note on the lower bound of least common multiple

Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2013 Article ID 218125 4 pageshttpdxdoiorg1011552013218125

Research ArticleNote on the Lower Bound of Least Common Multiple

Shea-Ming Oon

Institute of Mathematical Sciences University of Malaya 50603 Kuala Lumpur Malaysia

Correspondence should be addressed to Shea-Ming Oon oonsmumedumy

Received 1 July 2012 Revised 24 December 2012 Accepted 10 January 2013

Academic Editor Pekka Koskela

Copyright copy 2013 Shea-Ming Oon This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

Consider a sequence of positive integers in arithmetic progression 119906119896 = 1199060 +119896119903with (1199060 119903) = 1 Denote the least commonmultipleof 1199060 119906119899 by 119871119899 We show that if 119899 ⩾ 1199032 + 119903 then 119871119899 ⩾ 1199060119903

119903+1(119903 + 1) and we obtain optimum result on 119899 in some cases for such

estimate Besides for quadratic sequences 1198982 + 119888 (119898 + 1)2 + 119888 1198992 + 119888 we also show that the least common multiple is at least2119899 when119898 ⩽ lceil1198992rceil which sharpens a recent result of Farhi

1 Introduction

Integer sequences in arithmetic progressions constitute arecurrent theme in number theory The most notable resultin this new century is perhaps the existence of arbitrary longsequences of primes in arithmetic progressions due to Greenand Tao [1]

The bounds of the least common multiple for the finitesequences in arithmetic progressions also attract some atten-tion The prime number theorem assures that the least com-mon multiple of the first 119899 positive integers is asymptoticallyupper bounded by (119890 + 120576)119899 and lower bounded by (119890 minus 120576)119899for any prefixed 120576 As for effective uniform estimate Hanson[2] obtained the upper bound 3119899 about forty years ago byconsidering Sylvester series of one Nair [3] gave 2119899 as alowerbound in a simple proof ten years later in view of obtaininga Chebyshev-type estimate on the number of prime numbersas in a tauberian theorem due to Shapiro [4]

Recently some results concerning the lower bound of theleast commonmultiple of positive integers in finite arithmeticprogressions were obtained by Farhi [5] Some other resultsabout the least common multiple of consecutive integers andconsecutive arithmetic progression terms are given by Farhiand Kane [6] and by Hong and Qian [7] respectively If1198860 119886119899 are integers we denote their least commonmultipleby [1198860 119886119899] Consider two coprime positive integers 1199060 and119903 and put 119906119896 = 1199060 + 119896119903 119871119896 = [1199060 119906119896] A recent resultof Hong et al (cf [8 9]) shows that for any positive integers119886 120572 119903 and 119899 such that 119886 ⩾ 2 120572 119903 ⩾ 119886 119899 ⩾ 2120572119903 we have119871119899 ⩾ 1199060119903

120572+119886minus2(119903 + 1)

119899

Recently Wu et al [10] improved the Hong-Kominerslower bound A special case of Hong-Kominers result tells usthat if 119899 ⩾ 119903(119903 + 3) (or 119899 ⩾ 119903(119903 + 4)) if 119903 is odd (or even)then 119871119899 ⩾ 1199060119903

119903+1(119903 + 1)

119899 In this note we find that the Hong-Kominers lower bound is still valid if 119899 isin [119903(119903 + 1) 119903(119903 + 1199030))with 1199030 = 3 or 4 if 119903 is odd or even That is we have thefollowing result

Theorem 1 Let 119899 1199060 119903 isin N with (1199060 119903) = 1 One puts for any119896 isin 0 119899 119906119896 = 1199060 + 119896119903 and 119871119899 = [1199060 119906119899] Then for any119899 ⩾ 119903(119903 + 1)

119871119899 ⩾ 1199060119903119903+1(119903 + 1)

119899 (1)

Furthermore if 1199060 gt 119903 or 1199060 lt min3 119903 the same estimateholds when 119899 = 1199032 + 119903 minus 1

In 2007 Farhi [11] showed that [12 + 1 22 + 1 1198992 +1] ⩾ 032(1442)

