 1(3)97 = 47(2) + 1(3) 107 = 52(2) + 1(3)

101 = 49(2) + 1(3) 109 = 53(2) + 1(3)103 = 50(2) + 1(3) 113 = 55(2) + 1(3)

The whole infinite file of natural numbers consists from infinite quantity 2 and 3, which are―units ‖ all numbers. 

N = (2) + (3) = (1)

N = 1 =  3

12  

Alone meanwhile ―units‖ they are even and odd multiplicity ―unity‖ 1 =3

12 1(2), 1(3).

In this way was proved mathematically indirectly that all numbers descend from one, becausethey consist from ―units ‖. 

P = 13

12

3

1

p pe. g. 179 = 1

3

11792

3

1179

 

p = [ 2(k) – 2] + 3 727 = [ 2(363) – 2] + 3 = (726 – 2) + 3

As to that indivisibility. Euler announced, that possesses algebraically proof on existence

God. His form looked so:  xn

ban

, hence God exists. If in place of algebraically signs to

substitute three first numbers, then for mathematician equation this can to be proof on

indivisibility number 3 13

123

. Philosopher can tell, that only plurality can to get unite.

Theologian meanwhile it will say: Father and Son with triple only Holy Spirit it is indivisibleHoly trinity, hence exists one God in three persons. And all are right, because plurality is theform of unity.

See this on example primes, which despite that they consist from many individuals, they existas individual indivisible numbers.

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9

6 7

2 3 4 5 1 =  

  

 

3

121 =

 

  

 

3

12 

1 =

 

 

 

 

3

121 =

 

 

 

 

3

121 =

 

 

 

 

3

121 =

 

 

 

 

3

12 

1 =  

  

 

3

121 =

 

  

 

3

121 =

 

  

 

3

121 =

 

  

 

3

121 =

 

  

 

3

121 =

 

  

 

3

12 

0 - - - - - - - - - - - 1 =  

  

 

3

12------------- 1 =

 

  

 

3

12--------------------- 1 =

 

  

 

3

12 

1 =  

  

 

3

121 =

 

  

 

3

121 =

 

  

 

3

121 =

 

  

 

3

121 =

 

  

 

3

121 =

 

  

 

3

12 

1 =  

  

 

3

121 =

 

  

 

3

121 =

 

  

 

3

121 =

 

  

 

3

12 

1 =   

  

312 1 =

  

  

312  

Three, as all odd numbers possesses symmetry creator "centre‖. Centre of three is two,quintuple the three, seven the five etc, hence with 2 and 3 consist all natural numbers and thethree the state "centre‖ of all odd numbers, and all natural numbers are the quotient of three.

Every plurality is the plurality of „units‖. 

N = 1  

  

 

3

12 

From those "units" be folded the whole cosmos, world of minerals, plants, animals and human

existences. "Man was created on range and similarity of only God's‖, it tells us so many belief, "all consists from the smallest and indivisible particles‖ - natural sciences teach so.Both on their way struggle about formulating one and the same truth: Such Unity is, fromwhich whole plurality descends.

0 : 1 = 03

0  

13

3

3

12

 

3

12

3

12

= 23

6  

312

312 + 3

39

312  

3

12

3

12

+

3

12

3

12

= 4

3

12  

3

12

3

12

+

3

12

3

12

+ 5

3

15

3

12

 

3

12

3

12

+

3

12

3

12

+

3

12

3

12

= 6

3

18  

3

12

3

12

+

3

12

3

12

+

3

12

3

12

+ 7

3

21

3

12

 

3

12

3

12 +3

12

3

12 +3

12

3

12 +3

12

3

12 = 83

24  

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10

3

12

3

12

+

3

12

3

12

+

3

12

3

12

+

3

12

3

12

+ 9

3

27

3

12

 

3

12

3

12

+

3

12

3

12

+

3

12

3

12

+

3

12

3

12

+

3

12

3

12

= 10

3

30  

If sum two following numbers prime form n and n + 2, it is divisible by 12, then they are

surely then twin numbers.

p + (p`+ 2) =12

)2( p p 

12 24 36 60 84  / \ / \ / \ / \ / \ 

5 + 7 11 + 13 17 + 19 29 + 31 41 + 43"Twin ‖ call pair of numbers prime between which steps out the even number divisible by 3,e.g. 5-6-7, 11 -12- 13, 17-18-19, 29-30- 31, 41 -42- 43, 59-60- 61,. it but not pair 131 -132-133, or 10 000 037 -10 000 038- 10 000 039, it because 2,3,5 number can was take apart onprime factors 133 = 7(19), 10 000 039 = 7(1 428 577), 10 000 037 = 43(232 559).

Divide the sum of twin pair by 12, we will find out near which following even number

divisible by 3, came into being numbers prime. 502312

3013930137

 because 5023 · 6

=30138/3During when sequence of the reciprocal of primes is divergent / with reason of growing space

(n)6 /   pprim p

1, sequence of the reciprocal of all twin numbers is convergent / because they

near mutually on distance 2/ 

 

  

 

 prim p p p2 2

11< ∞, and his exact value be well-known!

The six- wide array further helps to demonstrate the otherwise still unproven conjecture thatthere must be infinitely many twin primes.In the six- wide rectangular array, the consecutive multiples of each number higher than threelay on a straight line from zero to that number and beyond, and on periodic parallels to thatline further ―down‖ if we begin writing the numbers from the ―top‖ of the array. Soon after this ―factor line‖ leaves the array rectangle on one side, a parallel to it re - enters it on the otherside, farther down in the array at the next such multiple. Each so broken factor line thuscascades in evenly spaced stripes down the layers of the array. Whenever the factor lines fromall the primes above a given layer in the six- wide array happen to miss the two spaces beforeand after the 6n column in that layer, the entries there are not multiples of any among those

prior primes. They are therefore primes themselves and from a pair of twin primes, asillustrated in following table. This approach to the way Euclid suggested to multiply all the

 primes, up to a supposedly ―largest‖ one, with each other. He imagined this equally unfeasiblemultiplication to show that the result plus or minus one is either a prime, or else the productof two or more primes larger than the previously ―largest‖. By this method, he proved thatthere always exists a prime larger than any allegedly ―largest‖ one, and that there must thus bean infinite quantity of them.

It is from in pairs twin numbers similarly. Always the foundling oneself the larger pair of twinnumbers from allegedly "largest‖, and by then sequence their has not the end.  

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11

0  1  2  3 

4  5  6  7  8  9 

10  11  12  13  14  15 

16  17  18  19  20  21 

22  23  24  25  26  27 

28  29  30  31  32  33 

34  35  36  37  38  39 

40  41  42  43  44  45 

46  47  48  49  50  51 

52  53  54  55  56  57 

58  59  60  61  62  63 

64  65  66  67  68  69 

70  71  72  73  74  75 

76  77  78  79  80  81 82  83  84  85  86  87 

88  89  90  91  92  93 

94  95  96  97  98  99 

100  101  102  103  104  105 

106  107  108  109  110  111 

112  113  114  115  116  117 

118  119  120  121  122  123 

124  125  126  127  128  129 

130  131  132  133  134  135 136  137  138  139  140  141 

142  143  144  145  146  147 

148  149  150  151  152  153 

154  155  156  157  158  159 

160  161  162  163  164  165 

166  167  168  169  170  171 

172  173  174  175  176  177 

178  179  180  181  182  183 

184  185  186  187  188  189 

190  191  192  193  194  195 

196  197  198  199  200  201 

202  203  204  205  206  207 

208  209  210  211  212  213 

214  215  216  217  218  219 

The sum the pair of twin numbers equals sum of first three successive the pair as thetriangular multiplicities number 12, and the next different multiplicities in dependence fromthis, which they in turn are the pair with infinite set of numbers.

)2(

)(12)6,3,1(12 p p

 N n   n p p 12

)2(  

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12

Theorem: They twin numbers prime, place oneself before and after even number divisibleby 3, when sum of their ciphers of units equal 4, 10 or 16.

11 + 12/3 + 13 17 + 18/3 + 19 29 + 30/3 + 31 2087 + 2088/3 + 20891 + 3 = 4 7 + 9 = 16 9 + 1 = 10 7 + 9 = 16

Only primes, which even components are even, create the not only that is to say, of twinnumber e.g. 5 and 7, 11 and 13, form n and n + 2, but once even number ‖triplets‖: 3, 5, 7,form n and n + 2 and n + 4, in which this even components are even : -1 -3 -5 =2It exist also one peer of successive prime 2 and 3 which are not "twins‖ yet only "successive‖. 

p, (p +2), 5, (5 + 2), 11, (11 + 2), 29, (29 + 2), 107, (107 + 2)

511

1319

31

37

41

43

47

61

59

67

71

7379

97

101

107

109

127

131

137

139

151

157167

181

179

2

3 7

17

2329 53 8389 103113149 163 173

1 2 3

7

13

19

31

37

43

61

67

735

11

17

23

29

41

47

53

59

71

6

12

18

24

30

36

42

48

54

60

66

72

0%

20%

40%

60%

80%

100%

1

2 3 4 56

78

910

11

12

13

14

15

16

17

18

19

20

21

22

2324

25

26

27

2829

3031

3233

343536373839404142

4344

4546

47

48

49

50

5152

53

54

55

56

57

58

59

60

61

62

63

6465

6667

6869

70 71 72 73

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13

12 24 36 60 84

  / \ / \ / \ / \ / \ pd =12

)2( nn 

5 + 7 11 + 13 17 + 19 29 +31 41 + 43

p, (p +2), 5, (5 + 2), 11, (11 + 2), 29, (29 + 2), 107, (107 + 2)

511

1319

31

37

41

43

47

61

59

67

71

7379

97

101

107

109

127

131

137

139

151

157167

181

179

2

3 7

17

2329 53 8389 103113149 163 173

  _1 + _3 = 4 _7 + _9 = 16 _9 + _1 = 10

1237

13

19

31

3743

61

67

73

5 11

17

23

29

41

47

53

59

71

6 12

18

24

30

3642

48

54

60

66

72

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All natural numbers congruent to me according to module n‘ – n 0 mod 6.

Crossing through prism light, it appears as rainbow of colors. Goes out with unity of numbernatural put on shape six waves about length 6. Congruent to me according to module 6numbers they divide on three groups of even numbers and odd / 2, 4, 6 / 3 -2- 5 -2- 7 / keeping among me solid space 2 and 6 in every group 2/8, 3/9, 4/10, 5/11, 6/12, 7/13.

(p - 1) + (p + 1) /2 = p (53 - 1) + (53 + 1) /2 = 53

0

1

2

3

4

5

6

7

8

10

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17

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50

52

53

54

n' - n = 0 mod 6

1 2 2 2 2 24 4 4 4 428

1420

2632

3

915

2127

33

4

10

16

22

28

34

5

11

17

23

29

35

6

12

18

24

30

36

7

13

19

25

31

37

0

50

100

150

200

250

Serie8 7 13 19 25 31 37

Serie7 6 12 18 24 30 36

Serie6 5 11 17 23 29 35

Serie5 4 10 16 22 28 34

Serie4 3 9 15 21 27 33

Serie3 2 8 14 20 26 32

Serie2 4 4 4 4 4

Serie1 1 2 2 2 2 2

1 2 3 4 5 6 7 8 9 10 11

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78 + 80 /2 = 79 7(8) /3 - 7(9) - 8(0) /17/ 82 + 84 /2 = 83 8(2) - 8(3) - 8(4) /3 /9/ 96 + 98 /2 = 97

9(6)/3 - 9(7) - 9(8) /21/

83

89

101

107

131

137 84

90

96

102

108

114

120

126

132

138

144

113

79

97

103

109

127

139

150 + 152 / 2 = 151 /3/ 17(9) - 18(0) / 3 - 18(1) /10/ 19(1) - 19(2) / 3 - 19(3) /6/

167

173

179

191

197

156

162

168

174

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186

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216

151

157

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181

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211

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From first ten numbers prime rise for them four characteristic the number of unity

n n + 2 n + 6 n + 8k + 1 k + 3 k + 7 k + 9

11 13 17 19

222 + 224 /2 = 223 /9/ 22(7) - 22(8) /3 - 22(9) /24/ 23(9) - 24(0) /3 - 24(1) /10/

227

233

239

251

257

263

269

281

228

234

240

246

252

258

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270

276

282

288

223

229

241

271

277

283

306 + 308 /2 = 307 /21/ 31(1) - 31(2) /3 - 31(3) /6/ 358 + 360 /2 = 359 /17/

311

317

347

353

359

312

318

324

330

336

342

348

354

360307

313

331337

349

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and they step out in tens which number after deduction 1 is divisible by 3 e.g. 10-1 = 9:

Every almost prime numbers we can introduce as sum of 2 and 3 keeping definite proportions.

In almost prime numbers, which are multiplicity of number 5 the relation of 2 to 3 amount 1 1because 5 = 3 + 2

25 = 3 + 3 + 3 + 3 + 3 + 2 + 2 + 2 + 2 + 2

25 = 5(2) + 5(3) 5(5) „p― = )3()2( nn 55 = 11(2) + 11(3) 5(11)

65 = 13(2) + 13(3) 5(13) 85 = 17(2) + 17(3) 5(17)95 = 19(2) + 19(3) 5(19) 115 = 23(2) + 23(3) 5(23)

125 = 25(2) + 25(3) 5(25) 145 = 29(2) + 29(3) 5(29)155 = 31(2) + 31(3) 5(31) 175 = 35(2) + 35(3) 5(35)185 = 37(2) + 37(3) 5(37) 205 = 41(2) + 41(3) 5(41)625 = 125(2) + 125(3) 5(5)(25) 875 = 175(2) + 175(3) 5(7)(25)

In almost prime numbers, which are multiplicity of number 7 the relation of 2 to 3 amount 2 1because 7 = 2(2) +3

35 = 3 + 3 + 3 + 3 + 3 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2

35 = 10(2) + 5(3) 7(5) „p― = )3()2(2 nn 49 = 14(2) + 7(3) 7(7)

77 = 22(2) + 11(3) 7(11) 91 = 26(2) + 13(3) 7(13)119 = 34(2) + 17(3) 7(17) 133 = 38(2) + 19(3) 7(19)161 = 46(2) + 23(3) 7(23) 203 = 58(2) + 29(3) 7(29)

2401 = 686(2) + 343(3) 7(7)(49) 2695 = 770(2) + 385(3) 7(7)(55)

In almost prime numbers, which are multiplicity of number 11 the relation of 2 to 3 amount

4 1 because 11 = 4(2) + 3 

121 = 44(2) + 11(3) 11(11) „p― = )3()2(4 nn 143 = 52(2) + 13(3) 11(13)

275 = 100(2) + 25(3) 11(25) 385 =140(2) + 35(3) 11(35)

In almost prime numbers, which are multiplicity of number 13 the relation of 2 to 3 amount

5 1 because 13 = 5(2) + 3

169 = 65(2) + 13(3) 13(13) „p― = )3()2(5 nn 221 = 85(2) + 17(3) 13(17)

637 = 245(2) + 49(3) 13(49) 715 = 275(2) + 55(3) 13(55)

In almost prime numbers, which are multiplicity of number 17 the relation of 2 to 3 amount

7 1 because 17 = 7(2) + 3

289 = 119(2) + 17(3) 17(17) „p― = )3()2(7 nn 323 = 133(2) + 19(3) 17(19)

1105 = 455(2) + 65(3) 17(65) 1309 = 539(2) + 77(3) 17(77)

In almost prime numbers, which are multiplicity of number 19 the relation of 2 to 3 amount

8 1 because 19 = 8(2) + 3 

361 = 152(2) + 19(3) 19(19) „p― = )3()2(8 nn 437 = 184(2) + 23(3) 19(23)

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In almost prime numbers, which are multiplicity of number 23 the relation of 2 to 3 amount

10 1 because 23 = 10(2) + 3

529 = 230(2) + 23(3) 23(23) „p‖ = 10n(2) + n(3) 575 = 250(2) + 25(3) 23(25) 

In almost prime numbers, which are multiplicity of number 29 the relation of 2 to 3 amount13 1 because 29 = 13(2) + 3 

841 = 377(2) + 29(3) 29(29) „p‖ = 13n(2) + n(3) 899 = 403(2) + 31(3) 29(31) 841 = 754 + 87 899 = 806 + 93

In almost prime numbers, which are multiplicity of number 31 the relation of 2 to 3 amount

14 1 because 31 = 14(2) + 3 

In almost prime numbers, which are multiplicity of number 7 the relation of 2 to 3 amount 2 1because 7 = 2(2) +3

961 = 434(2) + 31(3) 31(31) „p‖ = 14n(2) + n(3) 1147 = 518(2) + 37(3) 31(37)961 = 868 + 93 1147 = 1036 + 111

In almost prime numbers, which are multiplicity of number 37 the relation of 2 to 3 amount

17 1 because 37 = 17(2) + 3 1369 = 629(2) + 37(3) 37(37) „p‖= 17n(2) + n(3) 25271 = 11611(2) + 683(3) 37(683)1369 = 1258 + 111 25271 = 23222 + 2049

It the whole infinite set of natural numbers consists with infinite quantity of 2 and 3, whichare "units‖ of all numbers. 

N = (2) + (3) = (1)Proof  N = 1 =  

3

12  

Alone meanwhile "units‖ they are even and odd multiplicity "unity‖ 1(2), 1(3), 1 =3

12

In this way was proved indirectly the basic theorem of algebraically theory of numbers, thatall numbers descend from one, because they consist from "units‖. Only plurality can to get unite, and primes as only they possess this ability, because they areindivisible.Why a number is prime? Because could be written as two smaller numbers multipliedtogether. That is, it is not possible to represent a prime as the product of two integers a x b with a, b > 1. Let q and r be the quotient and remainder of the division of n by d . That is, foreach n and d, let n = d q + r, where r and q are positive integers and 0 ≤ r < d .

Because all prime numbers contain in me one 3, it was not possible divide here by two.Superiority meanwhile 2 it causes, that they don´t divide by three a lso. So they are indivisibleby all different numbers, and on this depends the complete primality certificate! p = n(2) + 3

2 = 1(2) + 0 3 = 0(2) + 3 5 = 2 + 3 7 = 2(2) + 3 11 = 4(2) + 3 13 = 5(2) + 3

17 = 7(2) + 3 19 = 8(2) + 3 23 = 10(2) + 3 233 = 115(2) + 3 251 = 124(2) + 3

p = 1312

31

p p e.g. 179 = 1

311792

31179

 

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p = [ 2(k) – 2] + 3 727 = [ 2(363) – 2] + 3 = (726 – 2) + 3

2 1127 = 170 141 183 460 469 231 731 687 303 715 884 105 727

3170 141 183 460 469 231 731 687 303 715 884 105 724

34 279 974 696 877 740 253 374 607 431 768 211 4573

34 279 974 696 877 740 253 374 607 431 768 211 454 

The natural numbers in scheme of 2 and 3.

If  p ≥ 2 and  p‘ ≠ 0, are whole numbers not having common divisor, than such arithmeticalsequence contains in me all natural numbers.

2, 3, n(2), 2 + 3, n(3), 3 + n(2), n(2), n(3), n(2), ... n(2) + n(3)

2, 3, 4, 5, 6, 7, 8, 9, 10, .... 10 + 15 = 25

P(n) = p, p‘, n(p), p + p‘, n(p‘), p‘+ n(p), .... n(p) + n(p‘), 

n(2) + n(3) = N

2 2

3 3

2(2) 4

2 3 5

2(3) 6

2(2) 3 7

4(2) 8

3(3) 9

5(2) 10

4(2) 3 11

4(3) 12

5(2) 3 13

7(2) 14

5(3) 15

8(2) 16

7(2) 3 17

6(3) 18

8(2) 3 19

10(2) 20

7(3) 21

11(2) 2210(2) 3 23

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8(3) 24

5(2) 5(3) 25

13(2) 26

9(3) 27

14(2) 28

13(20 3 29

10(3) 30

14(2) 3 31

16(2) 32

11(3) 33

17(2) 34

7(2) 7(3) 35

12(3) 36

17(2) 3 37

19(2) 38

13(3) 3920(2) 40

19(2) 3 41

14(3) 42

20(2) 3 43

22(2) 44

15(3) 45

23(2) 46

And here how with two primes 2 and 3 come into being all natural numbers.

1

2 2

3 3

4 2

5 3 2

6 27 3 4

8 2

9 3 6

10 2

11 3 8

12 2

13 3 10

14 2

15 3 12

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16 2

17 3 14

18 2

19 3 16

20 2

21 3 18

22 2

23 3 20

24 2

25 3 22

26 2

27 3 24

28 2

29 3 26

30 2

31 3 28

32 2

33 3 30

34 2

35 3 32

p = 3 + n(2) n (2) =n= n (3) "p" = n (2) + n (3)

12 3 4 5 67

89

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1617

18192021

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2 22

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22222

2222

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3333

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3

3

33

3

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The scheme of natural numbers

And so harmoniously develop natural numbers in support about principle "larger about one‖on the base of 2 and 3 in 360 ° the circle.

The proprieties of natural numbers repeat oneself periodically, what six numbers according to

pattern of primes.Proof: 1 + 2 + 3 = 6 p + 6  = p‘ n + 6  = n‘ „p‘― –  „p‖= 6  

012345678910

1112

1314

1516

17

18

19

20

2122

23 24 25 2627

28

29

30

31

32

33

343536

p + 6 = p'

2 1719

2329

31

37

41

43

47

53

5961

67

71

73

79

83

891311753

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With discovery of regularity in sequence of primes, that what 6 numbers repeat oneself thesame proprieties, was decoded together pattern how be distributed primes and the basing onhim periodicity of natural numbers.

1 =3

12  

Two first numbers / 1 + 2 / added to me and divided by third next number / 3 /, it equals / 1 / that is to say, again the same first number from three taking part in this working. Three firstnext numbers added to me give perfect and triangular number 6, defining length of period inwhat will repeat oneself the same proprieties in whole sequence of natural numbers.

Tres faciunt collegium, then it means three numbers they decide about whole scheme of natural numbers. It 2 (3) = 6, was can introduce all numbers from here, as sum of ones (+ 1),the twos (+ 2) and the threes (+ 3). The periodical scheme of natural numbers is so perfect, asperfect is first perfect number 6, him untouched basis.

1 + 2 + 3 = 6 = 2 · 3Ranked according to propriety natural numbers create 6 groups. Propriety of numbers in fourcentral groups repeat oneself in turn periodically, what 6 numbers. Primes create here two therows the complementary to two rows of group sixth the almost prime numbers.

Periodical scheme of natural numbers.

n 1 n 2 n 3 n 4 n 5 n 6

2

3 4

5 6

7 8 9 10

11 12

13 14 15 16

17 18

19 20 21 22

23 24 26 25

27 28

29 30

31 32 33 34

36 35

37 38 39 40

41 4243 44 45 46

47 48 50 49

51 52

53 54 56 55

57 58

59 60

61 62 36 64

66 65

67 68 69 70

71 72

73 74 75 7678 77

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24

79 80 81 82

83 84 86 85

87 88

89 90

91 92 93 94

96 9597 98 99 100

101 102

103 104 105 106

107 108

109 110 111 112

113 114 116 115

117 118 119

120 122 121

123 124

126 125

Sieve of Eratosthenes.

In the six- wide rectangular array, the consecutive multiples of each number higher than threelay on a straight line from zero to that number and beyond, and on periodic parallels to thatline further ―down‖ if we begin writing the numbers from the ―top‖ of the array. In six groups of numbers we have 3 group of even numbers (II, IV, VI), and 3 odd (I, III, V).Her multiplicities for prime number 5 on left have lain cascade, until after number almostprime 25 = 5(5).Next multiplicities for prime number 7 on right have lain her cascade, among which we havesecond almost prime number 35 = 7(5). Parallel line by her runs factor 5 falling on left in pit,

until to fourth almost prime number 55 = 5(11).The parallel line factor 7 falls from the multiplicity number 7(7) = 49 in right, until to lying inV group of almost prime number 77 = 7(11).Parallel line factor 5 falling on left in pit it crosses out their 13(5) = 65 and 15(5) = 85multiplicity.In this way they the parallel lines factors 5 and 7 cross out all almost prime numbers in I andV the group of numbers.