119899 Note that Qian et al [12] obtained someresults on the least common multiple of consecutive terms ina quadratic progression We can now state the second resultof this paper

Theorem 2 Let 119888 119898 119899 isin N be such that 0 lt 119898 lt 119899 Supposethat119898 ⩽ lceil1198992rceil Then one has

[1198982+ 119888 (119898 + 1)

2+ 119888 119899

2+ 119888] ⩾ 2

119899 (2)

This theorem improves the result in [11]

2 Abstract and Applied Analysis

2 Proof of the First Theorem

Let 119909 119910 isin Rwith119910 = 0We say that 119909 is amultiple of119910 if thereis an integer 119911 such that 119909 = 119910119911 As usual lfloor119909rfloor denotes thelargest integer not larger than 119909 and lceil119909rceil denotes the smallestinteger not smaller than 119909

We will introduce the two following results The firstis a known result which tells us that 119871119899 is a multiple of(1199060 sdot sdot sdot 119906119899119899) This is can be proved by considering a suitablepartial fraction expansion (cf [11]) or by considering theintegral int1

01199091199060119903minus1

(1 minus 119909)119899119889119909 (cf [13])

For ℓ isin 0 119899 with a slight modification of notation as in[11] we put 119871119899ℓ = [119906119899minusℓ 119906119899] and

119861119899ℓ =

119906119899minusℓ sdot sdot sdot 119906119899

ℓ

(3)

Clearly we have 119871119899 = 119871119899119899 and for any ℓ isin 0 119899 119871119899 ⩾ 119871119899ℓThis result can be restated as

Lemma 3 For any ℓ isin [0 119899] one can find a positive integer119860119899ℓ such that 119871119899ℓ = 119860119899ℓ119861119899ℓ

Our modification aims to emphasize the estimate of theterms 119860119899ℓ and 119861119899ℓ that will give us some improvement If119899 ⩾ 1199060 then we see that the first term 1199060 does not play animportant role as it will be a factor of the term 1199060 + 1199060119903 =1199060(1 + 119903) Hence the behaviour should be different when 119899 islarge It will be more interesting to give a control over the lastℓ terms

Note that 119861119899ℓ+1 = (119906119899minusℓminus1(ℓ + 1))119861119899ℓ thus

119861119899ℓ+1 ⩽ 119861119899ℓ lArrrArr 1199060 + (119899 minus (ℓ + 1)) 119903 ⩽ ℓ + 1 lArrrArr ℓ + 1

⩾

119899119903 + 1199060

119903 + 1

(4)

We will put ℓ119899 = minlfloor119906119899(119903 + 1)rfloor 119899 The followinglemma tells us that keeping the first smaller terms canincrease at least the power 119899 in the estimate of lower bound insuch a way (cf also [13])

Lemma 4 One has 119861119899ℓ119899 ⩾ 1199060(119903 + 1)119899

Proof We can just proceed by mathematical inductionIf 119899 = 0 then ℓ119899 = 0 and 11986100 = 1199060 and it holds In fact if

119899 ⩽ 1199060 then we have ℓ119899 = 119899 and

119861119899119899 = 1199060 (

1199060

1

+ 119903) sdot sdot sdot (

1199060

119899

+ 119903) ⩾ 1199060(1 + 119903)119899 (5)

When 119899 ⩾ 1199060 we have (119899119903+1199060)(119903+1) ⩽ (119899119903+119899)(119903+1) ⩽119899 and ℓ119899 = lfloor(119899119903 + 1199060)(119903 + 1)rfloor It remains to derive the resultfor the case 119899 + 1 from that of 119899 ⩾ 1199060

It is obvious that ℓ119899 ⩽ ℓ119899+1 ⩽ ℓ119899 + 1 and 119906119899(119903 + 1) minus 1 ⩽ℓ119899 ⩽ 119906119899(119903 + 1)

If ℓ119899 = ℓ119899+1 then (119899 + 1)119903 + 1199060 ⩽ (ℓ119899 + 1)(119903 + 1) and119906119899minusℓ119899

= 1199060 + (119899 minus ℓ119899)119903 ⩽ ℓ119899 + 1Thus

119906119899+1

119906119899minusℓ119899

=

119906119899minusℓ119899+ (ℓ119899 + 1) 119903

119906119899minusℓ119899

= 1 +

ℓ119899 + 1

119906119899minusℓ119899

119903 ⩾ 1 + 119903 (6)

Hence

119861119899+1ℓ119899+1=

119906119899+1

119906119899minusℓ119899

sdot 119861119899ℓ119899⩾ 1199060(119903 + 1)

119899+1 (7)

If ℓ119899+1 = ℓ119899 + 1 then

119861119899+1ℓ119899+1=

119906119899+1

ℓ119899 + 1

sdot 119861119899ℓ119899⩾ 1199060(119903 + 1)

119899+1 (8)

In either case the principle of mathematical inductionassures the result

We can complete our proof now Suppose that 119899 ⩾ 119903(119903+1)Then

ℓ119899 = lfloor

119906119899

119903 + 1

rfloor ⩾ lfloor1199032+

1199060

119903 + 1

rfloor ⩾ 1199032 (9)

By considering the first 119903multiples of 119903 we have 119903119903+1 | ℓ119899Since (119903 1199060) = 1 we deduce that for all 119896 isin 0 119899 (119903 119906119896) = 1Writing the result of Lemma 3 for ℓ = ℓ119899 as

ℓ119899 sdot 119871119899ℓ119899= 119860119899ℓ119899

sdot 119906119899minusℓ119899sdot sdot sdot 119906119899 (10)

we conclude that 119903119903+1 | 119860119899ℓ119899 Using the Lemma 4 we obtain

119871119899 ⩾ 119871119899ℓ119899= 119860119899ℓ119899

119861119899ℓ119899⩾ 1199060119903119903+1(119903 + 1)

119899 (11)

which is our conclusionConsider the case 119899 = 1199032 + 119903 minus 1 If 1199060 gt 119903 then we still

have ℓ119899 ⩾ 1199032

Supposing now that 119903 ⩾ 2 and 1199060 ⩽ min2 119903minus1 we shallprove that for 1198990 = 119903

2+ 119903 minus 1 it is still possible to choose

ℓ1015840

1198990= 1199032 and 1198990 minus ℓ

1015840

1198990= 119903 minus 1 so that 1198611198990 ℓ10158401198990 ⩾ 1199060(119903 + 1)

1198990 On the one hand when 119899 = 1199032 minus 119903 minus 1 we have

ℓ119899 = lfloor

1199060 + 119903 (1199032minus 119903 minus 1)

119903 + 1

rfloor

= lfloor

1199032(119903 + 1) minus 2119903 (119903 + 1) + 119903 + 1199060

119903 + 1

rfloor

= 1199032minus 2119903 + lfloor

119903 + 1199060

119903 + 1

rfloor

= 1199032minus 2119903 + 1

(12)

Thus 119899 minus ℓ119899 = 119903 minus 2 and119906119903minus2 sdot sdot sdot 1199061199032minus119903minus1

(1199032minus 2119903 + 1)

⩾ 1199060(119903 + 1)1199032minus119903minus1 (13)

On the other hand we write

1198611198990 ℓ1015840

1198990

=

119906119903minus1 sdot sdot sdot 1199061199032+119903minus1

1199032

=

119906119903minus2 sdot sdot sdot 1199061199032minus119903minus1

(1199032minus 2119903 + 1)

sdot

1199061199032minus1199031199061199032

119906119903minus2 (1199032minus 119903 + 1)

sdot

119903minus1

prod

119904=1

1199061199032minus119903+1199041199061199032+119903minus119904

(1199032minus 2119903 + 119904 + 1) (119903

2minus (119904 minus 1))

(14)

Abstract and Applied Analysis 3

It suffices to show that for any 119904 isin 1 119903 minus 11199061199032minus119903+1199041199061199032+119903minus119904

(1199032minus 2119903 + 119904 + 1) (119903


⩾ (119903 + 1)2 (15)