So of the sieve Eratosthenes is situated less than 100 numbers 25 primes.

2 3  5 7 9 1113 15 17 19 21 2325 27 29 31 33 3537 39 41 43 45 4749 51 53 55 57 5961 63 65 67 69 7173 75 77 79 81 8385 87 89 91 93 95 97 99

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I II III IV V VI

All natural numbers congruent to me according to module.n‘ – n 0 mod. 6

2n = p – 3

(2n - 1)

=p(n) 2n = p – 3

p = 2n + 3

"p"= n(2)+n(3) 

2n = p(n)

= p±1 

p = 2n + 3

"p"= n(2)+n(3) 

5 - 3 = 2 3 7 - 3 = 4 2 +3 = 5 3(2) = 6 2(2) + 3 = 7

11 - 3 = 8 3(3) = 9 13 - 3 = 10 2(4) + 3 = 11 3(4) = 12 2(5) + 3 = 13

17 -3 = 14 3(5) = 15 19 - 3 = 16 2(7) + 3 = 17 3(6) = 18 2(8) + 3 = 19

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23 - 3 = 23 3(7) = 21 22 2(10)+3 = 23 3(8) = 24 5(2 + 3) = 25

29 - 3 = 26 25 + 2 = 27 31 - 3 = 28 2(13)+3 = 29 3(10) = 30 2(14)+3 = 31

32 35 - 2 = 33 34 7(2 + 3) = 35 3(12) = 36 37

41 - 3 =38 3(13) = 39 43 - 3 = 40 41 3(14) = 42 43

47 - 3 = 44 3(15) = 45 46 47 48 14(2)+7(3)=4953 - 3 = 50 49 + 2 = 51 52 53 54 11(2 + 3) = 55

59 - 3 = 56 55 + 2 = 57 61 - 3 = 58 59 60 61

62 65 - 2 = 63 64 13(2 + 3) = 65 66 67

68 3(23) = 69 70 71 72 73

74 75 + 2 = 77 76 22(2)+11(3)=77 78 79

80 3(27) = 81 82 83 84 17(2 + 3) = 85

86 85 + 2 = 87 88 89 90 26(2)+13(3)=91

92 95-2=91+2 94 19(2 + 3) = 95 96 97

98 3(33) = 99 100 101 102 103

2n - 1 = 6n - 3 9 = 6(2) - 3 15 = 6(3) - 3 21 = 6(4) - 3 27 = 6(5) - 3 33 = 6(6) -3 39 = 6(7) - 3 45 = 6(8) - 3 2n = 6n - 4 2 = 6(1) - 4 8 = 6(2) - 4 14 = 6(3) - 4 20 =6(4) - 4 26 = 6(5) - 4 32 = 6( 6)- 4 38 = 6(7) - 4 p = 6n - 7 5 = 6(2) - 7 11 = 6(3) - 717 = 6(4) - 7 23 = 6(5) - 7 29 = 6(6) - 7 p = 6n - 5 7 = 6(2) - 5 13 = 6(3) - 52n = 6n - 6 6 = 6(2) - 6 12 = 6(3) - 6 18 = 6(4) - 6 24 = 6(5) - 6 30 = 6(6) - 6 36 = 6(7) - 642 = 6(8) - 62n = 6n - 8 4 = 6(2) - 8 10 = 6(3) - 8 16 = 6(4) - 8 22 = 6(5) - 8 28 = 6(6) - 834 = 6(7) - 8 40 = 6(8) - 8And all runs according to pattern of prime numbers which seems, that they be scattered howsavagely growing weeds among natural numbers, but only there where they create fertile soilgiving the infinite quantity of natural numbers.

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

37

38

39

40

41

42

43

44

45

46

47

48

49

50

51

52

53

54

55

56

57

58

59

60

61

62

63

64

65

66

67

68

69

70

71

72

73

74

75

76

77

78

79

80

81

82

83

84

85

86

87

88

89

90

91

92

93

94

95

96

97

98

99

100

101

102

103

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From first ten prime numbers can possess four characteristic ends: - 1, - 3, - 7, - 9, resultingwith rhythm 2/4 in what 11 +(2), 13 +(4), 17 + (2), 19 step out and they repeat oneself what21 places in tens divisible by 3. 

Table of tens in which prime numbers step out. 

I II III IV V VI VII VIII IX X XI XII XIII XIV ..|E  – 1 | - 1 : 3 | : 3 | - 1 : 3 | : 3 | - 1 : 3 | - 1 : 3 | : 3 | -1 : 3 | : 3 | - 1 : 3 |: 3 | -1 : 3 | : 3 | - 1 : 3 || | 1 | 3 | 4 | 6 | 7 | 10 | | 13 |15 | | 18 | 19 | 21 | || | | 24 | 25 | 27 | 28 | 31 | 33 | | | | | 40 | 42 | 43 || 21x2 | | | 46 | | 49 | 52 | 54 | | 57 | | 60 | | 63 | 64 || | | 66 | 67 | 69 | 70 | 73 | 75 | 76 | | | 81 | 82 | | || | | | 88 | | 91 | 94 | | 97 | 99 | |102| 103 |105 | 106 || E- 7 | - 1 : 3 | : 3 | - 1 : 3 | : 3 | : 3 |- 1 : 3 | : 3 |- 1 : 3 | : 3 | - 1 : 3 | : 3 | - 1 : 3 | : 3 | - 1 : 3 |

| | 1 | 3 | 4 | 6 | 9 | 10 | 12 | 13 | 15 | 16 | | 19 | | 22 || | | | 25 | 27 | 30 | 31 | 33 | 34 | 36 | | 39 | | | || 21x2| | 45 | 46 | 48 | | | 54 | 55 | 57 | 58 | 60 | 61 | 63 | 64 || | | | 67 | | 72 | | 75 | | 78 | 79 | | 82 | | .| | | 87 | 88 | 90 | 93 | 94 | 96 | 97 | | | | | | |

. I II III IV V VI VII VIII IX X XI XII XIII XIV XV .|E- 3 | - 1 : 3 |+1: 3|-1 : 3|+1: 3|-1 : 3 |+1: 3|-1: 3|+1: 3|-1: 3|+1: 3|-1: 3|+1: 3|-1: 3|+1:3 |- 1: 3 || | 1 | 2 | 4 | 5 | 7 | 8 | 10 | 11 | | | 16 | 17 | 19| | 22 || | | 23 | 25 | 26 | 28 | 29 | 31 | | | 35 | 37 | 38 | | | 43 ||21x3 | | 44 | 46 | | | 50 | 52 | | | 56 | | 59 | 61| | 64 |

| | | 65 | 67 | 68 | | | 73 | 74 | | 77 | | | 82| | 85 || | | 86 | 88 | | 91 | | | 95 | | 98 | | 101| 103| | 106 ||E- 9 | - 1 : 3 |+1: 3 |-1 : 3|+1:3 |-1 : 3 |+1:3 |-1: 3|+1: 3|-1: 3|+1: 3|-1: 3|+1: 3|-1: 3|+1:3 |- 1: 3 || | 1 | 2 | | 5 | 7 | 8 | 10 | | 13 | 14 | | 17 | 19 | 20 | 22 || | | 23 | | 26 | | | | | 34 | 35 | 37 | 38 | 40 | 41 | 43 ||21x3| | 44 | | 47 | 49 | 50 | | | | 56 | | 59 | 61 | | || | | 65 | | | 70 | 71 | 73 | | 76 | | | 80 | 82 | 83 | || | | | | | 91 | 92 | | | | | 100 | 101| 103| 104 | 106 |

The table of tens in which step out prime numbers betrays us sure regularity what it reigns insequence of prime numbers. Not accidentally has written down in this table of ending of 

prime numbers in this way 1 - 7 = 6 = 3 - 9. This shows that the regularity what 6 numbersfrom what can step out prime numbers, crosses over on the whole sequence the naturalnumbers, which of propriety what they 6 numbers repeat oneself in six groups. Primes with ending 1 - 7 create XIV ranks, in which their endings repeat oneself what 21 and42 place, and with ending 3 - 9 create XV ranks, in which their endings repeat oneself what21, 42 or 63 places and they in both cases are then divisible numbers by 7, which will befurther great meaning.Prime, even and odd numbers they create "twelve segmental cycles‖. 

5 + 7 = 12 = 2 + 4 + 6 = 12 = 3 + 9

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28

Periodical scheme of prime numbers results with principle the "twelve of segmental cycles‖ in360 numbers which be comprises 30. Multiply thirty by unitary length of period (7) primes inwhat step out 30 · 7 = 210 - receive decimal length of period of prime numbers.

3

1'

3

1''210

3

1

3

1'

p p p p 

11 - 3931 i 17 - 4217

+

1/3

- 1

 /3

+

1/3

-

1/3

+

1/3

+

1/3 - 1/3

+

1/3

-

1/3

+

1/3

-

1/3

+

1/3

-

1/3

+

1/3

11 31 41 61 71 101 131 151 181 191 211

241 251 271 281 311 331 401 421 431

461 491 521 541 571 601 631 641

661 671 691 701 731 751 761 811 821

881 911 941 971 991 1021 1031 1051 1061

1091 1151 1181 1201 1231

1291 1301 1321 1361 1381 1451 1471 1481

1511 1531 1571 1601 1621

1721 1741 1801 1811 1831 1861 1871 1901

1931 1951 2011 2081 2111

2131 2141 2161 2221 2251 2281 2311

2341 2351 2371 2381 2411 2441 2521 2531

2551 2591 2621 2671 2711 2731 2741

2791 2801 2851 2861

2971 3001 3011 3041 3061 3121

3181 3191 3221 3251 3271 3301 3331 3361 3371

3391 3461 3491 3511 3541 3571 3581

3631 3671 3691 3701 3761

3821 3851 3881 3911 3931

17 37 47 67 97 107 127 137 157 167 197 227

257 277 307 317 337 347 367 397

457 467 487 547 557 577 587 607 617 637 647

677 727 757 787 797 827

877 887 907 937 947 967 977 997

1087 1097 1117 1187 1217 1237 1277

1297 1307 1327 1367 1427 1447 1487

1567 1597 1607 1627 1637 1657 1667 1697

1747 1777 1787 1847 1867 1877 1907

1987 1997 2017 2027 2087

2137 2207 2237 2267 2287 2297

2347 2357 2377 2417 2437 2447 2467 2477

2557 2617 2647 2657 2677 2687 2707

2767 2777 2797 2837 2857 2887 2897 2917 2927 2957

3037 3067 3137 3167

3187 3217 3257 3407 3347

3407 3457 3467 3517 3527 3547 3557

3607 3617 3637 3677 3697 3797

3847 3877 3907 3917 3947 3967 40074027 4057 4127 4157 4177 4217

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29

13 - 3793 i 19 - 4409

+1/3 -1/3 +1/3 -1/3 +1/3 -1+ 1

: 3- 1

: 3+ 1

: 3- 1

: 3+ 1

: 3- 1

: 3+ 1

: 3- 1

: 3

13 23 43 53 73 83 103 113 163 173 193 223

233 253 263 283 293 313 353 373 383 433

443 463 503 523 563 593 613 643

653 673 683 733 743 773 823 853

863 883 913 953 983 1013 1033 1063

1093 1103 1123 1153 1163 1193 1213 1223

1283 1303 1373 1423 1433 1453 1483

1493 1523 1543 1553 1583 1613 1663 1693

1723 1733 1753 1783 1823 1873

1913 1933 1973 1993 2003 2053 2063 2083 21132143 2203 2213 2243 2273 2293

2333 2383 2393 2423 2473 2503

2543 2593 2633 2663 2683 2693 2713

2753 2803 2833 2843 2903 2953

2963 3023 3083 3163

3253 3313 3323 3343 3373

3413 3433 3463 3533 3583

3593 3613 3623 3643 3673 3733 3793

19 29 59 79 89 109 139 149 179 199 209 229

239 269 349 359 379 389 409 419 439

449 479 499 509 569 599 619659 709 719 739 769 809 829 839

919 929 1009 1019 1039 1049 1069

1109 1129 1229 1249 1259 1279

1289 1319 1399 1429 1439 1459 1489

1499 1549 1559 1579 1609 1619 1669 1699

1709 1759 1789 1879 1889

1949 1979 1999 2029 2039 2069 2089 2099

2129 2179 2239 2269 2309

2339 2389 2399 2459 2539

2549 2579 2609 2659 2689 2699 2719 2729 2749

2789 2819 2879 2909 2939

2969 2999 3019 3049 3079 3089 3109 3119 3169

3209 3229 3259 3299 3319 3329 3359

3389 3469 3499 3529 3539 3559

3659 3709 3719 3739 3769 3779

3889 3919 3929 3989

4019 4049 4079 4099 4129 4139 4159 4219

4229 4259 4289 4339 4349 4409

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30

Spiral twelve segmental cycles of primes.

2 5 11 17 23 29 41 47 53 59 71

3 7 13 19 31 37 43 61 67 73 79

2 83 89 101 107 113 131 137 149

2 97 103 109 127 1393 151 157 163 191 197

3 167 173 179 181 193 199 211

4 227 233 239 251 257 263 269 281 293

4 223 229 241 271 277 283

5 307 313 331 337 349 367 373 379

5 311 317 347 353 359 383

6 397 409 421 433 439 457 463

6 389 401 419 431 443 449 461

7 487 499 503 509 521

7 467 479 491 523 541

8 563 569 587 593 599 6178 547 571 577 601 607 613 619

9 641 647 653 659 677 683

9 631 643 661 673 691

10 701 719 743 761 773

10 709 727 733 739 751 757 769

11 797 809 821 827 839

11 787 811 823 829 853

12 857 863 881 887 911 929

12 859 877 883 907 919

13 941 947 953 971 977 983

13 937 967 991 997 1009

2 + 3 = 5 + 7 = 12

23571113 83 151

227223

307

311

389

467

547

701

857859

937

171989 157

233

229

313

317397631

709

787

863 941

23

97

163

167239

401

479

94729

31

101 103

173

241

409487563

641

643

719

797

877

953

37

107 109

179

251

331

491

569

571647

727

881

88341 43

113

257

337

421

419499

577653

733

809

811

887967

47

181

263

347

503

659

661

739

971

53

127

191

269

349

353433

431

509

587

743

821

823

977

59

61197

193

271

359

439

593

673

751

827

829

907

983

67131

199

281

277

443

521

523599

601

677

757

911

991

71

73

137 139

283

367

449

607

683

761

839

919997

79

211

293

373

457

613

691

769

149

379

383

463

461541617619

773

853

929

1009

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31

Spiral scheme of natural numbers.

With spiral arrangement of primes and almost prime result spiral arrangement of all naturalnumbers, what we see in following table.

2 3 45 6 7 8 9 10 11

12 13 14 15 16 17 18

19 20 21 22 23 24 25

26 27 28 29 30 31 32

33 34 35 36 37 38 39

40 41 42 43 44 45 46

47 48 49 50 51 52 53

54 55 56 57 58 59 60

61 62 63 64 65 66 67

68 69 70 71 72 73 74

75 76 77 78 79 80 81

82 83 84 85 86 87 88

89 90 91 92 93 94 95

96 97 98 99 100 101 102

103 104 105 106 107 108 109

110 111 112 113 114 115 116

117 118 119 120 121 122 123

124 125 126 127 128 129 130

131 132 133 134 135 136 137

138 139 140 141 142 143 144

145 146 147 148 149 150 151152 153 154 155 156 157 158

159 160 161 162 163 164 165

166 167 168 169 170 171 172

173 174 175 176 177 178 179

180 181 182 183 184 185 186

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The spiral sequence of natural numbers / primes and almost prime /.

Spiral of primes.

Spiral of almost prime.

a = b mod 17

0 1 2345

67

891011

12

13

14

1516

17 1819

20

21

22

23

242526

27

28

29

30

31

32

3334 35

36

37

38

39

40

41

424344

45

46

47

48

49

5051

52

53

54

55

56

57

58

5960

61

62

63

64

65

66

6768

69

70

71

72

73

74

75

7677

78

79

80

81

82

83

84

8586

87

88

89

90

91

92

9394

95

96

97

98

99

100

101

p' - p= 0 mod 17 = 19 - 2

0 23

5

711

13

1719

23

29

31 37

41

43

47

53

59

61

67

71

73

79

83

89

97

101

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33

It is true in spirals primes and almost prime congruent according to different modules,however difference between them is common module all natural numbers 23 -17 = mod 6, what show above mentioned graphs. Module 40 = 17 + 23 arranges natural numbers ininfinite spiral. 

Congruence according to module 6 shine numbers with all colors of rainbow.

"p" - p = mod 23 = 25 - 2

0 225

35 55

65

7785

91

95

115

119

121

125

133 143

145

155

161

169

175

185

187

203

205 209

215

217

221

235

245

247

253

259

265

275

287

289

295

299

301

305

323

325

329

335

341

343

355

361

365

371

377

385

391

395

403

407

413

415

425

427

437

445

451

455

473

475

481

485

493

497

2 + 3 + 5 + 11 + 19 = mod 40

012 345678910

11121314

151617181920212223242526

272829

303132333435

3637 3839404142 4344

4546

47484950

5152

5354

5556

57585960616263

6465

6667

68

69

70

71

72

73

7475

7677

78 79 80 81 8283

8485

86

87

88

89

90

91

92

93

94

95

9697

9899100101102

103104

105

106

107

108

109

110

111

112

113

114

115

116117

118119 120 121 122

123124

125

126

127

128

129

130

131

132

133

134

135

136

137138

139140141142

143

144

145

146

147

148

149

150

151

152

153

154

155

156

157158

159 160 161162

163

164

165

166

167

168

169

170

171

172

173

174

175

176

177

178179180181182

183

184

185

186

187

188

189

190

191

192

193

194

195

196

197

198199

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34

The cross of numbers is bases on number 6, appointed by primes 2 and 3, which squeeze outhis brand on whole scheme of natural numbers.

a = b mod 6

06121824303642485460

667278

84

90

96

102

108114

120126132138

144

150

156

162

168

174

171319253137434955

6167

73

79

85

91

97

103

109

115121127

133

139

145

151

157

163

169

175

281420263238445056

6268

74

80

86

92

98

104

110

116122128

134

140

146

152

158

164

170

176

3915212733394551

5763

69

75

81

87

93

99

105

111

117123129

135

141

147

153

159

165

171

177410162228344046

5258

64

70

76

82

88

94

100

106

112

118

124130

136

142

148

154

160

166

172

178

51117232935 4147

53

59

65

71

77

83

89

95

101

107

113

119

125131

137

143

149

155

161

167

173

179

3 + 3 = 6 = 2 x 3

123

56

7

1112

13

1718

19

2324

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35

It comes from structures of cross of natural numbers from congruence of primes and almostprime according to module 2(2)2. 

+

0 235

7

1113

17

19

23 25

29

31

3537

41

43

47 49

53

55

5961

65

67

71 73

77

79

8385

89

91

95 97

101

103

107109

113

115

119 121

125

127

131133

137

139

143

+

0

2

35

7

1113

17

19

23 25

29

31

3537

41

43

47 49

53

55

5961

65

67

71 73

77

79

8385

89

91

95 97

101

103

107109

113

115

119 121

125

127

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36

Sequence of primes and almost prime in five groups about the same endings are possible thethanks their congruence according to module 5.

3 + 2(2)2 = 11 + 2(2)2 = 19 + 2[2(2)2] = 35 + 2(2)2 = 43 + 2[2(2)2] = 59 + 2(2)2 = 67

0

17 25 41 49 65 73 89 97 113 121

2

311

19

35

43

59

67

83

91

107

1155

13

29

37

53

61

77

85

101

109

125

7 23 31 47 55 71 79 95 103 119 127

0%

10%

20%

30%

40%

50%

60%

70%

80%

90%

100%

  p   '  +   p

   (  p   )  p   =   p

   "

Serie8 7 23 31 47 55 71 79 95 103 119 127

Serie7Serie6 5 13 29 37 53 61 77 85 101 109 125

Serie5

Serie4 3 11 19 35 43 59 67 83 91 107 115

Serie3 2

Serie2 17 25 41 49 65 73 89 97 113 121

Serie1 0

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

- 9 - 3 - 7 - 1 - 5

0

23

5

1317

19

23

25

29

35

3743

47

49

53

55

59

65

7377

79

83

85

89 91

95

97

7

1131

41

61

67

71

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37

So looks clocks of primes measuring it in rhythm 2/4 (6).

According to this rhythm of primes flow away us days in four times year.

(p-1) + (p+1) / 2 = p2 3

7

13

19

31

37

43

61677379

91

97

103

109

127

5

11

17

23

29

41

47

53

5971

83

89

101

107

113

131139137

2 + 3 = 5 x 73 = 3652 3

7

13

19

31

3743

61

67

735

11

17

23

29

41

47

53

59

71 46

89

10

12

14

15

16

18

20

21

22

24

25

26

27

28

30

323334353638394042

4445

46

48

49

50

51

52

54

55

56

57

58

60

62

63

64

65

66

6869

7072

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Binary and Ternary Goldbach's Conjecture, equation Pythagoras and great Fermat'stheorem.

Creative process in mathematics begins from conjecture. Mathematical conjecture really thenit becomes theorem, when we have on his truth irrefutable proof.

Theorem: Even numbers are "larger about 1‖ from one's odd, prime or almost primepredecessor, and so they are duplication different natural number.

Proof: (2n – 1) 1 \ „p‖ 1 = 2n 

p 1 / 

(2n – 1) + 1 = 2n 1 + (2n – 1) = 2n = 3p - p = p + p‘7 – 5 mod 2 5 –  3 p‘- p = n/2    p‘ + p = 2n 6 – 4 mod 2 10 – 8

3(2) – 2 = 4 3(3) – 3 = 6 3 + 5 = 8 3(5) – 5 = 10 5 + 7 = 12 3(7) – 7 = 14 11 + 5 = 16

This theorem proves the just truth Binary Goldbach's Conjecture, that every even largernumber than 2 is the sum two primes. Both even numbers how and primes congruent to meaccording to modules 2, that is to say differences between them divisible they are by 2. Fromhere simple conclusion, if differences this and sum two primes divisible they are by 2, as evennumbers.

2 + 3 = 5 x 73 = 365

237

13

19

31

3743

61

67

73

5 11

17

23

29

41

47

53

59

71

4 6 891012141516

18202122

24252627

28

3032

3334

353638394042

444546

484950

5152

54

55

56

57

58

60

62

63

64

65

66

6869

70

72

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39

It will permit then us on formulating polynomial describing the solution of Binary Goldbach‘sConjecture.

P(2n) = (p + p‘) (2+2) (3 + 3) (3 + 5) (5 + 5) (5 + 7) (7 + 7) (5 + 11) (5 + 13) (7 + 13) (5 + 17) 

2p = 2n = p + p', 2 + 2 = 4, 3 + 3 = 6, 3 + 5 = 8, 5 + 5 = 10, 5 + 7 = 12, 7 + 7 = 14, 5 + 11 = 16,

7 + 11 = 18, 7 + 13 = 20, 5 + 17 = 22, 7 + 17 = 24, 13 + 13 = 26, 11 + 17 = 28, 13 + 17 = 30, 13 +

19 = 32, 17 + 17 = 34

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 1819 20 21 22 23 24 25 26 27 28 29

24

68 10 12

1416

1820

2224

2628

3032

3436

3840

4244

4648

5052

5456

58

23

5

7

11

13

17

19

23

29

0

20

40

60

80

100

120

140

Serie3 2 3 5 7 11 13 17 19 23 29

Serie2 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 54 56 58

Serie1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29

p + p' = 2n 2 + 2 = 4 3 + 3 = 6 3 + 5 = 8 17 + 19 = 36 37 + 41 =78 59 + 61 = 120 67 +

71 = 138 101 + 103 = 204 109 + 113 = 222 163 + 167 = 330 193 + 197 = 390 227 + 229 = 456

197; 193

113; 109

71; 67

349; 359

307; 317

167; 163

41; 37

269; 257

227; 229

101; 103

59; 61

17; 19

3; 5

311; 313

-100

-50

0

50

100

150

200

250

300

350

400

450

-5 0 5 10 15 20 25 30 35

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40

Theorem: Every odd number larger than 5, is sum three primes, because difference amongodd and even number is always 3.

Proof: 2n = p + p (2n – 1) – (p + p) = p 2n –  1 = p + p + p‖ 

4 = 2 + 2 7-

(2 + 2) = 3 7 = 2 + 2 + 3

2n = p + p' 4 = 2 + 2 6 = 3 + 3 8 = 3 + 5 10 = 5 + 5 12 = 5 + 7 14 = 7 + 7 16 = 5 + 11

2 3 35 5

75

7 7

11 1113

1113 13

17 1719

2

35

57

711

1113

1113

1317

1719

1719

19

4

6

8

10

12

14

16

18

20

22

24

26

28

3032

34

36

38

0

10

20

30

40

50

60

70

80

Serie3 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38

Serie2 2 3 5 5 7 7 11 11 13 11 13 13 17 17 19 17 19 19

Serie1 2 3 3 5 5 7 5 7 7 11 11 13 11 13 13 17 17 19

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

p' + (p + p) = 2n - 1 2n = p + p (2n - 1) - (p + p) = p 3 + (2 + 2) = 7 4 = 2 + 2 7 - (2 + 2) = 3

3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 34

68

1012

14 16 1820

2224

2628

3032

3436

3840

4244

4648

79 11

1315

1719

2123

2527

2931

3335

3739

4143

4547

4951

0

20

40

60

80

100

120

Serie3 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51

Serie2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48

Serie1 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

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41

Theorem: Difference among two successive square numbers is always odd number.a(a) - b(b) = (2n - 1) 25 - 16 = 2(5) - 1

Next numbers from infinite file of odd numbers added to square minuend create alwayssquare subtrahend.

p + p' = p" 2 + 3 = 5 p + p = 2n 5 + 3 = 8 p'' + p' + p = 2n - 1 7 + 5 + 3 = 15

12

34

56

78

910

1112

1314

1516

1718

1920

2122

2324

2526

2728

29

23

23

23

23

23

23

23

23

23

23

23

23

23

23

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 10

5

10

15

20

25

30

35

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29

a(a) - b(b) = (a - b)(a + b) 5(5) - 3(3) = (5 - 3)(5 + 3) 25 - 9 = 2(8)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 253 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49

1

4964

100121

144

169

196225

256

289

324

361

400

441

484

529

576

625

81

36251694

0

100

200

300

400

500

600

700

1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49

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42

z 222 x y  

That is to say, that equation x nnn  z y   from great theorem Fermat‘s, only near n = 2 hassolution, because when add only odd number to square, we receive next square number.

1 and 3 are  not square numbers, and above mentioned theorem despite this and on themchecks because primes are the multiplicity of number one and only me also. p = 1(p)

In other words, equation for n > 2 in infinite file of natural numbers does not possess nosolution, because only the square numbers create the ternary Pythagoras.

z 222 y x 25 – 9 = 16 ( y 2mod0)

2  

16 : 2 = 8 Rest 0

x x‘ 22  x x 2  

2 2 ! = 24

2 2 ! + 1 2 = 5 2 3+5+7+9 + 1 = 25

First ternary Pythagoras comes into being, when the sum of differences among successive

square numbers reaches value of faculty 2 2 !.It number 2 modulates so formation so square numbers how and ternary Pythagoras, that is tosay that differences among odd numbers and squares in ternary Pythagoras they are divisibleby 2, therefore squares how and ternary Pythagoras are product 2 factors.

3(3) + 4(4) = 5(5) [2(5) – 1] + 2(8) = 2(13) – 1 3[2(2) – 1] + 2(2)2(2) = 5[2(3) – 1]

1 + 3 = 4 + 5 = 9 + 7 = 16 + 9 = 25 + 11 = 36 + 13 = 49 + 15 = 64 + 17 = 81 + 19 = 100 + 21..  \ 2 /\ 2 /\ 2 /\ 2 /\ 2 /\ 2 /\ 2 /\ 2 /\ 2 /\ 2 / 

25 – 9 = 2(8) 169 – 25 = 2(72) 289 – 225 = 2(32) 625 – 49 = 2(288) 841 – 441 = 2(200)

It because 2 is the solid value of differences among two the successive odd numbers becomesshe the modules of differences with them square numbers in ternary Pythagoras, where withdifference among horizontal the and vertical length the side of triangle square creates, beingsimultaneously the proof on truth of equation the Pythagoras and the Fermat's conjecture.

z 22

24 z x

 xy

 

  

  z 2 = 2xy + x 2 - 2xy + y 2 z 2 = x 2 + y 2  

1 + 1 1

+ 3

2 4+ 5

3 9

+ 7

4 16

+ 9

5 = 24 25

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43

If the product of hypotenuse to be equal to sum two products of legs and product of theirdifference, that is to say, that the square of hypotenuse is the sum of squares of legs.

Theorem: Difference among every odd square in triples Pythagoras even square is, which

congruent according to modules y 2mod2  

Proof: z 222 y x 25 – 9 = 16 ( y 2mod0)

2  

16 : 2 = 8 Rest 0

[2(5) – 1] + 2(8) = 2(13) – 1The congruence the even square y 2mod0

2 signifies, that in quadrate of hypotenuse(z 2 ) 4triangles of the same hypotenuse be comprise (z), replenished about quadrate came into beingwith differences between horizontal and vertical length the sides 4 triangles. e.g.

y 2 + x 2 = z 2 y – x = n  42

 yx+ n 2 = z 2  

e.g. 4 2 + 3 2 = 5 2 4 – 3 = 1 4(6) + 1 2 = 25

12 2 + 5 2 = 13 2 12 – 5 = 7 4(30) + 7 2 = 169

8 2 + 15 2 = 17 2 15 – 8 = 7 4(60) + 7 2 = 289

24 2 + 7 2 = 25 2 24 – 7 = 17 4(84) + 17 2 = 625

20 2 + 21 2 = 29 2 21 – 20 = 1 4(210) + 1 2 = 84112 2 + 35 2 = 37 2 35 – 12 = 23 4(210) + 23 2 = 1369

40 2 + 9 2 = 41 2 40 – 9 = 31 4(180) + 31 2 = 1681

28 2 + 45 2 = 53 2 45 – 28 = 17 4(630) + 17 2 = 2809

60 2 + 11 2 = 61 2 60 – 11 = 49 4(330) + 49 2 = 3721

56 2 + 33 2 = 65 2 56 – 33 = 23 4(928) + 23 2 = 4225

84 2 + 13 2 = 85 2 84 – 13 = 71 4(546) + 71 2 = 7225

72 2 + 65 2 = 97 2 72 – 65 = 7 4(2340) + 7 2 = 9409

144 2 + 17 2 = 145 2 144 – 17 = 127 4(1224) + 127 2 =21025

180

2

+ 19

2

= 181

2

180 – 19 = 161 4(1710) + 161

2

= 37261

4xy/2+(x - y)^= z^ z^- 4xy/2= (x - y)^2xy+(x-y)(x-y)=z(z) 2xy+x(x)-2xy+y(y)=z(z)

x(x) = x^ y(y) = y^ z(z) = z^ ^ = 2

Y

X

Z

Y

X

+ =

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44

Ternary Pythagoras this square equation, and how there are all quadratic functions as graph of function is a parabola. The running by vertex axis of symmetry be shifted in them about 2 in

direction on line - x, and about (y – x) 2 in direction on line - y.

Number 2 in every semi stabile elliptic curve over rational numbers modular is.

(1+3+5+7+9)25 + (11+13+15+17+19+21+23+25)144 = 169 = 49(1+3+5+7+9+11+13) +  \2/\2/\2/\2/ \2/ \2/ \2/ \2/ \2/ \2/ \2/ 120(15 +17+19+21+23+25)

y = (y - x)^ = 1 x^ + y^ = z^ 9 + 16 = 25 (1 + 3 + 5) + (7 + 9) = 25 - 1 = 24 - 3 = 21 - 5 = 16 - 7 = 9 -

9 = 0

25

1

25

0

5

10

15

20

25

30

0 0,5 1 1,5 2 2,5 3 3,5

y = (y - x)^ =(12 - 5)^ = 49 5^ +12^ =13^ 25+144=169 (1+3+5+7+9) + (11+13+15+17+19+21+23+25)

=169 -1=168 - 3 =165 - 5=160 -7=153 - 9 =144 -11=133 -13=120 -15 = 105 -17 = 88 -19 = 69 -21= 48 -

23=25 - 25 = 0

169

49

169

0

20

40

60

80

100

120

140

160

180

0 0,5 1 1,5 2 2,5 3 3,5

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Theorem: The square of hypotenuse is equal the sum of squares of legs,when it is sum of such quantity of odd numbers how degree of square hypotenuse.

Proof: x(2n – 1) = x 2 y(2n ) = y 2 z(2n – 1) = z 2 x + y = z y = z – x x +(z – x) = z

x2

+ y2

= z2

  z2

= z(2n – 1) z2

- z(2n – 1) = 0

x(2n - 1) + y(2n - 1) = z^ - z(2n -1) = 0 (1 + 3 + 5) + (7 + 9) = 25 - 1 = 24 - 3 = 21 - 5 = 16 - 7 = 9 - 9 = 0

3^ + 4^ = 5^ x^ - [2xy + (y-x)^] + y^ = 0 9 - (24 + 1) + 16 = 0

9

7

5

3

1

3

5

7

9

-11

0

9

16

21

2425

24

21

16

9

0

-11

-15

-10

-5

0

5

10

15

20

25

30

Serie1

Serie2 9 7 5 3 1 3 5 7 9

Serie3 -11 0 9 16 21 24 25 24 21 16 9 0 -11

1 2 3 4 5 6 7 8 9 10 11 12 13

2xy +(y - x)^ = z^ 2(3)4+(4 -3)^ = 25 2(5)12+(12 -5)^ =169 2(15)8+(15 -8)^ =289 2(7)24+(24 -7)^ =625

2(21)20 + (21 - 20)^ =841

25 25169 169

625 625

841 841

2809 2809

149

289289 289

0

500

1000

1500

2000

2500

3000

3500

4000

4500

5000

1 2 3

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46

3[2(2) – 1] + 2(2)2(2) = 5[2(3) – 1]

1 + 3 + 5 = 9 = 3(3) 1 + 3 + 5 + 7 = 16 = 4(4) 1 + 3 + 5 + 7 + 9 = 25 = 5(5)

(1 + 3 + 5) + (1 + 3 + 5 + 7) = (1 + 3 + 5 + 7 +9) Σ(2n - 1) = n(a + z)/2 = n(n)

3(1 +5)/2 = 3(3) 4(1 + 7)/2 = 4(4) 5(1 + 9)/2 = 5(5) 3(3) + 4(4) = 5(5) 9 + 16 = 25

1  +  3  + 

1  +  3  5 

1  +  +  + 

3  +  +  5  + =  7  +

5  =  9  7  = 16  9  = 25 

Because the square of hypotenuse is sum of such quantity of successive odd numbers, as

degree of square of hypotenuse, equation Pythagoras was can write as fraction:

2

2

2

2

2

2

 z

 z

 z

 y

 z

 x  

The common square denominator confirmed that the square of hypotenuse is the sum of thesquares of legs.

3^+4 =̂ 5(2n - 1) =5^= (1+3+5)+(7+9) = 25 13(2n - 1) = 13^=(1+3+5+7+9)+(11+13+15+19+21+23+25)

=169 15^+8 =̂17(2n - 1) = 17 =̂(1+3+5+7+9+11+13+15+17+19+21+23+25+27+29)+(31+33)=289

25(2n - 1) = 25^= 625 29(2n - 1) = 29^=841

1 3 5 7 9

251 3 5 7 9

11

13 15 17 19 21 23 25

169

1 3 5 7 9

11

13 15 17 19 21 23 25

27

29 31 33

2891 3 5 7 911

13 15 17 19 21 23 25

27

29 31 33

35

37 39 41 43 45 47 49

625

1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57

841

2n - 1 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 841

2n - 1 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 625

2n - 1 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 289

2n - 1 1 3 5 7 9 11 13 15 17 19 21 23 25 169

2n - 1 1 3 5 7 9 25

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

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47

Theorem: If the product of difference and the sum two numbers square number is, then she issquare difference two square numbers.

Proof: (z – y)(z + y) = xn

= (znn

 y ) n = 2

-10

0

9

16

21

24 25 24

21

16

9

0

-10

0 2 4 6 8 10 12 14

(1 + 3 + 5) + (1 + 3 + 5 + 7) = (1 + 3 + 5 + 7 +9) Σ(2n - 1)= n(a + z)/2 = n(n)

3(1 +5)/2 = 3(3) 4(1 + 7)/2 = 4(4) 3(3) + 4(4) = 5(5) 9 + 16 = 25

0 0 00 0 00 0 00 0 0625

0625

289

1681 1681

961

3721 3721

2401

7225 7225

5041

1 2 3

2xy +(y-x)^= z^ 2(7)24+(24-7)^=25^=625 2(9)40+(40-9)^=41^=1681

2(11)60+(60-11)^=61^=3721 2(13)84+(84-13)^=85^=7225

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48

z2

  (z – y)(z + y) = x2

  x2

 

25 (5 – 4)(4 + 5) = 9 9

169 (13 – 12)(12 + 13) = 25

289 (17 – 15)(17 + 15) = 64

625 (25 – 24)(25 + 24) = 49

841 (29 – 21)(29 + 21) = 4001369 (37 – 35)(37 + 35) = 144

1681 (41 – 40)(41 + 40) = 81

2809 (53 – 45)(53 + 45) = 784

3721 (61 – 60)(61 + 60) = 121

4225 (65 – 33)(65 + 33) = 3136

4225 (65 – 63)(65 + 63) = 256

5329 (73 – 55)(73 + 55) = 2304

7225 (85 – 77)(85 + 77) = 1296

7225 (85 – 84)(85 + 84) = 169

7921 (89 – 39)(89 + 39) = 6400

9409 (97 – 65)(97 + 65) = 518410201 (101 –99)(101 +99)= 400

11881 (109-91)(109+91) = 3600

12769 (113-112)(113+112) = 225

15625 (125-117)(125+117) = 1936

18769 (137-105)(137+105) = 7744

21025 (145-143)(145+143) = 576

21025 (145-144)(145+144) = 289

22201 (149-51)(149+51) = 19600

Also prime numbers except 2 can introduce as product of difference and sum two naturalnumbers and they are then prime difference two square numbers.

3 = (2 – 1)(2 + 1) 5 = (3 – 2)(3 + 2) 7 = (4 – 3)(4 + 3) 11 = (6 – 5)(6 + 5)

13 = (7 – 6)(7 + 6) 17 = (9 – 8)(9 + 8) 19 = (10 – 9)(19 + 9) 23 = (12 – 11)(12 + 11).

p =  ))((]2

1[]

2

1[

22 baba p

 p p

  3 = )12)(12(]2

133[]

2

13[

22

 

5 =)23)(23(]

2

155[]

2

15[

22

7 =)34)(34(]

2

177[]

2

17[

22

 This is Great Fermat's theorem for all values of n proved, because he is for all prime values of n valid.Looking closer at the following graph, you will see that half of the following sums of twoprimes on a straight line parallel to the y - axis with real part ½ y lie. This means that thelinear Diophantine equation ax + by - c = 0, with given integer pairs not have common divisorCoefficient a, b, c, always in prime x, y is solvable.

1(2) + 1(3) – 5 = 0 1(3) + 1(7) – 10 = 0 1(5) + 1(13) – 18 = 0 1(11) + 1(19) – 30 = 0

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Natural numbers, divisibility and primes.

Central notion within of natural numbers concerns divisibility, and more far order of primes -natural number larger than 1, which has not natural divisor, that is to say, no different divisorexcept 1 or me alone. Sequence of primes has begun since 2, 3, 5, 7, 11, 13, 17, 19,

23,…Already Euclid proved before over 2 000 years, that this sequence does not end, and sothere is no the largest prime. Beyond 2 all primes are odd with characteristic endings - 1 - 7 -3 - 9. From second side is in force the main theorem of arithmetic: every natural number willgive oneself unambiguously to introduce as product of primes. Primes gain by this onmeaning for mathematics, as contribution to construction of all different numbers. The everynumber, which is not prime, will give oneself with these indivisible factors to to put together.Prime numbers 2 and 3 are components of all natural numbers really. Why, for example, can‘tall numbers be built simply by multiplying and adding together different combinations of theprimes 2 and 3. e.g. 4 = 2(2),5 = 2 + 3, 6 = 3 + 3, 7 = 2(2) + 3, 8 = 2(2)2, 9 = 3(3), 10 = 2(2) + 3 + 3, 11 = 2(2)2 + 3,12 = 3(3)+ 3, 13 = 2(2)+ 3(3), 14 = 2(2)2 + 3 + 3, 15 = 3(3)+ 3 + 3, 16 = 2(2)2(2),

17 =2(2)2+3(3), 18 =2(2)2(2)+2, 19 =2(2)2(2)+3, 20 =2(2)2(2)+2(2), 21 =2(2)2(2)+2+322 = 2(2)2(2)+ 3 + 3, 23 = 2(2)2(2)+ 2(2)+ 3, 24 = 2(2)2(2)+ 2(2)2, 25 = 2(2)2(2)+ 3(3),26 =2(2)2(2)+2(2)2+2, 27 =3(3)3, 28 =2(2)2(2)+3+3(3), 29 =2+3(3)3, 30 =3+3(3)3.All prime be built according to simple formula: p = n(2)+3, 5 =2+3, 7 =2(2)+3 11 =4(2)+3,13 = 5(2) + 3, 17 = 7(2) + 3, 19 = 8(2) + 3, 23 = 10(2) + 3, 29 = 13(2) + 3. Formulathis permits us to divide primes on two classes: they class of basic primes (2, 3, 5, 7), whichalone for me are the building material and these, which are already the multiplicity of number7. e.g.11 = 7+(4) 13 = 7+(6) 17 = 2(7)+(3) 19 = 2(7)+(5) 23 = 3(7) + (2) 29 = 4(7) + (1)And so we write new formula: p = n(7) + The rest (1,2,3,4,5,6)

0 0 02

0

32,5

3

75

5

13

9

11

0

19

15

2y+3y=5y/2=2+0,5=3-0,5 3y+7y=10y/2=3+2=7-2

5y+13y=18y/2=5+4=13-4 11y+19y=30y/2=11+4=19-4

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50

It was can sequence of primes and write so: 2, 3, 5,-2- 7,-4- 11,-2- 13,-4- 17,-2- 19,-4- 23. Inspaces among numbers notice hidden formula: - 2 - 4 - 2 - 4. These two last formulae, theywill play further decisive part.Are there formulas that produce some of the prime? Here you are! p = n(2) + 3

2 = 1(2) + 0 3 = 0(2) + 3 5 = 1(2) + 3 7 = 2(2) + 3 11 = 4(2) + 3 13 = 5(2) + 317 = 7(2) + 3 19 = 8(2) + 3 23 = 10(2) + 3 13(2) + 3 = 29 14(2) + 3 =31 233 = 115(2) + 3251 = 124(2) + 3.

Irrefutable proof.

Mathematicians knew, however, that proving the Riemann Hypothesis would be of far greatersignificance for the future of mathematics than knowing that Fermat‘s equation has nosolutions when n is bigger than 2. The Riemann Hypothesis seeks to understand the mostfundamental objects in mathematics – prime numbers.

The primes are those indivisible numbers that cannot be written as two smaller numbersmultiplied together. The primes are the jewels studded throughout the vast expanse of theinfinite universe of numbers that mathematicians have explored down the centuries.

Their importance to mathematics comes from their power to build all other numbers. Everynumber that is not a prime can be constructed by multiplying together these prime buildingblocks /2 and 3/. Mastering these building blocks offers the mathematician the hope of discovering new ways of charting a course through the vast complexities of the mathematicalworld.

Yet despite their apparent simplicity and principal character, prime numbers remain the most

mysterious objects studied by mathematicians. They question about distribution of primes

1 2 35

711

13

1719

2325

2931

3537

4143

4749

53

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 22 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 324

8 10

1416

2022

2628

3234

3840

4446

50

1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53

p = 3 + n(2)

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belonged to the most difficult. They were the long time then the question of plain theoreticalnature, however today found primes in different realms the use.

Suddenly the economic interest appears also the question, or proof the Riemann's hypothesiscans you something say about distribution of primes in world of numbers. If centuries of 

searching had failed to unearth some unknowing formula which would generate the list of prime numbers, perhaps it was time to adopt a different strategy. Look through a list of primenumbers, and you‘ll find that it‘s impossible to predict when the next prime will appear. Thelist seems chaotic, random, and offers no clues as to how to determine the next number. Canyou find a formula that generates the numbers in this list, some unknowing rule that will tellyou what the 10 000 000th prime number is?

Not the question about quantity of primes in given interval of numbers, but the observation of spaces between two primes, she directed me on sure regularity from what they appear. 2, 3,5,-2- 7,-4- 11,-2- 13,-4- 17,-2- 19,-4- 23 and so 2, 4, 2, 4, then the smallest space is amongtwo primes and the decisive structure, recognizable in whole does not end sequence of primes.

It after 23 number first 29 comes however in space 6 (23,- 2 -25,-4-29), because placebetween them is for first product of primes, number almost prime 25 = 5(5). Since then allalmost prime numbers, as product of primes will take free place in sequence of primes,keeping spaces - 2 - 4 - 2 - 4. Generations have sat listening to the rhythm of the primenumber drum as it beats out its sequence of numbers: two beats, followed by three beats, five,seven, eleven. As the beat goes on, it becomes easy to believe that random white noise,without any inner logic, is responsible. At the centre of mathematics, the pursuit of order,mathematicians could only hear the sound of chaos.

I do realize, that prime and almost prime numbers appear in interval two and fourth. If itwalks about finding formulae and order, then primes are not more unequalled challenge.Knowing in what space sequent prime or relatively prime will appear, we can easily wholetheir list take down. And when we to this have yet the hand, as to qualify in sequence thesequent number, or it is prime or almost prime numbers, then and list of primes does notappear us as chaotic and accidental. The List of primes is, the heartbeat of mathematics, but apulse wired regular in rhythm by multiplicity of seven in two – by – four steps.Fractions are the numbers whose decimal expansions have a repeating pattern. For example

1/7 = 0,142 857 142 857 

At bases of distribution of primes in sequence of numbers, lies decomposition of their

products on prime factors. According to Fermat small theorem numbers to power ( p - 1)minus one, they are divisible without the rest by prime. e.g. 106 - 1 = 999 999/7 = ↓- 142 857 

857 142Proof:

−1−1

≅ 0    if a ≠ p p ≥ 3 a ≥ 2 26  = 64 – 1 = 63/7 36 = 729 – 1 = 728/7

Similarly by fractions: 1/7 = 0,142 857 142 857 1 …2/7 = 0,2857 142 857 14 …3/7 = 0,42857 142 857 1 …

4/7 = 0,57 142857 142857 1..5/7 = 0,7 142587 142587 1..

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6/7 = 0,857 142587 142587..8/7 = 1,142 857 142857 9/7 = 1,2857 142 857 14..