1199061199032minus1199031199061199032

119906119903minus2 (1199032minus 119903 + 1)

⩾ (119903 + 1)2 (16)

Equation (15) is equivalent to

1199061199032minus119903+1199041199061199032+119903minus119904 ⩾ (1199032minus 2119903 + 119904 + 1) (119903

2minus (119904 minus 1)) (119903 + 1)

2

(17)

By expanding it remains to verify for any integer 119903 ⩾ 119904 +1 ⩾ 2 that

119891119904 (119903) = 211990601199033minus (4119904 minus 1) 119903

2+ 2119904 (119904 minus 1) 119903 + 119904

2+ 1199062

0minus 1 ⩾ 0

(18)

With simple computation we obtain for any real 119903 119904 gt 0with 119903 ⩾ 119904 + 1

119891119904 (119904 + 1) = 2 (1199060 minus 1) 1199043+ 6 (1199060 minus 1) 119904

2

+ 2 (31199060 minus 2) 119904 + 1199062

0+ 21199060 ⩾ 0

1198911015840

119904(119903) ⩾ 6119903

2minus 2 (4119904 minus 1) 119903 + 2119904 (119904 minus 1)

⩾ 2 (119903 minus 119904) (3119903 minus 119904 + 1) ⩾ 0

(19)

which allow to establish (15)Equation (16) is equivalent to

(1199060 + (1199032minus 119903) 119903) (1199060 + 119903

3)

⩾ (1199060 + (119903 minus 2) 119903) (1199032minus 119903 + 1) (119903 + 1)

2

(20)

or

(2 minus 1199060) 1199034+ (1199060 minus 1) 119903

3+ (1 minus 1199060) 119903

2

+ (2 minus 1199060) 119903 + 1199062

0minus 1199060 ⩾ 0

(21)

Such inequality holds for any positive integer 119903when 1199060 =1 or 2

Finally we still have

1198611198990 ℓ1015840

1198990

⩾ 1199060(119903 + 1)1198990 (22)

and the condition ℓ10158401198990⩾ 1199032 allows us to conclude

3 Proof of the Second Theorem

We shall start by proving the following lemma

Lemma 5 Let 119888 119898 119899 isin N be such that 0 lt 119898 lt 119899 Put

1198711015840

119898119899= [119898

2+ 119888 (119898 + 1)

2+ 119888 119899

2+ 119888] (23)

Then

1198711015840

119898119899⩾

prod119899

119896=119898radic1198962+ 119888

(119899 minus 119898)

(24)

Proof We shall denote 119909120484 = cos(log119909) + 120484 sin(log119909)Consider the integral of complex-valued function of a real

variable

int

1

0

119909119898minus1+radic119888120484

(1 minus 119909)119899minus119898119889119909 (25)

Firstly by integrating by parts (119899 minus 119898) times

int

1

0

119909119898minus1+radic119888120484

(1 minus 119909)119899minus119898119889119909

=

119909119898+radic119888120484

(1 minus 119909)119899minus119898

119898 + radic119888120484

1003816100381610038161003816100381610038161003816100381610038161003816

1

0

+

119899 minus 119898

119898 + radic119888120484

int

1

0

119909119898+radic119888120484

(1 minus 119909)119899minus119898minus1

119889119909

=

(119899 minus 119898)

prod119899minus1

119896=119898(119896 + radic119888120484)

int

1

0

119909119899minus1+radic119888120484

119889119909

=

(119899 minus 119898)

prod119899

119896=119898(119896 + radic119888120484)

(26)

Secondly by expanding

int

1

0

119909119898minus1+radic119888120484

(1 minus 119909)119899minus119898119889119909

= int

1

0

119899minus119898

sum

119896=0

(minus1)119896(

119899 minus 119898

119896)119909119898minus1+119896+radic119888120484

119889119909

=

119899minus119898

sum

119896=0

(minus1)119896(119899minus119898

119896)

119898 + 119896 + radic119888120484

(27)

On one hand put such complex number in Cartesianform and after multiplying it by 1198711015840

119898119899 we get a linear combi-

nation of 119909119896 + 119910119896radic119888120484 with integer coefficients where 0 ⩽ 119896 ⩽119899 minus 119898 and 119909119896 119910119896 isin Z On the other hand it is easy to seethat the number obtained by integrating by parts (119899 minus 119898)times is not zero So its modulus is not smaller than 1 andwe conclude that

1198711015840

119898119899⩾

prod119899

119896=119898radic1198962+ 119888

(119899 minus 119898)

(28)

Now we write furthermore

1198711015840

119898119899⩾

119899

(119898 minus 1) (119899 minus 119898)

= 119898(

119899

119898) (29)

Suppose now that 119898 ⩽ 1198981015840 = lceil1198992rceil then 1198711015840119898119899⩾ 1198711015840

1198981015840 119899⩾

1198981015840(119899

1198981015840 ) and 119899 = 21198981015840 or 21198981015840 minus 1

The following Stirling estimate

(

119899

119890

)

119899

radic2120587119899 ⩽ 119899 ⩽ (

119899

119890

)

119899

radic21205871198991198901(12119899) (30)


allows us to obtain

(

2119899

119899) ⩾

22119899

radic119899120587

119890minus1(6119899)

(31)

If 119899 = 21198981015840 then

1198981015840(

21198981015840

1198981015840 ) ⩾ 2

21198981015840

radic1198981015840

120587

119890minus1(6119898

1015840)

(32)

As the functionradic(119909120587)119890minus1(6119909) is increasing and its valueat 119909 = 4 is ⩾ 108 we conclude that for any integer1198981015840 ⩾ 4

1198981015840(

21198981015840

1198981015840 ) ⩾ 2

21198981015840

(33)

If 119899 = 21198981015840 minus 1 then we can conclude similarly for1198981015840 ⩾ 5

1198981015840(

21198981015840minus 1

1198981015840 ) ⩾ (2119898

1015840minus 1)(

2 (1198981015840minus 1)

1198981015840minus 1

)

⩾ 221198981015840minus11198981015840minus 12

radic1198981015840120587

119890minus1(6119898

1015840)

⩾ 221198981015840minus1

(34)

In brief for any integer 119898 119899 with 119899 ⩾ 8 and 119898 ⩽ lceil1198992rceilwe just establish that

1198711015840

119898119899⩾ 2119899 (35)

The case 119899 = 7 can be checked directly as 4 ( 74) ⩾ 27

For the case 119899 = 6 we have

[16 + 119888 25 + 119888 36 + 119888] ⩾ [25 + 119888 36 + 119888]

=

(25 + 119888) (36 + 119888)

gcd (25 + 119888 36 + 119888)

⩾

25 times 36

11

⩾ 64 = 26

(36)


[9 + 119888 16 + 119888 25 + 119888] ⩾ [16 + 119888 25 + 119888]

⩾

16 times 25

25 minus 16

⩾ 25

(37)

It is easy to check the cases 119899 = 2 3 and 4 as it involvesonly two terms and we can make our final conclusion

Acknowledgment

The author would like to thank the referees for their helpfulsuggestions that improved the presentation of this paper

References

[1] B Green and T Tao ldquoThe primes contain arbitrarily long arith-metic progressionsrdquo Annals of Mathematics vol 167 no 2 pp481ndash547 2008

[2] D Hanson ldquoOn the product of the primesrdquo Canadian Mathe-matical Bulletin Bulletin Canadien de Mathematiques vol 15pp 33ndash37 1972

[3] M Nair ldquoOn Chebyshev-type inequalities for primesrdquo TheAmerican Mathematical Monthly vol 89 no 2 pp 126ndash1291982

[4] H N Shapiro ldquoOn the number of primes less than or equal 119909rdquoProceedings of the American Mathematical Society vol 1 pp346ndash348 1950

[5] B Farhi ldquoMinorations non triviales du plus petit communmul-tiple de certaines suites finies drsquoentiersrdquo Comptes Rendus Math-ematique Academie des Sciences Paris vol 341 no 8 pp 469ndash474 2005