10/7 = 1,42857 142 857 …11/7 = 1,57 142 857 1428 …

12/7 = 1,7 142 857 14285 …13/7 = 1,857 142 857 142 …

where the quotient in decimal expansion from some place after comma begins repeating ininfinity six - digits numbers since 1, and finishing on 7. In practice this marks that every thethe six - digit combination of numbers e.g. (x x x x x x)/  7, (x y x y x y)/ 7, (y x y x y x)/ 7,(xyz xyz)/  7, (zxy zxy)/ 7, (yzx yzx)/  7, (zyx zyx)/  7, (yxz yxz)/  7, (xzy xzy)/  7, and theirmultiplicities divide without the rest by

7. 111 111 111 111 111 111 / 7 = 15 873 015 873 015 873

0  1  2  3  4  5  6  7 1  2  3  4  5  6  7  8 

2  3  4  5  6  7  8  9 

3  4  5  6  7  8  9  0 

4  5  6  7  8  9  0  1 

5  6  7  8  9  0  1  2 

6  7  8  9  0  1  2  3 

7  8  9  0  1  2  3  4 

This gets from here, that all numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, 9,.. they are congruent to meaccording to module 7, as this shows following radar graph

0 1

2

3

4

5

6

7

1 2

3

4

5

6

7

8

23

4

5

6

7

8

9

34

5

6

7

8

9

0

4

5

6

7

8

9

0

1

5

6

7

8

9

0

1

2

6

7

8

9

01

2

3

7

8

9

0

12

3

4

a - 7/7

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53

If difference among given number a, and prime is divisible by prime, then given number be

complex.  pV  p

 pa

e.g. 10

17

17187

 

Only difference among two primes divisible it is by 7, because p - (2,3,5,11,13,29) = n(7), and

all primes be distributed according to multiplicity of number 7 (23 - 2)/ 7 = 3 (17 - 3)/ 7 = 2(19 - 5)/ 7 = 2 (53 - 11)/ 7 = 6 (41 - 13)/ 7 = 4 (43 - 29)/ 7 = 2

Primes and almost prime follow after me in rhythm on two fourth.

1.  3. + 2 = 5.- 2  – 7 –  4 –  11  – 2 –  13  – 4 –  17- 2 –  19  – 4 – 23 - 2 –  25  – 4 – 29  – 2 –  31 -

Theorem: Odd numbers, the cannot be written as product of two smaller numbers a and b witha, b> 1, are prim and congruent to me modulo 7.

Proof: p p`mod p 11 53 mod 7 when 11 =1(7)+ 4 and 53 = 7(7)+ 4 then p and p`congruent according to mod p, and difference p`- p is multiplicity p. 53 - 11 = 42/7 = 6

This proof gives mathematics to instruction very quick procedure on qualification of primesabout any quantity of places. p - (2,3,5,11,13,29) = n7

p = n(7) + R(1,2,3,4,5,6)

0  1  2  3  4  5  6 

7  11  13 

17  19 

23  25 

29  31 

35  37  41 

43  47 

49  53  55 

59  61 

65  67 

71  73 

77  79  83 

85  89 

91  95  97 

101  103 107  109 

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54

113  115 

119  121  125 

127  131 

133  137  139 

143  145 

149  151 

155  157 

161  163  167 

169  173 

175  176  177  178  179  180  181 

182  183  184  185  186  187  188 

189  190  191  192  193  194  195 

196  197  198  199  200  201  202 

203  204  205  206  207  208  209 

210  211  212  213  214  215  216 217  218  219  220  221  222  223 

224  225  226  227  228  229  230 

231  232  233  234  235  236  237 

238  239  240  241  242  243  244 

245  246  247  248  249  250  251 

252  253  254  255  256  257  258 

259  260  261  262  263  264  265 

266  267  268  269  270  271  272 

273  274  275  276  277  278  279 280  281  282  283  284  285  286 

287  288  289  290  291  292  293 

294  295  296  297  298  299  300 

301  302  303  304  305  306  307 

308  309  310  311  312  313  314 

315  316  317  318  319  320  321 

322  323  324  325  326  327  328 

329  330  331  332  333  334  335 

336  337  338  339  340  341  342 

343  344  345  346  347  348  349 

350  351  352  353  354  355  356 

357  358  359  360  361  362  363 

364  365  366  367  368  369  370 

371  372  373  374  375  376  377 

378  379  380  381  382  383  384 

385  386  387  388  389  390  391 

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Discovery of rhythm beating the heart of mathematics storm the safety of system the RSA,any business selling prime numbers could realistically in support about this proof peddle theirwares under the banner ―satisfaction guaranteed or your money back‖, without too much fear of going bust. And so it turns out, that it 64 numerical factor with 129 numerical code is not

number prime, because divisible it is by 7.

3 490 529 510 847 650 949 147 844 619 903 898 133 417 764 638 493 387 843 990820577:7= 498 647 072 978 235 849 878 263 517 129 128 304 773 966 376 927 626 834 855 831 511

But second 65 numerical factor is prime (32769132993 266 709 549 961 988 190 834 461413177 642 967 992 942 539 798 288 533 – 5):7 = 4 681 304 713 323 815 649994569741547780201 882 520 423 998 991 791 399 755 504

Prime numbers are numbers that are divisible only by one and themselves. They are the atomsof arithmetic, for any number is either a prime or a product of primes. The first few primes are

2, 3, 5, 7, 11, 13, 17, 19, but despite their simple definition the prime numbers appear to bescattered randomly amid the integers.

There is simple way to tell if a number is prime  – than they cannot be written as product of two smaller numbers a and b, with a, b>1, and that is the basis for most modern encryptionschemes.

Solving the Riemann Hypothesis could lead to new encryption schemes and possibly providetools that would make existing schemes, which depend on the properties of prime numbers,more vulnerable.

0

10

20

30

40

50

60

70

8090

100

1 2 3 4 5 6 7 8 9 10 11 12 13 14

07

0 0 0

35

0

49

0 0 0

77

0

91

0 0 0 0

29

0

43

0 00

71

0

85

0

20 0

23

0

37

0 0 0

65

0

79

0 0

30

17

0

31

0 0 0

59

0

73

0 0 0

0

11

0

25

0 0 0

53

0

67

0 0 0

95

0

19

0 0 0

47

0

61

0 0 0

89

0

0

13

0 0 0

41

0

55

0 0 0

83

0

97

p'- p = n(7) 19 - 5 = 2(7) 47 - 19 = 4(7) 61 - 47 = 2(7) 89 - 61 = 4(7)

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Primes do not possess except 1 and only number no factors, but number almost prime arealmost so good, because they have only two factors prime. 23 is prime,  but 25 (5 · 5) it isalmost prime. So alone numbers 35(5·7), 49(7·7), 55(5·11), 65(5·13), 77(7· 11), 85(5 · 17) 

01

2

34

5

6

7

11

13

17

19 23

25

29

31

35

37

41 43

47

49

53

55

59

61 65

67

71

73

77

79

83 85

89

91

95

97

p, p', p" = 2n + p' 2, 3, 5 = 2 + 3, 7 = 4 + 3, p'" = n(7) + Rest(1,2,3,4,5,6), 11 = 7 + 4, 13 = 7 +

6, 17 = 2(7) + 3, 19 = 2(7) + 5, 23 = 3(7) + 2, 29 = 4(7) + 1, 2, 3, 5, -2- 7, -4- 11, -2- 13, -4- 17, -2- 19,

-4- 23,

2337

79

107

149163

1731

5973

101

157

11

5367

109

137

151

179

1947

61

89

103

131

173

13

41

83

97

139

167

181

0

50

100

150

200

250

300

350

400

13 41 83 97 139 167 181

5 19 47 61 89 103 131 173

11 53 67 109 137 151 179

3 17 31 59 73 101 157

2 23 37 79 107 149 163

7 29 43 71 113 127

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Number almost prime built with prime numbers larger than three, they develop how splendidfan in infinity.

Sequences of numbers almost prime.

(10)(20)

25

35 (14)(28)49

"p" = n(3) + (1,2) 25 = 8(3) + 1 35 = 11(3) + 1 49 =16(3) + 1 55 = 18(3) +1 65 = 21(3) + 2

77 = 25(3) + 2 85 = 28(3) + 1 91 = 30(3) + 1

275

215

299

235

319

91

133

175

217

259

301

155

323

221

305

115

325

265

119

161

203

245

287

329

143

185

121

205247

289

145

187

125

209251

293

335

9585

169253

295

5,7,11,13,17,19,23,29,31,37,41,43,(p) = "p"

5 711

13

1719

23252931

35

37

4143 25 35

55

65

85

95

115125145

155

175

185

205

215

4977

91

119

133

161175203

217

255

259

287

301

121

143

187

209

253

275319

341

385

407

451

473

169

221

247

299

325377

403

455

481

559

289

323

391

425493

361

437

475

0

100

200

300

400

500

600

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58

55

65

77

85

95 91

115 119 (22)(44)

125 121

133

145 143

155 (52)(26)

161 169

175

185 187

205 203 209

215 217221

235

245 247

259 253

265

275 (34)(68)

287 289

295 299

305 301

319

325 329 323

335

343 341

355 (76)(38)

365 361

371 377

385

395 391

407 403

415 413

425 427

437

445

455 451

469

475 473

485 481

497 493

505

515 511 517 (46)(92)

527 529

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59

In interval what 30 numbers (10-40) on three numbers with ending 5 (15, 25, 35), two of themare almost prime. Primes and almost prime follow after me in interval what 2(p) and 4(p).

25 –  2(5) – 35 –  4(5) – 55 49 –  4(7) – 77 –  2(7) – 91 121 –  2(11) - 143 –  4(11) – 187

17 - 4(5) – 37 –  2(5) – 47 –  4(5) – 67- 2(5) – 77 –  4(5) – 97 – 2(5) – 107 – 4(5) – 127 –  2(5)..

Triangle of almost prime.

5 7 11 13 17 19 23 25 29 31 35 37 41 43 47 49 53 55 59

5 25 35 55 65 85 95 115 125 145 155 175 185 205 215 235 245 265 275 295

7 49 77 91 119 133 161 175 203 217 245 259 287 301 329 343 371 385 413

11 -22 -44 121 143 187 209 253 275 319 341 385 407 451 473 517 539 583 605 649

13 -52 -26 169 221 247 299 325 377 403 455 481 533 559 611 637 689 715 767

17 -34 -68 289 323 391 425 493 527 595 629 697 731 799 833 901 935 1003

19 -76 -38 361 437 475 551 589 665 703 779 817 893 931 1007 1045 1121

23 -46 -92 529 575 667 713 805 851 943 989 1081 1127 1219 1265 1357

25 100 -50 625 725 775 875 925 1025 1075 1175 1225 1325 1375 1475

29 -58 116 841 899 1015 1073 1189 1247 2363 1421 1537 1595 1711

31 124 -62 961 1085 1147 1271 1333 1457 1519 1643 1705 1829

35 -70 140 1225 1295 1435 1505 1645 1715 1855 1925 2065

37 148 -74 1369 1517 1591 1739 1813 1961 2035 2183

41 -82 164 1681 1763 1927 2009 2173 2255 2419

43 172 -86 1849 2021 2107 2279 2365 2537

47 -94 188 2209 2303 2491 2585 2773

49 196 -98 2401 2597 2695 2891

53 106 212 2809 2915 3127

55 220 110 3025 3245

59 118 236 3481

"p" + 2(p) "p' " + 4(p) 25 + 2(5), 35 + 4(5), 55 + 2(5).. 49 + 4(7), 77 + 2(7), 91 + 4(7), 119 +

2(7), 133 + 4(7), 121 + 2(11), 143 + 4(11)

23 43 53 73 83 103 113 163 173

17

37 47 67 97

107

127

137

157 167

31

41

61

71 101

151

25 55 85 17535 65 95 125 15549

77 91

133 161121 143 169

181

131

115

145119

0%

20%

40%

60%

80%

100%

2 5 11 169

2 5 11 121 143

2 5 11 49 77 91 119 133 161

2 5 11 35 65 95 125 155

2 5 11 25 55 85 115 145 175

2 5 11 31 41 61 71 101 131 151 181

3 7 13 17 37 47 67 97 107 127 137 157 167

3 13 23 43 53 73 83 103 113 163 173

19 29 59 79 89 109 139 149 179

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25 + 2(5) = 35 + 4(5) =55 + 2(5) = 65,… 49 + 4(7) = 77 + 2(7) = 91 + 4(7) = 119,..

25 35

55 65 49

85 95 77 91

115 125 121 119145 155 133 143

175 185 169 161

205 215 203 187 209

235 245 221 217

265 275 289 253 247

295 305 319 287 301 299

325 335 323 329

355 365 361 343 341

385 395 377 391 371

415 425 407 403 413

445 455 437 451 427475 485 481 473

505 515 497 511 493 517

535 545 527 533 539

565 575 553 551

595 605 601 583 581

625 635 623 611 629

655 665 649 637

685 695 679 671 667 689

715 725 707 721 703 713 697

745 755 737 731 749

775 785 767 781 763 779

805 815 799 793 803 791

835 845 841 833 817

865 875 871 851 847 869

895 905 889 901 893 899

925 935 917 931 913 923

955 965 949 961 943 959

985 995 979 973 989

1015 1025 1007 1003 1001

1045 1055 1037 1043 1027

1075 1085 1067 1081 1073 1057 1079

1105 1115 1099 1111

1135 1145 1127 1141 1133 1121 1139

1165 1175 1159 1157 1147 1169

1195 1205 1189 1183 1177 1199

1225 1235 1219 1211 1207

1255 1265 1247 1261 1243 1253 1241

1285 1295 1273 1271 1267

1315 1325 1309 1313

1345 1355 1339 1337 1351 1333 1343 1331 1349

1375 1385 1369 1363 1357 1379

1405 1415 1397 1411 1393 1403 1391 1387

1435 1445 1441 1421 1417

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61

1465 1475 1457 1463 1469

1495 1505 1501 1493 1477

1525 1535 1519 1517 1513 1507 1529

1555 1565 1547 1561 1541 1537

1585 1595 1577 1591 1573 1589

1615 1625 16031645 1655 1639 1651 1633 1643 1631 1649

1675 1685 1681 1673 1661 1679

1705 1715 1711 1703 1691 1687

1735 1745 1729 1727 1717 1739

1765 1775 1757 1771 1751 1769

1795 1805 1793 1781 1799

5 7 11 13 17 19

5 29 23 25

37 31 35

7 43 47 41 49

59 53 55

67 61 65

11 79 73 71 77

89 83 85

13 97 95 91

109 103 107 101

17 113 115 119

11 127 125 121

19 139 137 131 133

13 149 145 143

157 151 155

13 23 163 167 161 169

179 173 175

17 181 185 187

199 193 197 191

19 29 205 203 209

31 211 215 217

17 229 223 227 221

239 233 23519 241 245 247

23 37 257 251 259 253

269 263 265

277 271 275

17 41 283 281 287 289

23 293 295 299

43 307 305 301

29 313 317 311 319

19 47 325 329 323

337 331 335

31 49 349 347 343 341359 353 355

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19 367 365 361

29 53 379 373 371 377

389 383 385

23 397 395 391

31 37 409 401 407 403

59 419 415 41361 421 425 427

23 439 433 431 437

449 443 445

41 457 455 451

67 463 467 461 469

43 479 475 473

37 487 485 481

29 71 499 491 497 493

509 503 505

47 73 515 511 517

Theorem: Odd numbers, the can be written as product of two primes, are almost prime andcongruent to me modulo 3.

Proof: "p" "p`" mod p 49 85 mod 3 when 49 =16(3)+1 and 85=28(3)+1, then "p" and "p`"congruent according mod p, and difference "p`"-"p" is multiplicity p. 85 - 49 = 36/3 = 12

„p‖- (5,7,11,13) = n(3)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

250

49 55

0 0

85 91

0

115121133145

0

169175

0 0

205

0

217

0

235247253265

0

289295 301

0

35

0 0

65 77

0

95

0

119125

0

143155161

0

185

0

203209215221

0

245259

0

275287299305

"p" ≡ "p" mod 3, 91 - 49 = 14(3), 119 - 77 = 14(3), 161 - 143 = 6(3), 169 -

133 = 12(3), 299 - 287 = 4(3), 301 - 289 = 4(3),

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350 351 352 353 354 355 356 355 (76)(38)

357 358 359 360 361 362 363 361

364 365 366 367 368 369 370 365

371 372 373 374 375 376 377 377

378 379 380 381 382 383 384

385 386 387 388 389 390 391 391

392 393 394 395 396 397 398 395

399 400 401 402 403 404 405 403

406 407 408 409 410 411 412 407

413 414 415 416 417 418 419 415

420 421 422 423 424 425 426 425

427 428 429 430 431 432 433

434 435 436 437 438 439 440 437

441 442 443 444 445 446 447 445

448 449 450 451 452 453 454 451

455 456 457 458 459 460 461

462 463 464 465 466 467 468

469 470 471 472 473 474 475 475 473

476 477 478 479 480 481 482 481

483 484 485 486 487 488 489 485

490 491 492 493 494 495 496 493

497 498 499 500 501 502 503

504 505 506 507 508 509 510 505

511 512 513 514 515 516 517 515 517

518 519 520 521 522 523 524 (46)(92) 

525 526 527 528 529 530 531 527 529 

532 533 534 535 536 537 538 535 533

539 540 541 542 543 544 545 539

546 547 548 549 550 551 552

553 554 555 556 557 558 559

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It will suffice only to look on table of primes and almost prime, to oneself about this toconvince, what order and rules reign here. This would border about absurdity, these justfundamental elements of well ordered world of mathematics, if would they behave savagelyand unforeseeable.

2617 2615 2611

2621 2627 2623 26292633 2635 2639

2647 2645 2641

2659 2657 2651 2653

2663 2665 2669

Above mentioned table shows, that primes and almost prime ranked according to 4 basicnumbers unity 9  –  3  –  7  –  1, divide number almost prime about number unity 5 on twocomplementary parts. Fact this is the denial so far general opinion, about irregularities of occurrence of primes and them the alleged decreasing on further of up growth of sequencenumbers. Smaller or larger gaps among primes 2-4-6-8-14-18-20-22-24-26-30-34-36-44-52-

60-72-86-96-112-114-118-132-148-154-180-210-220-222-234-248-250-282-288-292-320-336-354-382-384-394-456-464-468-474-486-490-500-514-516-532-534-540-582-588-602-652-674-716-766-778-804-806-906, … be full with numbers almost prime, so as appearingprimes in interval 2 and 4.Among primes 1327 and 1361 his place occupies 10 successive numbers divisible by primes,that is almost primes.

1327 + 4 = 1331/11 + 2 = 1333/31 + 4 = 1337/7 + 2 = 1339/13 + 4 = 1343/17 + 2 = 1345/5 +4 = 1349/19 + 2 = 1351/7 + 4 = 1355/5 + 2 = 1357/23 + 4 = 1361  – 1327 = 34

Similarly is among primes 8467 and 8501. Among 370261 and 370373 we have gap about

length 112. For p < N the largest at present well-known maximal gap equal m =1442, p =804 212 830 686 677 669.

"p"+n(7)="p' " 35+ 2(7)=49+ 4(7)= 77 + 2(7) = 91 + 4(7) = 119 + 2(7) = 133 + 4(7) = 161 + 2(7) = 175 +

4(7) = 203 + 2(7) = 217 + 4(7) = 245 + 2(7) = 259 + 4(7) = 287

25

3549

55

77

6585

91

95

119133

125

121

145

175

169

209

205

217

215 221

247

265

275

287

295

299

7

115143

161

155

187

185

203

235

245

253

259

289

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10 000 019+2 = 10 000 021/97+4 = 10 000 025/5+2 = 10 000 027/37+4 = 10 000 031/227 + 2

=10 000 033/397+4 = 10 000 037/43+2 = 10 000039/7+4 = 10 000 43/2089+2 = 10 000 045/5

+ 4=10000049/47+2 =10 000 051/73+4 =10 000 055/5+2=10 000 57/79+4= 10 000 061/19+2

=10 000 063/17+4 = 10 000 067/7+2 = 10 000 069/181+4 = 10 000 073/31+ 2 = 10 000 075/5

+ 4 = 10 000 079 – 10 000 019 = 60

2677 2671 2675

2689 2683 2687 2681

2699 2693 2695

2707 2705 2701

2719 2713 2711 2717

2729 2725 2723

2731 2735 27372749 2741 2747 2743

2753 2755 2759

2767 2765 2761

n  p  p  n  P  P 

2  3  5  292  1453168141  1453168433 

4  7  11  320  2300942549  2300942869 

6  23  29  336  3842610773  3842611109 

8  89  97  354  4302407359  4302407713 

14  113  127  382  10726904659  10726905041 

18  523  541  384  20678048297  20678048681 

20  887  907  394  22367084959  22367085353 

22  1129  1151  456  25056082087  25056082543 

30  13063  13093  464  42652618343  42652618807 

34  1327  1361  468  127976334671  127976335139 

36  9551  9587  474  182226896239  182226896713 

44  11633  11677  486  241160624143  141160624629 

52  19609  19661  490  297501075799  297501076289 

60  100000019  100000079  500  303371455241  303371455741 

72  31397  31469  514  304599508537  304599509051 

86  155921  156007  516  416608695821  416608696337 

96  360653  360749  532  461690510011  461690510543 

112  370261  370373  534  614487453523  614487454057 

114  492113  492227  540  738832927927  738832928467 

118  1349533  1349651  582  1346294310749  1346294311331 

132  1357201  1357333  588  1408695493609  1408695494197 

148  2010733  2010881  602  1968188556461  1968188557063 

154  4652353  4652507  652  2614941710599  2614941711251 180  17051707  17051887  674  7177162611713  7177162612387 

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Theorem about congruence odd number permits faultlessly to distinguish primes from anotherdivisible numbers, that is almost prime. e. g. prime confirms the legitimacy of formula:

p = 1 + n(7) p = 2 + n(7) p = 3 + n(7) p = 4 + n(7) p = 5 + n(7) p = 6 + n(7)

2 89 - 1 = 618 970 019 642 690 137 449 562 111- 3

618 970 019 642 690 137 449 562 108/7 = 88 424 288 520 384 305 344 937 444

(3 203 000 719 597 029 781 – 3) : 7 = 457 571 531 371 004 254

(810 433 818 265 726 529 159 – 5) : 7 = 115 776 259 752 246 647 022 andalmost prime with numerous iterations inside, as and in quotient of formula ―p‖= 2 + n(3)

7 · 20408163265306122449 = 142 857 142 857 142 857 143- 2

142 857 142 857 142 857 141/3 = 476 190 476 190 476 190 47

We happen in second factor of following expression sure unusual prime:

10 31+ 1 = 11· 909 090 909 090 909 090 909 090 909 091 =10 000 000 000 000 000 000 000 000 000 001

Decomposition on primes her product, it lies at bases of iteration in this number.From this, that 1001 = 7 · 143 = 11 · 91 = 13 · 77 and 10 001 = 73 · 137 create followingiterations. Products:

7 · 1001 = 7007 11 · 1001 = 11011 13 · 1001 = 13013 77 · 1001 = 7707791 · 1001 = 91091 143 · 1001 = 143143 73 · 1001 = 73073 137 · 1001 = 137137and 999 multiplicity 1001 e.g. 323 · 1001 = 323 323 and number 10 001, 43 ·10001= 43004329 · 430 043 = 124 7 124 7 3 · 12 471 247 = 37 41 37 41

We see noteworthy iterations in prime 9 090 909 091 and her square, and so number almostprime 826 644 628 100 826 446 281 and prime 82 644 628 099 173 553 719, in which exceptiteration see two peers of numbers in reflection mirror.

210  20831323  20831533  716  13829048559701  13829048560417 

220  47326693  47326913  766  19581334192423  19581334193189 

222  122164747  122164969  778  42842283925351  42842283926129 

234  189695659  1899695893  804  90874329411493  90874329412297 

248  191912783  191913031  806  171231342420521  171231342421327 

250  387096133  387096383  906  218209405436543  218209405437449 

282  436273009  436273291  1132  1693182318746371  1693182318747503 

288  1294268491  1294268779  1308  749565457554371299  749565457554372607 

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On radar graph number almost prime are visible on black background.

9999907 9999901 9999905

9999913 9999911 9999917 9999919

9999929 9999925 9999923

9999931 9999937 9999935

9999943 9999949 9999947 9999941

9999955 9999959 9999953

9999961 9999965 9999967

9999973 9999971 9999977 9999979

9999985 9999983 9999989

9999991 9999997 9999995

10000003 10000001 10000007 10000009

10000019 10000015 10000013

10000021 10000025 10000027

10000039 10000037 10000031 10000033

10000045 10000043 10000049

10000051 10000055 10000057

10000069 10000067 10000061 10000063

10000079 10000073 10000075

10000085 10000081 10000087

10000097 10000091 10000099 10000093

Let‘s apply so well-known us a formula to constructing successive primes and almost prime,that could generate this kind of pattern.

29

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(1,2,3,4,5,6) + n(7) = p 2 + 3(7) = 23 (1,2) + n(3) = „p― 1 + 8(3) = 25 

9 999 901 = 1 428 557(7) + 2 9 999 905 = 3 333 301(3) + 29 999 907 = 1 428 558(7) + 1 9 999 913 = 3 333 304(3) + 1

9 999 929 = 1 428 561(7) + 2 9 999 923 = 3 333 307(3) + 29 999 931 = 1 428 561(7) + 4 9 999 925 = 3 333 308(3) + 19 999 937 = 1 428 562(7) + 3 9 999 935 = 3 333 311(3) + 29 999 943 = 1 428 563(7) + 2 9 999 949 = 3 333 316(3) + 19 999 971 = 1 428 567(7) + 2 9 999 977 = 3 333 325(3) + 29 999 973 = 1 428 567(7) + 4 9 999 985 = 3 333 328(3) + 19 999 991 = 1 428 570(7) + 1 9 999 997 = 3 333 332(3) + 1

10 000 019 = 1 428 574(7) + 1 10 000 015 = 3 333 338(3) + 110 000 079 = 1 428 582(7) + 5 10 000 085 = 3 333 361(3) + 2

9999907 9999901 9999905

9999911 9999917 9999913 9999919

9999929 9999925 9999923

9999937 9999931 9999935

9999943 9999941 9999947 9999949

9999955 9999953 9999959

9999965 9999961 9999967

9999973 9999971 9999977 9999979

9999985 9999983 9999989

9999991 9999995 9999997

10000001 10000007 10000003 10000009

10000019 10000015 10000013

10000025 10000021 10000027

10000031 10000037 10000033 10000039

10000045 10000043 10000049

10000055 10000051 10000057

10000061 10000067 10000063 10000069

10000079 10000075 10000073

10000085 10000081 10000087

10000091 10000097 10000093 10000099

Here are the primes amongst the 100 numbers either side of 10 000 000. For example in the100 numbers immediately before 10 000 000 since 9 999 901 to 9 999 991 there are 9 primes,but look now at how few there are in the 100 numbers above 10 000 000: only 2 primessince10 000 001 to 10 000 099.

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Arithmetical sequence of primes and almost prime are sequence line and helical growing.

Helical sequence of primes and almost prime.

3 2

7 5

11 13

f(p"p") = p, p', p' + d, p" + 2d, d = 2, f(p"p") = 2, 3, 3 + 2, 5 + 2, 7 + 4, 11 + 2, 13 + 4, 17 + 2,

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2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 223 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 33

0

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1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53

P(p) = (2,3,5,11,13,29) + n(7) 59 = 3 + 8(7) P("p") = (5,7,11,13) + n(3) 55 = 7 + 16(3)

1 2 2 2 2 2 2 2 2 2 24 4 4 4 4 4 4 4 425

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   2

 ,   3 ,  -   2  -   5 ,  -   2  -   7 ,  -   4  -   1   1 ,  -   2  -   1   3 ,  -

   4

  -   1   7 , . .

Serie4 3 7 13 19 25 31 37 43 49 55 61

Serie3 2 5 11 17 23 29 35 41 47 53 59

Serie2 4 4 4 4 4 4 4 4 4

Serie1 1 2 2 2 2 2 2 2 2 2 2

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

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Theorem: It sequence of primes and relatively primes is the twin sequence frominitial term p = 2 and constantly difference d = 6 (- 2  – 4).

Therefore though in Riemann‘s conjecture function of location of primes π (x), is the gradualfunction about high irregularity, as helical arithmetical twin sequence of primes and almostprimes, which the difference the d = 6 is constant, it shows the amazing smoothness.

From here sequence of prime numbers is not similar to accidental sequence of numbers, but towell ordered structure. So basic numbers does not be definite per nature the method of accidental throw with coin. Accident and chaos they are for mathematician simply cruelty.

2 + 3 = 5 -2- 7 -4- 11 -2- 13 -4- 17 -2- 19 -4- 23 -2- 25

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Serie4 3 9 15 21 27 33 39 45 51

Serie3 6 12 18 24 30 36 42 48 54

Serie2 2 5 8 11 14 17 20 23 26 29 32 38 41 44 47 50 53

Serie1 1 7 13 19 31 37 43

1 2 3 4 5 6 7 8 91

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Twin sequences of prime and almost prime numbers a congruent to me according to algebraicmodule 72.

Distribution of primes and almost prime according to rules of congruence of modules 7 and 3is the reason, why these folded with two and threes numbers be place on straight line of line,which confirms the legitimacy of the Riemann hypothesis. The uniformity from what rises thegraph of primes e.g.: by 100 000, he owes not quantity of primes to number N what can

express with logarithmic function, but proportionate distributing, resulting from congruenceof according to modules 7.

P(p) = (2, 3 + R2 + R4), 2, 3+ 2 = 5, 3 + 4 = 7= 5 + 2, 7 + 4 = 11 + 2 = 13 + 4 = 17 + 2 = 19 + 4 = 23 + 2 =

25 + 4 = 29 + 2 = 31 + 4 = 35 + 2 = 37 + 4 = 41 + 2 = 43 + 4 = 47 + 2 = 49 + 4 = 53 + 2 = 55

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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45

p'="p" mod p 65 = 13(2) + 13(3) 137 = 67(2) + 3 p' - "p" = n/p 137 - 65 = 72/2

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2 3

751311

269275281

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„Upon looking at these numbers, one has the feeling of being in the presence of theinexplicable secrets of creation.‖ /D. Zagier/ Are the primes distributed chaotically or can we find some means for computing them?Will it ever be possible to predict with arbitrary accuracy where the next one lies? Yes, hereyou are! p + 6(7) = p‘, or almost prime ―p‖ 

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The prime numbers are distributed not chaotically. All prime and almost prime numbers to becongruent modulo 7.Because the smallest gap between their equal 2 + 4 = 6, and 6(7) = 42than is possible to predict with arbitrary accuracy that the next one lies what 42 gap.

Primes and almost prime can settle according to their quantity. Such in a row creates fourteenthe vertical groups and the innumerable amount of horizontal rows / periods / primes andalmost prime. Length period 42 = 7(6) it is product of length of period all natural numbers and

seven units about what grow primes and almost prime. In third and eighth group exceptingprime 7, have only almost prime numbers.I II III IV V VI VII VIII IX X XI XII XIII XIV

5 7 11 13 17 19 23

25 29 31 35 37 41 43

47 49 53 55 59 61 65

67 71 73 77 79 83 85

89 91 95 97 101 103 107

109 113 115 119 121 125 127

131 133 137 139 143 145 149

151 155 157 161 163 167 169

173 175 179 181 185 187 191

193 197 199 203 205 209 211

215 217 221 223 227 229 233

235 239 241 245 247 251 253

257 259 263 265 269 271 275

277 281 283 287 289 293 295

299 301 305 307 311 313 317

319 323 325 329 331 335 337341 343 347 349 353 355 359

p + 7(6) = p' 5 -42- 47 -42- 89 -42- 131 -42- 173 -84- 257 -126- 383 -84-467 -42- 509 -,,, "p"+ 7(6)

= "p' " 35 -42- 77 -42- 119 -42- 161 -42- 203 - 42- 245 -42- 287 -42- 329 -42- 371 -42- 413 -42- 455 -

42- 497 -42- 539 ..

0 23

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Number 19, is prime in that case the sum 19 + 42 = 61 is prime too. Number 9 091, is primein that case the sum 9 091 + 42 = 9 133 is prime too. 9 091  – 19 = 9 072 : 42 = 216Number 909 091, is prime in that case the sum 909 091 + 42 = 909 133 is prime too.Number 909 090 909 090 909 090 909 090 909 091, is prime so I can predict with arbitrary

accuracy that the next one lies in gap 42 = 909 090 909 090 909 090 909 090 909 133 – 43 = 909 090 909 090 909 090 909 090 909 090 : 42 = 21 645 021 645 021 645 021 6 45 021 645

909 1, is prime + 42 = 9 133 too,

909 091, is prime + 42 = 909 133 too,

9 090 909 091 = 11 · 23 · 4093 · 8779 

909 090 909 091 = 859 · 1 058 313 049

9 090 909 090 909 091 = 103 · 4013 · 21 993 833 369

909 090 909 090 909 091, is prime

9 090 909 090 909 090 909 091, is prime

909 090 909 090 909 090 909 091, is prime

9 090 909 090 909 090 909 090 909 091 = 59(154 083 204 930 662 557 781 201 849)

909 090 909 090 909 090 909 090 909 091, is prime too. They are 4,6,18, 22,24, and 30 digitsprimes. One from 100 and 1000 million digits prime are 9.090909091e99 999 999and 9.090909133e999 999 999.

8 264 462 809 917 355 371 900 826 446 281 They are 32 digits 90 909 090 909 090 909 090 909 090 909 091 is dividable by 11 and

e38―p― = e26 + e10 + e2 105 831 304 899 989 415 869 510 001 058 313 049 38 digits numbers 90 909 090 909 090 909 090 909 090 909 090 909 091 : 859

e32 „p― = e22 + e9 + e1 8 264 462 809 917 355 371 900 826 446 281 

+ 82 644 628 099 173 553 719 008 264 462 8190 909 090 909 090 909 090 909 090 909 091

e99 999 998 ―p― = e4 545 454(22) + e9 + e1 = 9.090909091e99 999 997

e1000 000 000 „p― = e38 461 538(26) + e10 + e2 = 9.090909091e999 999 999 

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2 + 3 + 5 + 7 + 11,..+ 29 + 31 + 37 = 222 35 + 41 + 43 + 47 + 49,..+ 71 + 73 + 79 = 798

77 + 83 + 85,.. + 113 + 115 + 121 = 1386 119 + 125 + 127,.. + 155 + 157 + 163 = 1986

222 – 48(12) – 798 – 49(12) -1386 – 50(12) – 1986 ,.. 14n + [14n + n(12)],..

They in this way grow with 14 of primes and almost prime built-up terms, arrange inexquisite mosaic illustrating their row in intervals 2 and 4 in arrangement of sevens.There are two facts about the distribution of prime numbers of which I hope to convince youso overwhelmingly that they will be permanently engraved in your hearts. The first is that,despite their simple definition and role as the building blocks of the natural numbers, theprime numbers same for me a balding blocks, that is to say every prime bigger than 3 the sumtheir predecessor 2, 3, 5, 11, 13, and 29 is, and n-the multiplicity of prime 7. They grow notlike weeds among the natural numbers, seeming to obey no other law than that of chance, andnobody can predict where the next one will sprout. The second fact is even more astonishing,for it states just the opposite: that the prime numbers exhibit stunning regularity, that there are

laws governing their behavior, congruence laws modulo 7, and that they obey these laws withmilitary precision. To support the first of these claims, let me begin by showing you a list of the prime up to 100. I hope you will agree that there is apparent reason why one number isprime and another not.2, 3, 2 + 3 = 5 5 + 6(7) = 47

5 + 2 = 7 11 + 6(7) = 532(2) + 7 = 11 3 + 8(7) = 592(3) + 7 = 13 5 + 8(7) = 613 + 2(7) = 17 11 + 8(7) = 675 + 2(7) = 19 29 + 6(7) = 712 + 3(7) = 23 3 + 10(7) = 73

1 + 4(7) = 29 2 + 11(7) = 793 + 4(7) = 31 13 + 10(7) = 83

p + 2(7) = p' + 4(7) = p" 3 + 2 = 5 + 2(7) = 19 + 4(7) = 47 + 2(7) = 61 + 4(7) = 89 + 2(7) = 103 + 4(7) =

131 + 2(7) = 145 + 4(7) = 173

519

47 61 89 103 131 145 173

1341 55 83 97 125 139 167

7

35

49

77

91

119

133

161

175

29

43

71

85

113

127

155

169

23

37

65

79

107

121

149

16317

31 59 73 101 115 143 15711

25 53 67 95 109 137 151 179

2

181

3

185

0%

20%

40%

60%

80%

100%

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27

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2 + 5(7) = 37 5 + 12(7) = 8913 + 4(7) = 41 13 + 12(7) = 9729 + 2(7) = 43

I  II  III  IV  V  VI  VII  VIII  IX  X  XI  XII  XIII 2  3  5  7  4  6  8  9 

11  13  10  12  14  15 

17  19  16  18  20  21 

23  25  22  24  26  27 

31  29  28  30  32  33 

37  35  34  36  38  39 

41  43  40  42  44  45 

47  49  46  48  50  51 

53  55  52  54  56  57 59  61  58  60  62  63 

67  65  64  66  68  69 

73  71  70  72  74  75 

79  77  76  78  80  81 

83  85  82  84  86  87 

89  91  88  90  92  93 

97  95  94  96  98  99 

101  103  100  102  104  105 

107  109  106  108  110  111 

113  115  112  114  116  117 

121  119  118  120  122  123 

127  125  124  126  128  129 

The numbers 2 and 3 are building blocks all natural numbers. Even indivisible by 2 and 3prime and almost prime numbers can you from n(2) and n(3) to put together e.g. 2 + 3 = 52(2) + 3 = 7 4(2) + 3 = 11 5(2) + 3 = 13 7(2) + 3 = 17 8(2) + 3 = 19 10(2) + 3 = 235(2) + 5(3) = 25 9(3) = 27The periodical table of natural numbers distinguishes 13 groups of even and odd numbers. Incolumns I - VII we have prime numbers appearing what n(7). e.g. 5 + 2(7) = 19 + 4(7) = 47 +

2(7) = 61 + 4(7) = 89 + 2(7) = 103 + 4(7) = 131 + 6(7) = 173 …In VI column except 7 we have free places on stepping out what n(7) almost prime numbers.e.g. 35 + 2(7) = 49 + 4(7) = 77 + 2(7) = 91 + 4(7) = 119 + 2(7) = 133 + 4(7) = 161 which arein VIII and IX column.25 + 2(5) = 35 + 4(5) = 55 + 2(5) = 65 + 4(5) = 85 + 2(5) = 95 + 4(5) = 115 + 2(5) = 125 …121 + 2(11) = 143 + 4(11) = 187 + 2(11) = 209 + 4(11) = 253 + 2(11) = 275 + 4(11) = 319 …

In tenth and twelfth column we have even numbers, and in XI and XIII column even and oddnumbers divisible by 3, following what 2(3).

101, 1 001=11(91), 100 001=11(9091), 10 000 001=11(909 091), 1.000 001E+99 999 999

103, 1 003=17(59), 100 003, 1 000 003, 1.000 003E+12,+18,+19,+99 999 999,+999 999 999

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107, 1 007=19(53), 100 007, 1 000 007, 1.000 007E+99 999 999, E+999 999 999

109, 1 009, 10 009, 100 009, 1 000 009, 1.000 009E+99 999 999, E+999 999 999

113, 1 013, 10 013, 100 013, 1 000 013, 1.000 013E+99 999 999, E+999 999 999

115, 1 015, 10 015, 100 015, 1 000 015, 1.000 015E+99 999 999, E+999 999 999

119, 1 019, 10 019, 100 019, 1 000 019, 1.000 019E+99 999 999, E+999 999 999

121, 1 021, 10 021, 100 021, 1 000 021, 1.000 021E+99 999 999, E+999 999 999

125, 1 025, 10 025, 100 025, 1 000 025, 1.000 025E+99 999 999, E+999 999 999

127, 1 027=13(79), 100 027, 1 000 027, 1.000 027E+99 999 999, E+999 999 999

131, 1 031, 10 031, 100 031, 1 000 031, 1.000 031E+99 999 999, E+999 999 999

133, 1 033, 10 033, 100 033, 1 000 033, 1.000 033E+99 999 999, E+999 999 999

137, 1 037=17(61), 10 037, 100 037, 1 000 037, 1.000 037E+14, E+16, E+99 999 999

139, 1 039, 10 039, 100 039, 1 000 039, 1.000 039E+13, E+99 999 999, E+999 999 999

2, 3, 5, 11, 13, 29 + n(7) = p

2 + 15(7) = 107 3 + 14(7) = 101 5 + 14(7) = 103

11 + 14(7) = 109 13 + 18(7) = 139 29 + 12(7) = 113

3 + 148(7) = 1 039 29 + 1430(7) = 10 039 5 + 142 862(7) = 1 000 039

5 + 142 857 142 862(7) = 1 000 000 000 039

5 + 142 857 142 862e99 999 999(7) = 1.000 000 039E+100 000 000 

5 + 142 857 142 862e999 999 999(7) = 1.000 000 039E+1000 000 000

3 + 1(7) = 10

2 + 14(7) = 100

6 + 142(7) = 1 000

4 + 1 428(7) = 10 000

5 + 14 285(7) = 100 000

1 + 142 857(7) = 1,00E+06

3 + 1 428 571(7) = `1,00E+07

2 + 14 285 714(7) = 1,00E+08

6 + 142 857 142(7) = 1,00E+09

4 + 1 428 571 428(7) = 1,00E+10

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5 + 14 285 714 285(7) = 1,00E+11

1 + 142 857 142 857(7) = 1,00E+12

3 + 1 428 571 428 571(7) = 1,00E+13

2 + 14 285 714 285 714(7) = 1,00E+14

6 + 142 857 142 857 142(7) = 1,00E+15

4 + 1 428 571 428 571 428(7) = 1,00E+16

5 + 14 285 714 285 714 285(7) = 1,00E+17

4 + 1,428 571 428e99(7) = 1,00E+100

4 + 1,428 571 428e999(7) = 1,00E+1000

4 + 1,428 571 428e99 999 999(7) = 1,00E+100 000 000

4 + 1,428 571 428e999 999 999(7) = 1,00E+1000 000 000

Factorization of almost primes in prime factors.

Factorise large numbers on factors prime, it was in last 2000 years difficult problem. Majoritymathematicians‘ is opinion, that factorisation numbers is fundamental extraordinarycomputational problem. One of main reasons, why the factorise numbers is so difficult; shewas alleged fortuity of occurrence of primes.We know since, that primes and almost prime are present not accidentally, but according torules of congruence of modules 7 and 3, we have also the way on factorise their products.It with theorems about congruence odd numbers results brightly, or number is prime or almostprime, and we for help of binomial formula easily will take apart every odd number on factors

prime. We know, that difference among two successive square numbers state always oddnumber, we have such number from here to write down in some way as difference twosquares and take out the common factor.Difference of 2 squares:

a(a) – b(b) = (a – b)(a + b)p is common to both terms. Put this common factor outside the brackets.

―p‖ = p(p) ―p‖ = p(p + p`) 25 = 5(2 + 3)„p― = p(p´) = [(p + p‗)/2 –  {(p + p‗)/2 –   p}][(p + p‗)/2 + {(p + p‗)/2 – p}]

147 573 952 589 676 412 927 = 193 707 721(761 838 257 287) =[(193 707 721 + 761 838 257 287)/2 – {(193 707 721 + 761 838 257 287)/2 – 193 707 721}][(193 707 721 + 761 838 257 287)/2 + {(193 707 721 + 761 838 257 287)/2 – 193 707 721}]

(381 015 982 505 – 380 822 274 783)( 381 015 982 505 + 380 822 274 783)

35 = 6(6) – 1(1) = (6 – 1)(6 + 1) = 5(7) 55 = 8(8) – 3(3) = (8 – 3)(8 + 3) = 5(11)

143 = 12(12) – 1(1) = (12 – 1)(12+ 1) = 11(13) 221 = 15(15)-2(2) =(15-2)(15 + 2) = 13(17)

253 = 17(17) – 6(6) = (17 – 6)(17+ 6) = 11(23) 247 = 16(16)-3(3) =(16-3)(16 + 3) = 13(19)If the difference between one number a, and a prime number is divisible by a prime number,then the number is complex.

)1'(  p p pa e. g. 40

77287    p(p‘) = (p‘- 1)p + p 7(41) = (41 – 1)7 + 7

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The straight is dividing by 3 given number and to subtract from her rounded quotient withoutthe rest. Then we level rounded quotient to the closest third multiplicity of prime numbers oralmost prime. About the same number we round off received previously the difference to n -the multiplicity of the same prime numbers or almost prime. And so we receive prime factorson what factorizing is number almost prime.

319 : 3 = 106 - 19 = 87 : 3 = 29 343 : 3 = 114 + 33 = 147 : 3 = 49

319 - 106 = 213 + 19 = 232 : 8 = 29 343 - 11 4 = 229 - 33 = 196 : 4 = 49

319 = 11(29) 343 = 7(49)

8051 : 3 = 2683 – 2392 = 291 : 3 = 978051 – 2683 = 5368 + 2392 = 7760 : 80 = 97

8051 = 83(97)

9 090 909 091 : 3 = 3 030 303 030 – 550 964 187 = 2 479 338 843 : 3 = 826 446 2819090909091 – 3030303030 =6060606061 + 550964187 = 6611570248 : 8 = 826446281

9 090 909 091 = 11(826 446 281)

909 090 909 091 : 3 = 303 030 303 030 – 299 855 363 883 = 3174939147 : 3 = 1 058 313 049909090909091 – 303030303030 = 606060606061 + 299855363883 = 905 915 969 944 : 856=1 058 313 049 909 090 909 091 = 859(1 058 313 049)

9090909090909091 : 3 = 3030303030303030 – 2765519270373639 = 264783759929391 : 3=88261253309797

9090909090909091 – 3030303030303030 = 6060606060606061 + 2 765 519 270 373 639 =8 826 125 330 979 700 : 100 

9 090 909 090 909 091 = 103(88 261 253 309 797)

9  090 909 090 909 090 909 090 909 091 : 3 = 3 030 303 030 303 030 303 030 303 030- 2 568 053 415 511 042 629 686 697 483

462 249 614 791 987 673 343 605 547/ 3 = 154 083 204 930 662 557 781 201 849

9090909090909090909090909091 – 3030303030303030303030303030 = 6 060 606 060 606 060 606 060 606 061

+ 2 568 053 415 511 042 629 686 697 483

8 628 659 476 117 103 235 747 303 544 : 56 = 154 083 204 930 662 557 781 201 8499 090 909 090 909 090 909 090 909 091 = 59(154 083 204 930 662 557 781 201 849)

8 051 = 90(90) – 7(7) = (90 – 7)(90 + 7) = 83(97) 493=23(23)-6(6)=(23-6)(23+6) = 17(29)

341 = 21(21)-10(10) = (21-10)(21+10) = 11(31) 391 = 20(20)-3(3) =(20-3)(20+3)= 17(23)

529 = 23(20+3) 497 = 17(68+3) 1105 = 17(62 + 3) 1309 = 17(74 + 3) 1147 = 31(34 + 3)

1369 = 37(34 + 3) 25271 = 37(680 + 3) 734 591 = 11(66778 + 3)

8453= 79(107) 11111 = 41(271) 120481 = 211(571) 526313=281(1873) 322577= 163(1979)

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434779=197(2207) 353357=307(1151) 10 000 043=2089(4787) 10 000 127= 167(59881)

370 267 = 479(773) 370 283 = 379(977) 370 289 = 349(1061) 370 297 = 353(1049)

370 303 = 367(1009) 370 319 = 547(677) 370 327 = 107(3461) 370 339 = 199(1861)

370 351 = 179(2069) 370 361 = 383(967) 370 309 = 67(5527) 370 313 = 47(7879)

370 273 = 43(8611) 370 301 = 29(12769) 370 333 = 37(10009) 370 369 = 23(16103)

370 243 = 17(21779) 370 249 = 11(33659) 370 253 = 13(28481) 370 271 = 11(33661)

370 277 = 17(21781) 370 291 = 19(19489) 370 331 = 13(28487) 370 337 = 11(33667)

370 379 = 17(21787) 370 343 = 59(6277) 370 279 = 7(52897) 370 381 = 11(33671)

370 307 = 7(52901) 370 321 = 7(52903) 370 349 = 7(52907) 370 363 = 7(52909)

9 999 913 = 7(1428559) 9 999 917 = 23(434779) 9 999 941 = 7(1428563)

9 999 947 = 19(526313) 9 999 949 = 31(322579) 9 999 971 =13(769229)

10 000 001 = 11(909 091) 10 000 003 = 13(769 231) 10 000 007 = 941(10 627)

10 000 009 = 23(434 783) 10 000 013 = 421(23 753) 10 000 021 = 97(103 093)

10 000 027 = 37(270 271) 10 000 031 = 227(44 053) 10 000 037 = 43(232 559)

10 000 039 = 7(1 428 577) 10 000 033 = 397(25 189) 10 000 043 = 2 089(4 787)

10 000 049 = 47(212 767) 10 000 061 = 19(526 319) 10 000 067 = 7(1 428 581)

10 000 081 = 7(1 428 583) 10 000 091 = 251(39 841) 10 000 093 = 53(188 681)

10 000 097 = 17(588 241) 10 000 099 = 19(526 321) 10 000 111 = 11(909 101)

10 000 123 = 7(1 428 589) 10 000 127 = 167(59 881) 10 000 133 = 11(909 103)

10 000 129=89(112361) 10 000 171 = 271(36901) 10 000 187 = 41(243907) 4 294 967 297=6 700 417(638+3) 1000001=101(9901) 8 547 008 547(13) = 111 111 111 111

7 709 321 041 217 = 25 271(305 065 927)

7 709321041217=(152 545 599-152520 328)(152 545 599+152 520 328)=25271(305065927)

We have with same the also fast way on qualification of primes, necessary to construction of 

code the RSA. She in end was found hidden behind primes and almost prime full secrets

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structure, since ages in demand throughout mathematicians, and her music can write inaddition in infinity.

Who knows this basic interval two four, two four, knows also where what note will come withprime or almost prime numbers. We cannot already more now tell about their fortuity, but

more about them timeless and universal character.

No perceptible order and Riemann‘s Hypothesis.

Mathematicians since centuries listened intently in sound primes, and they heard unsettledtones only. These numbers resemble accidentally spilled notes on mathematical notes paper,without recognizable melody. Riemann sinusoidal waves what created right away zero Zetathey - showed scenery hidden harmony.

Mathematicians despite all could with sure probability to estimate, how many prime numbers

is in given interval. Only four in first ten are (2, 3, 5 and 7). It in first hundred is them 25, infirst thousand 168 their part comes down from 40 by 25 on 16,8 percentage.

Among smaller numbers from billion, 5% is the only just. To describe this down come of frequency of an occurrence in approximation the simple formula. From this however satisfiedmathematicians are not. They want to know how far real occurrence numbers first deviatesfrom counted frequency. Riemann in one's famous eight page paper ―On the Number of Prime Numbers less than a Given Quantity / "Über die Anzahl der Primzahlen unter einer gegebenen Größe" / he wrote: "The known approximating expression F( x) = Li(x) is therefore

valid up to quantities of the order x 2

1

and gives somewhat too large a value; But also the

increase and decrease in the density of the primes from place to place that is dependent on theperiodic terms has already excited attention, without however any law governing thisbehavior having been observed. In any future count it would be interesting to keep track of the influence of the individual periodic terms in the expression for the density of the primenumbers.‖ Real quantity of prime numbers differs from them counted frequency so aloneoften, as eagle near repeated throw with coin will fall out tails. Differently saying that isRiemann supposed, that occurrence prime numbers be subject to the rights of case. And hemade a mistake here, because prime numbers be subject to the rights of congruence of according to module p‘≡ p (mod.7).

He has written: ―One now finds indeed approximately this number of real roots within theselimits, and it is very probable that all roots are real. Certainly one would wish for a stricter

 proof here.‖ 

Riemann Hipothesis.

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Riemann zeta- function for s = 0,5 + i * t.

The Riemann hypothesis (also called the Riemann zeta-hypothesis), along with suitablegeneralizations, is considered by many mathematicians to be the most important unresolvedproblem in pure mathematics. First formulated by Bernhard Riemann in 1859, it haswithstood concentrated efforts from many outstanding mathematicians for 150 years (as of 2009).

The Riemann hypothesis (RH) is a conjecture about the distribution of the zeros of theRiemann zeta-function ζ(s). The Riemann zeta-function is defined for all complex numbers s ≠ 1. It has zeros at the negative even integers (i.e. at s = −2, s = −4, s = −6, ...). These are

called the trivial zeros. The Riemann hypothesis is concerned with the non-trivial zeros, andstates that:

The real part of any non-trivial zero of the Riemann zeta function is ½. 

Thus the non-trivial zeros should lie on the so-called critical line, ½ + it , where t  is a realnumber and i is the imaginary unit. The Riemann zeta-function along the critical line issometimes studied in terms of the Z-function, whose real zeros correspond to the zeros of thezeta-function on the critical line.

The Riemann hypothesis implies a large body of other important results. Most mathematicians

believe the Riemann hypothesis to be true, A $1,000,000 prize has been offered by the ClayMathematics Institute for the first correct proof.

Unsolved problems in mathematics:  Does every non-trivial zero of the Riemann zeta function have real part ½? 

Riemann mentioned the conjecture that became known as the Riemann hypothesis in his 1859paper On the Number of Primes Less Than a Given Magnitude, but as it was not essential tohis central purpose in that paper, he did not attempt a proof. Riemann knew that the non-trivial zeros of the zeta-function were symmetrically distributed about the line s = ½ + it , andhe knew that all of its non-trivial zeros must lie in the range 0 ≤ Re(s) ≤ 1. 

In 1896, Hadamard and de la Vallée-Poussin independently proved that no zeros could lie onthe line Re(s) = 1. Together with the other properties of non-trivial zeros proved by Riemann,

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this showed that all non-trivial zeros must lie in the interior of the critical strip 0 < Re(s) < 1.This was a key step in the first proofs of the prime number theorem. 

In 1900, Hilbert included the Riemann hypothesis in his famous list of 23 unsolved problems — it is part of Problem 8 in Hilbert's list, along with the Goldbach conjecture. When asked

what he would do if awakened after having slept for five hundred years, Hilbert said his firstquestion would be whether the Riemann hypothesis had been proven (Derbyshire 2003:197).The Riemann Hypothesis is one of the Clay Mathematics Institute Millennium PrizeProblems. 

In 1914, Hardy proved that an infinite number of zeros lie on the critical line Re(s) = ½.However, it was still possible that an infinite number (and possibly the majority) of non-trivial zeros could lie elsewhere in the critical strip. Later work by Hardy and Littlewood in1921 and by Selberg in 1942 gave estimates for the average density of zeros on the criticalline.

The Riemann hypothesis and primes

The zeta-function has a deep connection to the distribution of  prime numbers. Riemann gavean explicit formula for the number of primes less than a given number in terms of a sum overthe zeros of the Riemann zeta function. Helge von Koch proved in 1901 that the Riemannhypothesis is equivalent to the following considerable strengthening of the prime numbertheorem: for every ε > 0, we have

where π( x) is the prime-counting function, ln( x) is the natural logarithm of  x, and the Landaunotation is used on the right-hand side.[5]  A non-asymptotic version, due to LowellSchoenfeld, says that the Riemann hypothesis is equivalent to

The zeros of the Riemann zeta-function and the prime numbers satisfy a certain dualityproperty, known as the explicit formulae, which shows that in the language of  Fourier

analysis the zeros of the Riemann zeta-function can be regarded as the harmonic frequenciesin the distribution of primes.

The Riemann hypothesis can be generalized by replacing the Riemann zeta-function by theformally similar, but much more general, global L-functions. In this broader setting, oneexpects the non-trivial zeros of the global  L-functions to have real part 1/2, and this is calledthe generalized Riemann hypothesis (GRH). It is this conjecture, rather than the classicalRiemann hypothesis only for the single Riemann zeta-function, which accounts for the trueimportance of the Riemann hypothesis in mathematics. In other words, the importance of 'theRiemann hypothesis' in mathematics today really stems from the importance of thegeneralized Riemann hypothesis, but it is simpler to refer to the Riemann hypothesis only in

its original special case when describing the problem to people outside of mathematics.

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For many global  L-functions of  function fields (but not number fields), the Riemannhypothesis has been proven. For instance, the fact that the Gauss sum, of the quadraticcharacter of a finite field of size q (with q odd), has absolute value

is actually an instance of the Riemann hypothesis in the function field setting.

Consequences and equivalent formulations of the Riemann hypothesis

The practical uses of the Riemann hypothesis include many propositions which are stated tobe true under the Riemann hypothesis, and some which can be shown to be equivalent to theRiemann hypothesis. One is the rate of growth in the error term of the prime number theoremgiven above.

Riemann was particularly interested in feeding imaginary numbers into functions. Usually we

can draw a graph of a function where the input runs along the horizontal and the output is theheight of the graph. But a graph of an imaginary function consists of a landscape where theoutput is represented by the height above any point in the world of imaginary numbers.

An imaginary landscape

Riemann had found one very special imaginary landscape, generated by something called the zeta function, which he discovered held the secret to prime numbers. In particular, the pointsat sea-level in the landscape could be used to produce these special harmonic waves whichchanged Gauss's graph into the genuine staircase of the primes. Riemann used the coordinates

of each point at sea-level to create one of the prime number harmonics. The frequency of eachharmonic was determined by how far north the corresponding point at sea-level was, and howloud each harmonic sounded was determined by the east-west frequency.

Riemann‘s sinus –  waves what created with zero place Zeta  –  topography, they showedhidden harmony.

A prime number is a positive whole number greater than one which is divisible only by itself 

and one. The first few are shown above. If the definition doesn‘t mean much to you, think of prime numbers as follows:

If you are presented with a pile of 28 stones, you will eventually deduce that the pile can bedivided into 2 equal piles of 14, 4 equal piles of 7, 7 equal piles of 4, etc. However, if onemore stone is added to the pile, creating a total of 29, you can spend as long as you like, butyou will never be able to divide it into equal piles (other than the trivial 29 piles of 1 stone).In this way, we see that 29 is a prime number, whereas 28 is non-prime or composite.All composites break down uniquely into a product of prime factors: i.e. 28 = 2 x 2 x 7. Notethat 2 is the only even prime - all other even numbers are divisible by 2. 1 is neither prime nor

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composite by convention.

Here the sequence of primes is presented graphically in terms of a step function or counting

function which is traditionally denoted . (Note: this has nothing to do with the value

=3.14159...) The height of the graph at horizontal position x indicates the number of primes less than or equal to x. Hence at each prime value of x we see a vertical jump of oneunit. Note that the positions of primes constitute just about the most fundamental, inarguable,nontrivial information available to our consciousness. This transcends history, culture, andopinion. It would appear to exist 'outside' space and time and yet to be accessible to anyconsciousness with some sense of repetition, rhythm, or counting. The explanation in theprevious page involving piles of stones can be used to communicate the concept of primenumbers without the use of spoken language, or to a young child

By zooming out to see the distribution of primes within the first 100 natural numbers, we seethat the discrete step function is beginning to suggest a curve.

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Zooming out by another factor of 10, the suggested curve becomes even more apparent.Zooming much further, we would expect to see the "granular" nature of the actual graph

vanish into the pixilation of the screen.

Now zooming out by a factor of 50, we get the above graph. Senior Max Planck Institutemathematician Don Zagier, in his article "The first 50 million primes" [ Mathematical

 Intelligencer , 0 (1977) 1-19] states:"For me, the smoothness with which this curve climbs is one of the most astonishing facts inmathematics."(Note however that you are not looking at a smooth curve. Sufficiently powerfulmagnification would reveal that it was made of unit steps. The smoothness to which Zagierrefers is smoothness in limit .)The juxtaposition of this property with the apparent 'randomness' of the individual positionsof the primes creates a sort of tension which can be witnessed in many popular-mathematicalaccounts of the distribution of prime numbers. Adjectives such as "surprising", "astonishing",

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"remarkable", "striking", "beautiful", "stunning" and "breathtaking" have been used. Zagiercaptures this tension perfectly in the same article:

In 1896, de la Valee Poussin and Hadamard simultaneously proved what had been suspectedfor several decades, and what is now known as the  prime number theorem: 

In words, the (discontinuous) prime counting function is asymptotic to the (smooth)

logarithmic function x /log x. This means that the ratio of to x /log x can be made

arbitrarily close to 1 by considering large enough x. Hence x /log x provides an approximationof the number of primes less than or equal to  x, and if we take sufficiently large x thisapproximation can be made as accurate as we would like (proportionally speaking - verysimply, as close to 100% accuracy as is desired).The original proofs of the prime number theorem suggested other, better approximations. In

the above graph we see that x /log x, despite being asymptotic to , is far from being the

smooth function which suggests when we zoom out - there is plenty of room forimproving the approximation. These improvements turn out to be greatly revealing.

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The first improvement on x/log x we consider is the logarithmic integral function Li(x). Thisis defined to be the area under the curve of the function 1/log u between 2 and x, as illustratedin the left-hand figure. Gauss arrived at this from the empirical fact that the probability of 

 finding a prime number at an integer value near a very large number x is almost exactly 1/log x.

l'Hopital's rule can be used to show that the ratio of  x /log x to tends to 1

as x approaches infinity. Thus we may use either expression as an approximation to inthe statement of the prime number theorem.

In the right-hand figure we see that this function provides an excellent fit to the function. As Zagier states, "within the accuracy of our picture, the two coincide exactly."

Zagier goes on to state:

"There is one more approximation which I would like to mention. Riemann's research onprime numbers suggests that the probability for a large number  x to be prime should be evencloser to 1/log  x if one counted not only the prime numbers but also the  powers of primes,counting the square of a prime as half a prime, the cube of a prime as a third, etc. This leads tothe approximation

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or, equivalently,

The function on the right side . . . is denoted by R( x), in honor of Riemann. It represents an

amazingly good approximation to as the above values show."To be clear about this, it should be pointed out that the explicit definition for the the functionR( x) is

where are the Möbius numbers. These are defined to be zero when n is divisible by asquare, and otherwise to equal (-1)k  where k is the number of distinct prime factors in n. As 1

has no prime factors, it follows that (1) = 1.

It seems, then, that the distribution of prime numbers 'points to' or implies Riemann's functionR( x). This function can be thought of as a smooth ideal to which the actual, jagged, primecounting function clings. The next layer of information contained in the primes can be seen

above, which is the result of subtracting from R( x). This function relates directly to thelocal fluctuations of the density of primes from their mean density.

In their article "Are prime numbers regularly ordered?", three Argentinian chaos theoristsconsidered this function, treated it as a 'signal', and calculated its  Liapunov exponents. Theseare generally computed for signals originating with physical phenomena, and allow one todecide whether or not the underlying mechanism is chaotic. The authors conclude"...a regular pattern describing the prime number distribution cannot be found. Also, from aphysical point of view, we can say that any physical system whose dynamics is unknown butisomorphic to the prime number distribution has a chaotic behavior."A physicist shown the above graph might naturally think to attempt a Fourier analysis - i.e. tosee if this noisy signal can be decomposed into a number of periodic sine-wave functions. Infact something very much like this is possible. To understand how, we must look at the

Riemann zeta function. 

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We encountered this graph earlier. Recall that it shows us the deviations of the prime counting

function from the smooth approximating function R(x). It was hinted that this noisyfunction might somehow decompose into fairly simple component functions. Indeed, this isthe case.The usual process of Fourier analysis essentially decomposes "signals" such as this into

(periodic) sine wave functions. In this case, the component functions are quasi-periodic, basedon sine waves but with a particular kind of logarithmic deformation.Remarkably, the functions in question, the sum of which produces the function seen above,are intimately connected with the nontrivial zeros of the zeta function which we've just seen.

The difference function R( x) - seen earlier can be expressed as the infinite sum over the

set of zeros (both trivial and nontrivial) of the Riemann zeta function: 

This sum separates into sums over the trivial and nontrivial zeros respectively. The former is

the relatively simple function R( x-2

) + R( x-4

) + R( x-6

) + ...

The sum over the nontrivial zeros can be expressed as the sum of the sequence of functions {-Tk ( x)} where Tk  is defined as follows:

where the and are the k th pair of nontrivial zeta zeros, which we know must becomplex conjugates. The first four functions T1( x), T2( x),T3( x), and T4( x) are pictured above.

Our first apparent obstacle is that and are complex numbers. However, the function xk  

can be meaningfully extended from real k to complex k in a fairly straightforward way. Thismeans that the and are also complex- valued.

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This also initially seems like a problem, as the Riemann function R defined earlier as an

approximation of was clearly intended to act on real values only. However, by the sameprocess of analytic continuation discussed earlier, R can be extended to the entire complexplane, taking the form given by the Gram series:

Here ln x is the usual extension of the logarithm function to . Also note the role of theRiemann zeta function.

So we see that and can be given precise meanings, and will yield complexnumbers. Usefully, the imaginary parts of this pair of complex numbers can be shown tocancel, so that their sum which is Tk ( x) will always be a real-valued function.

Some numbers have the special property that they cannot be expressed as the product of twosmaller numbers, e.g., 2, 3, 5, 7, etc. Such numbers are called  prime numbers, and they playan important role, both in pure mathematics and its applications. The distribution of suchprime numbers among all natural numbers does not follow any regular pattern; however theGerman mathematician G.F.B. Riemann (1826 - 1866) observed that the frequency of primenumbers is very closely related to the behavior of an elaborate function

ζ (s) = 1 + 1/2s

+ 1/3s

+ 1/4s

+...

called the   Riemann Zeta function. The Riemann hypothesis asserts that all interesting solutions of the equation

ζ(s) = 0

lie on a certain vertical straight line. This has been checked for the first 1,500,000,000solutions. A proof that it is true for every interesting solution would shed light on many of themysteries surrounding the distribution of prime numbers.

Primes seem to be, at the same time very irregularly distributed among all numbers, and yet  –  if squinted at from a sufficiently far distance – they reveal an astoundingly elegant pattern.

Over 2,300 years ago Euclid proved that the number of primes is infinite, so two possiblequestions come to mind:

1. How many primes are there less than the number x?2. There are infinitely many primes, but how big of an infinity?

This document will focus on the first question.1.1. π(x) is the number of primes less than or equal to xLet x be a positive real number. The question "how many primes are there less than x?" hasbeen asked so frequently that its answer has a name:

[π (x) using the Greek letter pi] = π(x) = the number of primes less than or equal to x.

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The primes under 25 are 2, 3, 5, 7, 11, 13, 17, 19 and 23 so π(3) = 2, π(10) = 4 and π25) = 9.

Table 1. Values of π(x) 

x π(x) reference

10 4100 251,000 16810,000 1,229100,000 9,5921,000,000 78,49810,000,000 664,579100,000,000 5,761,4551,000,000,000 50,847,53410,000,000,000 455,052,511100,000,000,000 4,118,054,8131,000,000,000,000 37,607,912,01810,000,000,000,000 346,065,536,839100,000,000,000,000 3,204,941,750,802 [LMO85]1,000,000,000,000,000 29,844,570,422,669 [LMO85]10,000,000,000,000,000 279,238,341,033,925 [LMO85]100,000,000,000,000,000 2,623,557,157,654,233 [DR96]1,000,000,000,000,000,000 24,739,954,287,740,860 [DR96]10,000,000,000,000,000,000 234,057,667,276,344,607100,000,000,000,000,000,000 2,220,819,602,560,918,8401,000,000,000,000,000,000,000 21,127,269,486,018,731,928

10,000,000,000,000,000,000,000 201,467,286,689,315,906,290100,000,000,000,000,000,000,000 1,925,320,391,606,803,968,923

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

2 5 110

31 41

0

61 71

0 0

101

0 0

131

0

151

0 0

181 191

3 7 13 23

0

43 53

0

73 83

0

103 113

0 0 0 0

163 173

0

193

0 017

0

37 47

0

67

0 0

97 107

0

127 137

0

157 167

0 0

197

0 019 29

0 0

59

0

79 89

0

109

00

139 149

0 0

179

0

199

π(10) = 4 = 7(0,57143), π(100) = 25 = 7(3,57143), π(1 000) = 168

=7(0,57143)6(7)

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In 1859 the German mathematician Bernard Riemann proposed a way of understanding andrefining that pattern. Its main result is a suggestion, not rigorously proved, for a perfectlyprecise formula giving the number of primes less than a given quantity. His hypothesis haswide – ranging implications, and this day after 150 years of careful research and exhaustive

study we know it is correct.

N  π(x)  π(N)  π(x) = 4(n)7 4 = 8(0,5) 

1,00E+03  168  5,9524  168 = 8(0,5)6(7) 

7 

1,00E+06  78 498  12,739  78 498 = 8(0,5) 2803,5(7) 

7 

1,00E+09  50 847 534  19,667  50 847 534 = 8(0,5)1 815 983,3514286(7) 

7 

1,00E+12  37 607 912 018  26,59  37 607 912 018 = 8(0,5)1 343 1139 714,92857(7)  

7 

1,00E+15  29 844 570 422 669  33,507  29 844 570 422 669 = 8(0,5)1 065 877 515 095,32(7)  

7 

1,00E+18  24 739 954 287 740 860  40,42  24739954287740860 = 8(0,5)883569795990,745(7) 

7 

1,00E+21  21127269486018731928  47,332  21127269486018731928=4(754545338786383283,143)7  

There are 4 primes up to 10 (2, 3, 5, 7), because those they cannot be expressed as the productof two smaller numbers (4 = 2(2), 8 = 4(2), 9 = 3(3), 10 = 5(2). Between 1 and 100 there are

25 primes, and 168 primes up to 1 000. Why 168? Is there a rule, a formula, to tell me howmany primes there are less than a given number?The formula is simple: The ratio of half a given number by a given number N, is directlyproportional to the quotient of quantity prime numbers by its dual quantity.

½N : N = πx : 2(πx)  πx ∝ ½N ½N(2 πx) = N(πx) πx/(2 πx) = y = ½ 

2πx(½) = πx

N πx 

1,0 E+3 168

1,0 E+6 78498

1,0 E+9 508475341,0 E+12 37607912018

1,0 E+15 29844570422669

1,0 E+18 24739954287740860

1,0 E+21 2112726946018731928

A two – column table like this is an illustration of a function. The main idea of a function isthat some number depends on some other number according to some fixed rule or procedure.Another way to say the same thing is: a function is a way to turn (―maps‖) a number in toanother number.

The function πx ∝ ½N turns, or maps, the number 1000 in to the number 168 – again, by wayof some definite procedure. 500(336) = 1000(168)

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Therefore primes there are less than a given number sure do thin out, but are directlyproportional to the half a given number.We know that all prime numbers be congruent to me modulo 7, and this seven tell me howmany primes there are less than a given number. We show in table, that quantity primes thereare less than a given number is always product number 4 = 7(0,57143), and n  – the multiple

number 7. If so the formula is:π(x) = 4(n)7, 4 = 7(0,57143), 168 = 4(6)7

Ultimately, it is in the Riemann Hypothesis about the multiplicative basic building blocks of natural numbers to understand: the primes. Can their distribution in the sea of natural numbersmean? How long do you calculate until the next prime coming? Why is the next primenumber, such as accidental times already after a few steps, sometimes on the other hand, onlyafter great distance? Is there perhaps a hidden pattern?

2  0  3  0  5 

7  9  11 13  15  17 

19  21  23 

25  27  29 

31  33  35 

37  39  41 

43  45  47 

49  51  53 

55  57  59 

61  63  65 67  69  71 

0

100

200

300

400

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Serie1 2 11 23 41 59 73 97 109 137 157 179 197 227 241 269 283

Serie2 3 13 29 43 61 79 101 113 139 163 181 199 229 251 271 293

Serie3 5 17 31 47 67 83 103 127 149 167 191 211 233 257 277 307

Serie4 7 19 37 53 71 89 107 131 151 173 193 223 239 263 281 311

7(0,57143) = 4 = π (10), 4(6)7 = 168 = π (1 000)

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73  75  77 

79  81  83 

85  87  89 

91  93  95 

97  99  101 

103  105  107 

109  111  113 

115  117  119 

121  123  125 

127  129  131 

133  135  137 

139  141  143 

145  147  149 

151  153  155 

157  159  161 

163  165  167 

169  171  173 

175  177  179 

181  183  185 

187  189  191 

193  195  197 

199  201  203 

205  207  209 

211  213  215 217  219  221 

223  225  227 

229  231  233 

235  237  239 

241  243  245 

247  249  251 

253  255  257 

259  261  263 

265  267  269 

271  273  275 

277  279  281 

283  285  287 

289  291  293 

295  297  299 

301  303  305 

307  309  311 

313  315  317 

319  321  323 

325  327  329 

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589  591  593 

595  597  599 

601  603  605 

607  609  611 

613  615  617 

619  621  623 

625  627  629 

631  633  635 

637  639  641 

643  645  647 

649  651  653 

655  657  659 

661  663  665 

667  669  671 

673  675  677 

679  681  683 

685  687  689 

691  693  695 

697  699  701 

703  705  707 

709  711  713 

715  717  719 

721  723  725 

727  729  731 733  735  737 

739  741  743 

745  747  749 

751  753  755 

757  759  761 

763  765  767 

769  771  773 

775  777  779 

781  783  785 

787  789  791 

793  795  797 

799  801  803 

805  807  809 

811  813  815 

817  819  821 

823  825  827 

829  831  833 

835  837  839 

841  843  845 

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99

847  849  851 

853  855  857 

859  861  863 

865  867  869 

871  873  875 

877  879  881 

883  885  887 

889  891  893 

895  897  899 

901  903  905 

907  909  911 

913  915  917 

919  921  923 

925  927  929 

931  933  935 

937  939  941 

943  945  947 

949  951  953 

955  957  959 

961  963  965 

967  969  971 

973  975  977 

979  981  983 

985  987  989 991  993  995 

997  999  1001 

All zeros of ζ(s) in the half-plane Re(s) > 0 have real part ½. 

2  0  3 

0  7  0 

5  0  9 

0  13  0 

11  0  15 

0  19  0 

17  0  21 

0  25  0 

23  0  27 

0  31  0 

29  0  33 

0  37  0 

35  0  39 

0  43  0 41  0  45 

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100

0  49  0 

47  0  51 

0  55  0 

53  0  57 

0  61  0 

59  0  63 

0  67  0 

65  0  69 

0  73  0 

71  0  75 

0  79  0 

77  0  81 

0  85  0 

83  0  87 

0  91  0 

89  0  93 

0  97  0 

95  0  99 

0  103  0 

[p + (p+8)]/2 = 2n – 1 then 2(2n – 1) = [p + (p+8)] p = [2(2n – 1) – 8]/2[5 + (5 + 8)]/2 = 9 2(9) = [5 + (5 + 8)] 5 = [2(9) – 8]/2

[11 + (11 + 8)]/2 = 15 2(15) = [11 + (11 + 8)] 11 = [2(15) – 8]/2

The Riemann Hypothesis is about the distribution of primes in the sea of natural numbers.This sea is defined over the sum, because of numbers will always be number 1 add  – just thenormal process of counting. The primes, however, are about the multiplication defined, theyare about the factorization the prime multiplicative components of the natural numbers.The distribution of primes and the Riemann Hypothesis says something about the changingrelationship between addition and multiplication of natural numbers. Both are not problemsfor themselves, but both together are incredibly complex and still not fully penetrated, such asthe lack of evidence for the Riemann Hypothesis impressive displays.All these ideas are based on an analogy, which is easier to describe something like this lets:The primes are ―elementary particles‖, which are about the multiplication in interaction occur 

and so the composite numbers up. At the same time, ―the particles‖ are arranged through theaddition. In the zeta functions are now in the form of a sum  – relatively product formula bothaspects (additive/natural numbers and multiplicative/primes) linked.

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101

2 + 3 = 5 + 2 = 2(2) + 3 = 7 + 3 + 2 = 12 = 6(2)2 + 3 = 5 + 4(2) = 13 = 5(2) + 3 + 5 = 18 = 9(2)

4(2) + 3 = 11 + 4(2) = 19 = 8(2) + 3 + 11 = 30 = 15(2)7(2) + 3 = 17 + 4(2) = 25 = 5(2 + 3) + 17 = 42 = 21(2)

10(2) + 3 = 23 + 4(2) = 31 = 14(2) + 3 + 23 = 54 = 27(2)13(2) + 3 = 29 + 4(2) = 37 = 17(2) + 3 + 29 = 66 = 33(2)

2 

3  7  12 

5  13  18 

11  19  30 

17  25  42 

23  31  54 

29  37  66 

43  35  78 41  49  90 

47  55  102 

53  61  114 

59  67  126 

73  65  138 

71  79  150 

85  77  162 

83  91  174 

89  97  186 

103  95  198 

0

50

100

150

200

250

1 2 3 4 56 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

2 3 5 11 17 23 29

0

41 47 53 59

0

71

0

83 89

0

101 107 113

0 7 13 19 31 37 43

0 0

61 67 73 79

0 0

97 103 109

0 0

0 0 0 025

0 0 0

49 55

0 0 0 0

85 91

0 0 0

115 121

0 0 0 0 0 0 0

35

0 0 0 0

65

0

77

0 0

95

0 0 0

0 12 18 30 42 54 66 78 90 102 114 126 138 150 162 174186

198210

222234

2 + 3 = 5 + 4(2) = 13 = 5(2) + 3 + 5 = 18 = 9(2)4(2) + 3 = 11 + 4(2) = 19 = 8(2) + 3 + 11 = 30 = 15(2)

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103

359  367  726 

373  365  738 

379  371  750 

377  385  762 

383  391  774 

389  397  786 

403  395  798 

401  409  810 

407  415  822 

421  413  834 

419  427  846 

433  425  858 

431  439  870 

437  445  882 

443  451  894 

449  457  906 

463  455  918 

461  469  930 

467  475  942 

481  473  954 

487  479  966 

493  485  978 

491  499  990 

505  497  1002 503  511  1014 

509  517  1026 

523  515  1038 

521  529  1050 

535  527  1062 

541  533  1074 

547  539  1086 

553  545  1098 

559  551  1110 

557  565  1122 

563  571  1134 

577  569  1146 

583  575  1158 

589  581  1170 

587  595  1182 

593  601  1194 

599  607  1206 

Theorem: The quotient of half a given magnitude ½N, by a given magnitude N, is directlyproportional to the quotient of quantity prime numbers, by its dual quantity.

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Proof: ½N : N = πx : 2(πx)  πx ∝ ½N  ½N(2 πx) = N(πx) πx/(2 πx) = Re(s) = ½ 

5/10 = 4/8 = 3/6 = 2/4 = 1/2 8(1/2) = 4 2πx(½) = πx

 N 

 x N    22

1

 

In the interval of 10 numbers of 5 consecutive odd numbers is always one divisible by 3, sothat it can act like this at most 4 primes, so double large interval of 8 numbers claim ( 4·2 = 819 - 11 = 8 11, 12, 13, 14, 15, 16, 17 , 18, 19). Generally in the interval of 10 numbers eacheven and odd number 2 number interval claims, including in 4(2), 3(2), 2(2), 1(2), primes. If the interval first 10 numbers come before 4 primes, then in the interval of 100 numbers 25prime numbers occur, and each number in the interval, the proportion ½ is to keep. In otherwords, all zeros of ζ(s) in the half-plane Re(s) > 0 have real part ½ .

In mathematics, two quantities are said to be proportional if they vary in such a way that oneof the quantities is a constant multiple of the other, or equivalently if they have a constantratio. Proportion also refers to the equality of two ratios. In proportional quantities is thedoubling (tripling, halved) one quantity is always a double (triple, halve) connected to theother quantities.

2πx  πx N ½ N 2 2 2 2

8 4 10 5 2 3 5 7

16 8 20 10 11 13 17 19

20 10 30 15 0 23 0 29

24 12 40 20 31 0 37 0

30 15 50 25 41 43 47 0

34 17 60 30 0 53 0 59

38 19 70 35 61 0 67 0

44 22 80 40 71 73 0 79

48 24 90 45 0 83 0 89

50 25 100 50 0 0 97 0

58 29 110 55 101 103 107 109

The proportion of ½ means, that is involved in the creation of a half-block of numbers twicethe amount of prime numbers. 5/10 = 4/8 50/100 = 25/50 500/1000 = 168/336

1 = 3 –  2 5 = 3 + 2 7 = 5 + 2 9 = 7 + 2

11 = 9 + 2 13 = 11 + 2 15 = 13 + 2 17 = 15 + 2 19 = 17 + 2 21 = 19 + 2 23 = 21 + 225 = 23 + 2 27 = 25 + 2 29 = 27 + 2 31 = 29 + 2 33 = 31 + 2 35 = 33 + 2 37 = 35 + 239 = 37 + 2 41 = 39 + 2 43 = 41 + 2 45 = 43 + 2 47 = 45 + 2 49 = 47 + 2 51 = 49 + 253 = 51 + 2 55 = 53 + 2 57 = 55 + 2 59 = 57 + 2 61 = 59 + 2 63 = 61 + 2 65 = 63 + 267 = 65 + 2 69 = 67 + 2 71 = 69 + 2 73 = 71 + 2 75 = 73 + 2 77 = 75 + 2 79 = 77 + 281 = 79 + 2 83 = 81 + 2 85 = 83 + 2 87 = 85 + 2 89 = 87 + 2 91 = 89 + 2 93 = 91 + 295 = 93 + 2 97 = 95 + 2 99 = 97 + 2

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Assuming proportional functions graphically in a coordinate system, so you can see thatproportional functions are monotonically increasing. The properties of the zeroes out in thecomplex plane determine the properties of the primes! Riemann conjectured that all the

relevant zeroes have real part ½ . The statement that the equation πx/2(πx) = y = ½, is valid

for every x with real part equal ½, with the quotient on the right hand side converging, isequivalent to the Riemann hypothesis.

2 3 5 711 13

17 19 23

29 31

3741 43

4753

59 6167

71 73

7983

89

97

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

πx ∝ ½N ½N(2πx) = N(πx) 2πx(½) = πx 8 · 0,5 = 4 50 · 0,5 = 25

y = 0,5x

R² = 1

0

5

10

15

20

25

30

35

0 10 20 30 40 50 60 70

       Y

X

πx = 2πx(0,5) 4 = 8(0,5) 25 = 50(0,5) 168 = 336(0,5)

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Riemann for help of total numbers translated distribution of prime numbers in mathematicalscenery on two-dimensional plane (so called zeta-function). It topography of this scenerycontains near this general knowledge about prime numbers. It will suffice, so to know onlevel of sea points (zero places), to can reconstruct whole scenery.

2

3

5

7 11

13

17

1923

29

31

37

41

4347

53

59

61

6771

73

79 83

8997

101

103

πx ∝ ½N ½N(2πx) = N(πx) 2πx(½) = πx 8 · 0,5 = 4 50 · 0,5 = 25

20

50

11

0

17

0

23

0

29

0

35

0

41

0

47

0

53

0

59

0

65

0

71

0

77

0

83

0

89

00

7

0

13

0

19

0

25

0

31

0

37

0

43

0

49

0

55

0

61

0

67

0

73

0

79

0

85

0

91

0

97

3

0

9

0

15

0

21

0

27

0

33

0

39

0

45

0

51

0

57

0

63

0

69

0

75

0

81

0

87

0

93

0

0 5 10 15 20 25 30 35

[p + (p + 8)]/2 = 2n - 1 [5 + (5 + 8)]/2 = 9 [11 + (11 + 8)]/2 = 15 [ 17 +

(17 + 8)]/2 = 21 [23 + (23 + 8)]/2 = 27

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107

Because zero places contain all information about distribution of prime numbers. Riemanncreated concrete formula, to right away zero to regain distribution of prime numbers. Nearwhat every zero place is how source for spreading wave we which can introduce me asacoustic sound. Sounds of all zero places overlap on me in distribution of prime numbers.Near what zero place is about so many louder, if it lies further eastwards (in right from axis -

y), and her sound is about so many higher, if it lies further north (over axis - x).

They fill with the same gap in thousands theorems basing on legitimacy Riemann‘shypothesis. Because many mathematicians be obliged for its results such presumption toaccept. Primes betrayed their secret, and by this was proved Riemann‘s Hypothesis 

A solution of the Riemann Hypothesis are huge implications for many other mathematicalproblems. The transformation of hypothesis the Riemann in theorem, suddenly it proves allthe not proved results.

Riemann Hypothesis admits to receive, so that really every from infinitely of many, of zeroplaces lies on this straight line then it means, that all sounds in music of prime numbers arealike loud. This would mean, it that was can distribution of prime numbers to me reallyintroduce how even throw dice. Hexahedron dice after line of natural numbers rolls, whichwhat second and fourth wall shows next prime number or almost prime. 

5_7__11_13__17_19__23_25=5·5__29_31__35=5·7_37__41_43__47_49=7·7__53_55=11·5 

__59_61__65 = 13·5_67__71_73__77 = 11·7_79__83_85 = 17·5__89_91 = (13·7)_95 = 19·5 

y = 0,5x

R² = 1

-5E+17

0

5E+17

1E+18

1,5E+18

2E+18

2,5E+18

-2E+18 0 2E+18 4E+18 6E+18

     y

x

πx ∝ ½N ½N(2πx) = N(πx) 8 · 0,5 = 4 50 · 0,5 = 25

336 · 0,5 = 168 πx/(2 πx) = y = ½ 2πx(½) = πx

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109

½ N = πx + [½ N – (πx + N/6)] + N/6 50 = 25 + [50 – (25 +16)] + 16

2

0

5

0

11

0

17

0

23

0

29

0

35

0

41

0

47

0

53

0

59

0

65

0

71

0

77

0

83

0

89

0

95

0

101

0

0

3

0

9

0

15

0

21

0

27

0

33

0

39

0

450

510

570

630

690

750

810

870

930

990

105

7

0

13

0

19

0

25

0

31

0

37

0

43

0

49

0

55

0

61

0

67

0

73

0

79

0

85

0

91

0

97

0

103

0

109

0

0%

10%

20%

30%

40%

50%

60%

70%

80%

90%

100%

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

2

0

30

5

7

9

11

13

15

17

19

21

23 25

27

29

31

3335

37

39

41

43

45

47 49

51

53 55

57

59

61

6365

67

69

71

73

7577

79

81

83 85

87

89 91

9395

97

99

101

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110

Riemann hypothesis:The non-trivial zeros (lie zeros in the strip right next to the y-axis with real part of s from 0 to 1)are all on a line parallel to the y-axis with real part ½ .

Proof:

547 89 131 173 215 257 299 341 383 425 467 509 551 593 635 677 719 761 803 845 887 929971

7

49 91 133 175 217 259 301 343 385 427 469 511 553 595 637 679 721 763 805 847 889 931973

11

5395 137 179 221 263 305 347 389 431 473 515 557 599 641 683 725 767 809 851 893 935977

13

5597 139 181 223 265 307 349 391 433 475 517 559 601 643 685 727 769 811 853 895 937979

17

59101 143 185 227 269 311 353 395 437 479 521 563 605 647 689 731 773 815 857 899 941983

19

61103 145 187 229 271 313 355 397 439 481 523 565 607 649 691 733 775 817 859 901 943985

23

65107 149 191 233 275 317 359 401 443 485 527 569 611 653 695 737 779 821 863 905 947989

25

67109 151 193 235 277 319 361 403 445 487 529 571 613 655 697 739 781 823 865 907 949991

29

71113 155 197 239 281 323 365 407 449 491 533 575 617 659 701 743 785 827 869 911 953995

31

73115 157 199 241 283 325 367 409 451 493 535 577 619 661 703 745 787 829 871 913 955997

35

77119 161 203 245 287 329 371 413 455 497 539 581 623 665 707 749 791 833 875 917 9591001

37

79 121 163 205 247 289 331 373 415 457 499 541 583 625 667 709 751 793 835 877 919 96110034183 125 167 209 251 293 335 377 419 461 503 545 587 629 671 713 755 797 839 881 923 9651007

43 85 127 169 211 253 295 337 379 421 463 505 547 589 631 673 715 757 799 841 883 925 9671009

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

1/2N = πx + [1/2N - (πx + N/6)] + N/6 500 = 168 + [500 - (168 + 166)] +

166

0 2 33 7513111917

2523

3129

3735

4341

4947

5553

6159

6765

7371

79

1 2 3

y^- 5y + 6 = 0 x^= 1 y = x+ 1 = 2 y = 4x - 1 = 3+ 4= 7 y= 8 x- 3= 5+ 8 = 13

y=8x+3 = 11 + 8= 19 y=8x+9= 17

y=8x+15=23 + 8=31 y=8x+21=29+8=37 y= 8x+27=35 +8=43 y=8x+33=41

y=8x+39=47 y= 8x+45=53+8=61

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111

What is the solution of the equation y² - 5y + 6 = 0? The solution is 2 und 3. Another way to say,

this is that 2 and 3 are the zeros of the function y ² - 5y + 6 = 0.

y y‘ ½ y 

y² - 10y + 21 = 0 Solution 3 and 7 5

y² - 18y + 65 = 0 - „ - 5 and 13 9

y² - 30y + 209 = 0 - „ - 11 and 19 15

y² - 42y + 425 = 0 - „ - 17 and 25 = 5(5) 21

y² - 54y + 713 = 0 - „ - 23 and 31 27

y² - 66y +1073 = 0 - ― - 29 and 37 33

y² - 78y +1505 = 0 5(7 )= 35 and 43 39

y² - 90y +2009 = 0 - ― - 41 and 49 = 7(7) 45

y² - 102y +2585 = 0 47 i 55 = 5(11) 51

q.e.d.

2 0 00 2,50 03

05

0

7

05

0

9

0

13

0

11

0

15

0

19

0

17

0

21

0

25

0

23

0

27

0

31

0

29

0

33

0

37

0

35

0

39

0

43

0

41

0

45

0

49

0

47

0

51

0

55

0

53

0

57

0

61

0

59

0

63

0

67

0

65

0

69

0

73

0

71

0

75

0

79

0

77

0

81

0

85

0

83

0

87

0

91

0

89

0

93

0

97

0

95

1 2 3

(y + y')/2 = 1/2 y (2+3)/2=2,5 (3+7)/2=5 (5+13)/2=9 (11+19)/2=15(17+25)/2=21 (23+ 31)/2=27 (29+37)/2=33

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112

A zero off the critical line would induce a pattern in the distribution of the primes.

2

19

29

0 0

59

0

79

89

0

109

313

23

0

43

53

0

73

83

0

103

7 17

0

37

47

0

67

0 0

97

107

5 11

0

31

41

0

61

71

0 0

101

0 0 0 0 0 0 0 0 0 0 0

0 2 4 6 8 10 12

p + 5(6) = p' 7 + 5(6) = 37 + 5(6) = 67 + 5(6) = 97 + 5(6) = 127 + 5(6) = 157

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114

97  101  103 

107  109 

115  119  113 

121  125  127 

133  131 

143  137  139 

145  149  151 

155  157 

161  163  167 

169  173 

175  179  181 

187  185  191 

193  197  199 

205  203 

209  211 

217  215  223 

221  227  229 

235  233  239 

247  245  241 

253  251 

259  257  263 

265  269  271 

275  277 

287  281  283 289  293 

295  299 

301  305  307  311 

319  313  317 

325  323 

329  331 

343  335  337 

341  347  349 

355  353  359 

361  365  367 

The Riemann Hypothesis had been proved, and we are able, to answering the severity of theproblem of Goldbach to go, whether each grade number as the sum of two primes is representable. If proportionality factor all primes in a given quantity ½ is, but this means that the

equation πx/2πx = ½N/N is the answer to the problem of Goldbach. She says that every evennumber is composed of two primes.

Theorem: The quotient of quantity prime numbers by its dual quantity, is directly proportionalto the quotient of quantity even numbers by a given magnitude.

Proof: πx/2πx = 2n/N 25/50 = 50/100 = ½ 

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115

The proportion of ½ in the case of even numbers means that all even numbers in a block madeup of two primes. 2 + 2 = 4 3 + 3 = 6 3 + 5 = 8 5 + 5 = 10 That is to say 50 even numbers in a block of 100 numbers, is the sum of 4(25) primes, asshown in the diagram below.

-2

410 16 22 28 34 40 46 52 58 64 70 76 82 88 94 100

0

612 18 24 30 36 42 48 54 60 66 72 78 84 90 96 102

2

8 14 20 26 32 38 44 50 56 62 68 74 80 86 92 98 104

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

2n/N = πx/2πx 50/100 = 25/50

Serie1 Serie2 Serie3

2 55

11 11 17 17 23 23 2923 29 29

41 41 41

47

25 11 11 17 17 23 23 29 29 41 41 47 41 47 53

53

4

10 16 22 28 34 40 46 52 58 64 70 76 82 88 94

100

35 7

11 13 17 19 17 23 29 29 31 37 41 43 433

7 11 13 17 19 23 29 31 31 37 41 41 43 47 53

6

12 18 24 30 36 42 48 54 60 66 72 78 84 90 963

7 713 13

1913 19 19

31 31 37 37 4331 31

57 13 13 19 19 31 31 37 31 37 37 43 43 61 67

8 14 20 26 32 38 44 50 56 62 68 74 80 86 92 98

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

πx/2πx = 1/2N/N 2n = p + p 25/50 = 50/100 4 = 2 + 2

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116

How we see on above diagram, sum two prime numbers lies always on parallel straight line toaxis - y and it is even numbers that is consisting with two prime numbers.

0 002

02

02

0

4

0

3

0

3

0

6

0

3

0

5

0

8

0

5

0

5

0

10

0

5

0

7

0

12

0

7

0

7

0

14

0

5

0

11

0

16

0

7

0

11

0

18

0

7

0

13

0

20

0

11

0

11

0

22

0

11

0

13

0

24

0

13

0

13

0

26

0

11

0

17

0

28

0

11

0

19

0

30

0

13

0

19

0

32

0

17

0

17

0

34

0

17

0

19

0

36

0

19

0

19

0

38

0

17

0

23

0

40

0

19

0

23

0

42

0

13

0

31

0

44

0

23

0

23

0 0

0 0,5 1 1,5 2 2,5 3 3,5

n/p + p = 1/2 2n = p + p' 2+2=4 3+3=6 3+5=8 5+5=10 5+7=12

7+7=14 5+11=16 7+11=18 7+13=20...

2 23

0

7

0

3 3 3

0

9

0

3 3

5

0

11

0

3

5 5

0

13

0

5 5 5

0

15

0

5 5

7

0

17

0

5

7 7

0

19

0

5 5

11

0

21

0

5 5

13

0

23

0

7 7

11

0

25

0

7 7

13

0

27

0

1 2 3

2n-1=p+p´+p" 7= 2+2+3 9 = 3+3+3 11=3+3+5 13=3+5+5 15=5+5+5

17=5+ 5+7 19=5+7+7 19-16=17-14=15-12=13-10=11-8=9-6=7-4=3

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117

Legitimacy "strong‖ hypothesis Goldbacha proves legitimacy "weak‖ hypothesis Goldbacha -because 3 have will sufficed from given odd larger number since 7 to subtract and tointroduce received even number with strong hypothesis Goldbacha peaceably.

(2n - 1) - 3 = 2n = p + p' → 2n - 1 = p + p + p

Also, as you can see, every odd integer greater than 5 is the sum of 3 primes, because thedifference between odd and even numbers always of prime numbers 3 is.

4

0

4

0

7

0

6

0

6

0

9

0

8

0

8

0

11

0

10

0

10

0

13

0

12

0

12

0

15

0

14

0

14

0

17

0

16

0

16

0

19

0

18

0

18

0

21

0

20

0

20

0

23

0

22

0

22

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0

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0

24

0

27

0

26

0

26

0

29

0

28

0

28

31

0 0,5 1 1,5 2 2,5 3 3,5

2n + p = 2n - 1 = p + p + p' 2 + 2 + 3 = 7 3 + 3 + 3 = 9 3 + 5 + 3 = 11

5 + 5 + 3 = 13 5 + 7 + 3 = 15

0

2

4

6

8

10

12

14

16

17

16

14

12

10

8

6

4

2

0

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 0

p+p=2n+p=2n-1 2+2=4+3=7 3+3=6+3=9 3+5=8+3=11 5+5=10+3=13

5+7=12+3 =15 7+7=14+3=17

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In addition to the familiar question of whether there are infinitely many prime pairs withdifference 2 there. The six- wide array further helps to demonstrate the otherwise stillunproven conjecture that there must be infinitely many twin primes.Here are the reasons for this: if there are infinite primes, then twin pairs, with even number

divisible by 3 shares.

5 70

60

11 13

0

12

0

17 19

0

18

0

29 31

0

30

0

41 43

0

42

0

59 61

0

60

0

71 73

0

72

0

101 103

0

102

0

107 109

0

108

0

137 139

0

138

0

149 151

0

150

0

1 2 3

p + (p +2)/2 = 2n/3 5+7/2=6/3 11+13/2=12/3 17+19/2=18/3

29+31/2=30/3 41+43/2=42/3

0 0 0

2

0

32,5

3

7

5

5

13

9

11

0

19

15

2y+3y=5y/2=2+0,5=3-0,5 3y+7y=10y/2=3+2=7-2

5y+13y=18y/2=5+4=13-4 11y+19y=30y/2=11+4=19-4

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Looking closer at the above graph, you will see that half of the following sums of two primeson a straight line parallel to the y - axis with real part ½ y lie. This means that the linear Diophantine equation ax + by - c = 0, with given integer pairs not have common divisorCoefficient a, b, c, always in prime x, y is solvable.1(2) + 1(3) – 5 = 0 1(3) + 1(7) – 10 = 0 1(5) + 1(13) – 18 = 0 1(11) + 1(19) – 30 = 0

Still, then, we go on listing to that mysterious prime numbers beat: 2, 3, 5, 7, 11, 13, 17, 19.The primes stretch out into the far reaches of the universe of numbers, never running dry. Dowe really have to accept that, despite our desire for order and explanation, these fundamentalnumbers might forever remain out of reach?

Sale long we reflect upon with perspective Gauß and Riemann‘s and we should earlier alreadylook for different possibilities, to better to get know these full of secrets numbers. The primesbetrayed in end their secret, and remain not an unanswered riddle. I‘m who made the primessing.

The system of numbers is not the man‘s invention, because in distribution of primes and

almost primes uncover coded plan of building of not only nature, but whole universe. Thereality of existence of timeless plan motivates consideration, or also for events in time andspace that is our history does not hide transcendent guidance.

Number reveals divine think and order. It permits simultaneously to get to know the basicstructures of reality. Number assures insight in God‘s internal secret and the world‘s.  Who knows definite number, possesses power. Calculating man makes something like howonly God, when full arranging power over things: he distinguishes and it assigns, he definesborders and it unites together. He places also qualitative standards: what it lays on first place,is most valuable, important, and significant.

Then our perception called with science would bring us to confession with belief for Book Wisdom 11, 21: ―But you settled all according to measure, number and weight‖, and under -standing his deep meaning. Seemingly disarrays are regulated, for what God let will be thanksthat we need not what the least million years wait on understanding of mysterious nature of primes.

FOR THE GREATER GLORY OF GOD.

„AD MAJOREM DEI GLORIAM 

SOLI DEO HONOR ET GLORIA.

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2,3,5,11,13,29, + n(7) = p p(p) + 6(7) = p‗(p―) 7(7) + 6(7) = 13(7) 5(7) + 6(7) = 11(7)

2 3 5 7

17 19 13 11

23 29 25

37 31 35

47 41 43 49

59 53 55

61 67 65

79 73 71 77

89 83 85

97 91 95

107 101 103 109

113 119 115

127 121 125

131 139 137 133

149 143 145

157 151 155

163 167 169 161

173 179 175

181 185 187

191 199 193 197

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205 203 209

211 215 217

227 229 223 221

233 239 235

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257 251 259 253

269 263 265

271 277 275

283 281 289 287

293 299 295

307 305 301

317 311 313 319

325 323 329

331 337 335

349 347 341 343359 353 355

367 361 365

373 379 377 371

383 389 385

397 391 395

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443 449 445

457 455 451

467 461 463 469

479 475 473

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509 503 505

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535 533 539541 547 545

557 559 551 553

569 563 565

577 571 575

587 589 581 583

593 599 595

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647 643 641 649

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653 659 655

661 667 665

677 673 679 671

683 685 689

691 697 695

709 701 703 707

719 715 713

727 721 725

733 739 731 737

743 745 749

751 757 755

761 769 763 767

773 775 779

787 781 785

797 793 791 799809 805 803

811 817 815

821 829 823 827

839 835 833

841 845 847

857 859 853 851

863 865 869

877 871 875

887 881 883 889

895 893 899

907 901 905

919 911 913 917

929 925 923

937 931 935

947 941 943 949

953 955 959

967 961 965

971 977 973 979

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997 991 995

1009 1007 1001 1003

1013 1019 1015

1021 1025 1027

1031 1039 1033 1037

1049 1045 1043

1051 1057 1055

1069 1063 1061 1067

1075 1073 1079

1087 1081 1085

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1097 1091 1093 1099

1109 1103 1105

1117 1111 1115

1129 1123 1121 1127

1135 1133 1139

1141 1145 1147

1151 1153 1157 1159

1163 1165 1169

1171 1177 1175

1181 1187 1183 1189

1193 1195 1199

1201 1207 1205

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1223 1229 1225

1237 1231 12351249 1243 1241 1247

1259 1255 1253

1261 1265 1267

1277 1279 1271 1273

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1297 1291 1295

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1471 1477 1475

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1567 1565 1561

1571 1579 1573 1577

1583 1585 1589

1597 1591 1595

1601 1609 1607 1603

1619 1613 1615

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1645 1643 1649

1657 1651 1655

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1997 1993 1999 1991

2003 2009 2005

2011 2017 2015

2029 2027 2023 2021

2039 2035 2033

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2063 2069 2065

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2363 2369 2365

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2389 2383 2381 2387

2399 2393 2395

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3797 3793 3791 3799

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3833 3835 3839

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3853 3851 3857 3859

3863 3865 3869

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3997 3995 3991

4007 4001 4003 4009

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4025

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4105 4103 4109

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4127 4129 4123 41214139 4133 4135

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4315 4313 4319

4327 4321 4325

4337 4339 4333 4331

4349 4345 4343

4357 4351 4355

4363 4369 4361 4367

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4381 4385 43874391 4397 4393 4399

4409 4405 4403

4411 4415 4417

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4579 4577 4571 45734583 4585 4589

4597 4591 4595

4603 4601 4607 4609

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4771 4775 4777

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4831 4835 48374849 4847 4841 4843

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4951 4957 4955

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4993 4999 4997 4991

5003 5009 5005

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5021 5023 5029 50275039 5035 5033

5041 5047 5045

5059 5051 5053 5057

5065 5063 5069

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9323 9329 9325

9337 9335 93319341 9343 9349 9347

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11159 11153 11155

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11177 11173 11171 11179

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11207 11201 11209 11203

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12775  12773  12779 12781  12787  12785 

12791  12799 12793  12797 

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12817  12815  12811 

12829  12823  12821  12827 

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12841 12845  12847 

12853 12857  12851  12859 

12863  12869  12865 12877  12875  12871 

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12923 12929  12925 

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12967 12965  12961 

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13025  13027  13021 

13037 13033  13031  13039 

13043  13049 13045 13055  13051  13057 

13063 13067  13061  13069 

13079  13073  13075 

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13177 13171  13175 

13183  13187  13181  13189 

13199  13193  13195 13205  13201  13207 

13219  13217  13211  13213 

13229  13223  13225 

13235  13231  13237 

13249  13241  13247  13243 

13259 13253  13255 

13267  13265  13261 

13277  13271  13273  13279 

13289  13283  13285 

13291  13297  13295 

13309  13301  13307  13303 

13313  13319  13315 

13327  13325  13321 

13337  13331 13339  13333 

13343  13349  13345 

13355  13351  13357 

13367  13361  13363  13369 

13373  13375 

13381  13379 13385  13387 

13399 13397  13391  13393 

13403  13409  13405 

13417  13411  13415 

13421  13427  13423  13429 

13433  13439  13435 

13441 13445  13447 

13457 13451  13453  13459 

13463  13469 13465 

13477  13475  13471 

13487  13481  13483  13489 

13499 13493  13495 

13505  13501  13507 

13513 13517  13511  13519 

13523  13529  13525 

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13607  13609  13601  13603 

13613  13619  13615 13627  13621  13625 

13633  13637  13631  13639 

13649  13645  13643 

13651  13655  13657 

13669  13667  13663  13661 

13679 13675  13673 

13687  13681 13685 

13697  13693 13691  13699 

13709 13703  13705 13711  13715  13717 

13629  13723  13721  13727 

13739  13735  13733 

13741  13745  13747 

13757  13751 13759  13753 

13763 13765  13769 

13771  13775  13777 

13781  13789  13783  13787 

13799  13795  13793 

13807 13801  13805 

13817  13813  13811  13819 

13829  13825  13823 

13831  13837  13835 

13841  13843  13847  13849 

13859  13853  13855 

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13877 13879  13873  13871 

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14009  14003  14005 

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14071 14075  14077 

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E X P O S É. Jan Lubinaul. Korfantego 51/10Pl. 43-200 Pszczynae-mail: lubinaj7@gmail.com 

Personal informationI was born in Katowice/Poland on 01.05.1947. I'm no mathematician, but mathematics hasbecome increasingly intriguing to me. I find it extremely fascinating. As my earliermathematical research had not involved number theory.Mathematics is often referred to as the search for patterns; it is the language of nature.Everything around us can be represented and understood through numbers, and primenumbers are the building block of all numbers, the DNA of arithmetic.But the famous mathematician Euler had to say about the primes: "There are some mysteriesthat the human mind will never penetrate. To convince ourselves we have only to cast aglance at tables of primes and we should perceive that there reigns neither order nor rule."And a Hungarian mathematician Paul Erdös, who spent his career investigating primenumbers, said that it will be another million years, at least, before we understand the primes.Despite Euler´s and Erdös pessimism, I have found ways to understand the primes, and theirdistribution.Are there formulas that produce some of the prime? Here you are! p = n(2) + 32 = 1(2) + 0 3 = 0(2) + 3 5 = 1(2) + 3 7 = 2(2) + 3 11 = 4(2) + 3 13 = 5(2) + 3

17 = 7(2) + 3 19 = 8(2) + 3 23 = 10(2) + 3 233 = 115(2) + 3 251 = 124(2) + 3Why a number is prime? Because could be written as two smaller numbers multipliedtogether. That is, it is not possible to represent a prime as the product of two integers a x b with a, b > 1. Let q and r be the quotient and remainder of the division of n by d . That is, foreach n and d, let n = d q + r, where r and q are positive integers and 0 ≤ r < d .

Because all prime numbers contain in me one 3, it was not possible divide here by two.Superiority meanwhile 2 it causes, that they don´t divide by three also. So they are indi visibleby all different numbers, and on this depends the complete primality certificate!"God does not play dice with the universe." /A. Einstein/, also not with the prime numbers.The nature used not a dice to decide if the number N is prime, but rule of congruence modulo

 p‘≡ p mod 7. p = n(7) + (1, 2, 3, 4, 5, 6) 11 = (7) + 4 13 = (7) + 6 17 = 2(7) + 3

19 = 2(7) + 5 23 = 3(7) + 1 29 = 3(7) + 4(2) 31 = 4(7) + 3 37 = 5(7) + 2 p‘ – p = n(7) 19 – 5 = 2(7) 47 – 19 = 4(7) 61 – 47 = 2(7) 89 – 61 = 4(7) 103 – 89 = 2(7)So the next question is, can we understand how the primes are distributed? Can the primes befitted into a pattern in the way that the elements can be organized in the Periodic Table? Theanswer is yes!

And I find a good model for the way the primes are distributed. It looks like they have beenchosen with 6 sides dice on one side what two and four space is painted successive prime 2, 3,_5_7__11_13__17_19__23_25__29_31__35_37__41_43__47_49__53_55__59_61__65

_67__71_73__77_79__83_85, and almost prime / 25, 35, 49, 55, 65, 77, 85 /. The primenumbers are distributed not chaotically. All prime and almost prime numbers to be congruent

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modulo 7.Because the smallest gap between their equal 2 + 4 = 6, and 6(7) = 42 than ispossible to predict with arbitrary accuracy that the next one lies what 42 gap.

I II III IV V VI VII VIII IX X XI XII XIII XIV

5 7 11 13 17 19 23

25 29 31 35 37 41 43

47 49 53 55 59 61 65

67 71 73 77 79 83 85

89 91 95 97 101 103 107

109 113 115 119 121 125 127

131 133 137 139 143 145 149

151 155 157 161 163 167 169

173 175 179 181 185 187 191

193 197 199 203 205 209 211

215 217 221 223 227 229 233

235 239 241 245 247 251 253

257 259 263 265 269 271 275

277 281 283 287 289 293 295

299 301 305 307 311 313 317

319 323 325 329 331 335 337

341 343 347 349 353 355 359

Some numbers have the special property that they cannot be expressed as the product of two

smaller numbers, e.g., 2, 3, 5, 7, etc. Such numbers are called  prime numbers, and they playan important role, both in pure mathematics and its applications. The distribution of suchprime numbers among all natural numbers does not follow any regular pattern; however theGerman mathematician G.F.B. Riemann (1826 - 1866) observed that the frequency of primenumbers is very closely related to the behavior of an elaborate function

ζ (s) = 1 + 1/2s

+ 1/3s

+ 1/4s

+...

called the   Riemann Zeta function. The Riemann hypothesis asserts that all interesting solutions of the equation

ζ(s) = 0

lie on a certain vertical straight line. This has been checked for the first 1,500,000,000solutions. A proof that it is true for every interesting solution would shed light on many of themysteries surrounding the distribution of prime numbers.

Primes seem to be, at the same time very irregularly distributed among all numbers, and yet  –  if squinted at from a sufficiently far distance – they reveal an astoundingly elegant pattern. In1859 the German mathematician Bernard Riemann proposed a way of understanding andrefining that pattern. His hypothesis has wide  – ranging implications, and this day after 150

years of careful research and exhaustive study, we know it is correct.

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N  π(x)  π(N)  π(x) = 4(n)7 4 = 8(0,5) 

1,00E+03  168  5,9524  168 = 8(0,5)6(7) 

7 

1,00E+06  78 498  12,739  78 498 = 8(0,5) 2803,5(7) 

7 

1,00E+09  50 847 534  19,667  50 847 534 = 8(0,5)1 815 983,3514286(7) 

7 

1,00E+12  37 607 912 018  26,59  37 607 912 018 = 8(0,5)1 343 1139 714,92857(7)  

7 

1,00E+15  29 844 570 422 669  33,507  29 844 570 422 669 = 8(0,5)1 065 877 515 095,32(7)  

7 

1,00E+18  24 739 954 287 740 860  40,42  24739954287740860 = 8(0,5)883569795990,745(7) 

7 

1,00E+21  21127269486018731928  47,332  21127269486018731928=4(754545338786383283,143)7  

There are 4 primes up to 10 (2, 3, 5, 7), because those they cannot be expressed as the productof two smaller numbers (4 = 2(2), 8 = 4(2), 9 = 3(3), 10 = 5(2). Between 1 and 100 there are25 primes, and 168 primes up to 1 000. Why 168? Is there a rule, a formula, to tell me howmany primes there are less than a given number?We know that all prime numbers be congruent to me modulo 7, and this seven tell me howmany primes there are less than a given number. We show in table, that quantity primes thereare less than a given number is always product number 4 = 7(0,57143), and n  – the multiplenumber 7: π(x) = 4(n)7, 168 = 4(6)7.

Theorem: The quotient of half a given magnitude ½N, by a given magnitude N, is directly  proportional to the quotient of quantity prime numbers, by its dual quantity.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

2 5 110

31 41

0

61 71

0 0

101

0 0

131

0

151

0 0

181 191

3 7 13 23

0

43 53

0

7383

0

103 113

0 0 0 0

163 173

0

193

0 017

0

37 47

0

67

0 0

97 107

0

127 137

0

157 167

0 0

197

0 019 29

0 0

59

0

79 89

0

109

0 0

139 149

0 0

179

0

199

π(10) = 4 = 7(0,57143), π(100) = 25 = 7(3,57143), π(1 000) = 168

=7(0,57143)6(7)

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156

Proof: ½N : N = πx : 2(πx)  πx ∝ ½N ½N(2 πx) = N(πx) πx/(2 πx) = Re(s) = ½ 

5/10 = 4/8 = 3/6 = 2/4 = 1/2 8(1/2) = 4 2πx(½) = πx

 N 

 x N    22

1

 

In the interval of 10 numbers of 5 consecutive odd numbers is always one divisible by 3, sothat it can act like this at most 4 primes, so double large interval of 8 numbers claim (4·2 = 819 - 11 = 8 11, 12, 13, 14, 15, 16, 17 , 18, 19). Generally in the interval of 10 numbers eacheven and odd number 2 number interval claims, including in 4(2), 3(2), 2(2), 1(2), primes.If the interval first 10 numbers come before 4 primes, then in the interval of 100 numbers 25

 prime numbers occur, and each number in the interval, the proportion ½ is to keep. In other words, all zeros of ζ(s) in the half-plane Re(s) > 0 have real part ½ .

In mathematics, two quantities are said to be proportional if they vary in such a way that oneof the quantities is a constant multiple of the other, or equivalently if they have a constantratio. Proportion also refers to the equality of two ratios. In proportional quantities is thedoubling (tripling, halved) one quantity is always a double (triple, halve) connected to theother quantities.

2πx  πx N ½ N 2 2 2 2

8 4 10 5 2 3 5 7

16 8 20 10 11 13 17 19

20 10 30 15 0 23 0 29

24 12 40 20 31 0 37 0

30 15 50 25 41 43 47 034 17 60 30 0 53 0 59

38 19 70 35 61 0 67 0

44 22 80 40 71 73 0 79

48 24 90 45 0 83 0 89

50 25 100 50 0 0 97 0

58 29 110 55 101 103 107 109

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158

Riemann hypothesis:

The non-trivial zeros (lie zeros in the strip right next to the y-axis with real part of s from 0 to 1)

are all on a line parallel to the y-axis with real part 1 / 2.

Proof:

2

0

5

0

11

0

17

0

23

0

29

0

35

0

41

0

47

0

53

0

59

0

65

0

71

0

77

0

83

0

89

0

95

0

101

0

0

3

0

9

0

15

0

21

0

27

0

33

0

39

0

450

510

570

630

690

750

810

870

930

990

105

7

0

13

0

19

0

25

0

31

0

37

0

43

0

49

0

55

0

61

0

67

0

73

0

79

0

85

0

91

0

97

0

103

0

109

0

0%

10%

20%

30%

40%

50%

60%

70%

80%

90%

100%

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

0 2 33 7513111917

2523

3129

3735

4341

4947

5553

6159

6765

7371

79

1 2 3

y^- 5y + 6 = 0 x^= 1 y = x+ 1 = 2 y = 4x - 1 = 3+ 4= 7 y= 8 x- 3= 5+ 8 = 13

y=8x+3 = 11 + 8= 19 y=8x+9= 17

y=8x+15=23 + 8=31 y=8x+21=29+8=37 y= 8x+27=35 +8=43 y=8x+33=41

y=8x+39=47 y= 8x+45=53+8=61

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159

What is the solution of the equation y² - 5y + 6 = 0? The solution is 2 und 3. Another way to say,

this is that 2 and 3 are the zeros of the function y ² - 5y + 6 = 0.

y y‘ ½ y 

y² - 10y + 21 = 0 Solution 3 and 7 5

y² - 18y + 65 = 0 - „ - 5 and 13 9

y² - 30y + 209 = 0 - „ - 11 and 19 15

y² - 42y + 425 = 0 - „ - 17 and 25 = 5(5) 21

y² - 54y + 713 = 0 - „ - 23 and 31 27

y² - 66y +1073 = 0 - ― - 29 and 37 33

y² - 78y +1505 = 0 5(7 )= 35 and 43 39

y² - 90y +2009 = 0 - ― - 41 and 49 = 7(7) 45

y² - 102y +2585 = 0 47 i 55 = 5(11) 51

2 0 00 2,50 03

05

0

7

05

0

9

0

13

0

11

0

15

0

19

0

17

0

21

0

25

0

23

0

27

0

31

0

29

0

33

0

37

0

35

0

39

0

43

0

41

0

45

0

49

0

47

0

51

0

55

0

53

0

57

0

61

0

59

0

63

0

67

0

65

0

69

0

73

0

71

0

75

0

79

0

77

0

81

0

85

0

83

0

87

0

91

0

89

0

93

0

97

0

95

1 2 3

(y + y')/2 = 1/2 y (2+3)/2=2,5 (3+7)/2=5 (5+13)/2=9 (11+19)/2=15

(17+25)/2=21 (23+ 31)/2=27 (29+37)/2=33

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160

A zero off the critical line would induce a pattern in the distribution of the primes.

2

19

29

0 0

59

0

79

89

0

109

313

23

0

43

53

0

73

83

0

103

7 17

0

37

47

0

67

0 0

97

107

5 11

0

31

41

0

61

71

0 0

101

0 0 0 0 0 0 0 0 0 0 0

0 2 4 6 8 10 12

p + 5(6) = p' 7 + 5(6) = 37 + 5(6) = 67 + 5(6) = 97 + 5(6) = 127 + 5(6) = 157

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161

The second graph shows the values of the zeta function in the field The

x and the y -axis corresponds to real- and imaginary part of functional value. The coloring and the- axis give real and imaginary archetype of the back. The black curve corresponds to the critical

line

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163

00

2

30

5

0

7

0

9

0

11

0

13 0 15

0

17

0

19

0

21

0

23

0

25

0

270

29

0

31

033

0

35

0

370

39

0

41

0

43

0

45

047

0

49

0

510

53

0

55

0

57

0

59

061

0

63

0

650

67

0

69

0

71

0

73

075

0

77

0

790

81

0

83

0

85

0

87

089

0

91

0

930

95

0

97

0

99

0

101

0103

0

105

0

1070

109

0

111

65

107

149

191

233

275

31747

89

131

173215

257

299

67

109

151

193

235

277

319

49

91

133

175217

259

301

71

113

155

197

239

281

323

53

95

137

179221

263

305

73

115

157

199

241

283

325

55

97

139

181223

265

307

77

119

161

203

245

287

329

59

101

143

185227

269

311

79

121

163

205

247

289

331

61

103

145

187229

271

313

83

125

167

209

251

293

335

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