[6] B Farhi and D Kane ldquoNew results on the least common multi-ple of consecutive integersrdquo Proceedings of the American Math-ematical Society vol 137 no 6 pp 1933ndash1939 2009

[7] S Hong and G Qian ldquoThe least common multiple of consecu-tive arithmetic progression termsrdquo Proceedings of the EdinburghMathematical Society vol 54 no 2 pp 431ndash441 2011

[8] S Hong and Y Yang ldquoImprovements of lower bounds for theleast common multiple of finite arithmetic progressionsrdquo Pro-ceedings of the American Mathematical Society vol 136 no 12pp 4111ndash4114 2008

[9] S Hong and S D Kominers ldquoFurther improvements of lowerbounds for the least common multiples of arithmetic progres-sionsrdquo Proceedings of the American Mathematical Society vol138 no 3 pp 809ndash813 2010

[10] R Wu Q Tan and S Hong ldquoNew lower bounds for the leastcommon multiple of arithmetic progressionsrdquo Chinese Annalsof Mathematics In press

[11] B Farhi ldquoNontrivial lower bounds for the least common mul-tiple of some finite sequences of integersrdquo Journal of NumberTheory vol 125 no 2 pp 393ndash411 2007

[12] G Qian Q Tan and S Hong ldquoThe least common multiple ofconsecutive terms in a quadratic progressionrdquo Bulletin of theAustralian Mathematical Society vol 86 pp 389ndash404 2012

[13] S Hong and W Feng ldquoLower bounds for the least commonmultiple of finite arithmetic progressionsrdquo Comptes RendusMathematique Academie des Sciences Paris vol 343 no 11-12pp 695ndash698 2006

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2 Proof of the First Theorem

Let 119909 119910 isin Rwith119910 = 0We say that 119909 is amultiple of119910 if thereis an integer 119911 such that 119909 = 119910119911 As usual lfloor119909rfloor denotes thelargest integer not larger than 119909 and lceil119909rceil denotes the smallestinteger not smaller than 119909

We will introduce the two following results The firstis a known result which tells us that 119871119899 is a multiple of(1199060 sdot sdot sdot 119906119899119899) This is can be proved by considering a suitablepartial fraction expansion (cf [11]) or by considering theintegral int1

01199091199060119903minus1

(1 minus 119909)119899119889119909 (cf [13])

For ℓ isin 0 119899 with a slight modification of notation as in[11] we put 119871119899ℓ = [119906119899minusℓ 119906119899] and

119861119899ℓ =

119906119899minusℓ sdot sdot sdot 119906119899

ℓ

(3)

Clearly we have 119871119899 = 119871119899119899 and for any ℓ isin 0 119899 119871119899 ⩾ 119871119899ℓThis result can be restated as

Lemma 3 For any ℓ isin [0 119899] one can find a positive integer119860119899ℓ such that 119871119899ℓ = 119860119899ℓ119861119899ℓ

Our modification aims to emphasize the estimate of theterms 119860119899ℓ and 119861119899ℓ that will give us some improvement If119899 ⩾ 1199060 then we see that the first term 1199060 does not play animportant role as it will be a factor of the term 1199060 + 1199060119903 =1199060(1 + 119903) Hence the behaviour should be different when 119899 islarge It will be more interesting to give a control over the lastℓ terms

Note that 119861119899ℓ+1 = (119906119899minusℓminus1(ℓ + 1))119861119899ℓ thus

119861119899ℓ+1 ⩽ 119861119899ℓ lArrrArr 1199060 + (119899 minus (ℓ + 1)) 119903 ⩽ ℓ + 1 lArrrArr ℓ + 1

⩾

119899119903 + 1199060

119903 + 1

(4)

We will put ℓ119899 = minlfloor119906119899(119903 + 1)rfloor 119899 The followinglemma tells us that keeping the first smaller terms canincrease at least the power 119899 in the estimate of lower bound insuch a way (cf also [13])

Lemma 4 One has 119861119899ℓ119899 ⩾ 1199060(119903 + 1)119899

Proof We can just proceed by mathematical inductionIf 119899 = 0 then ℓ119899 = 0 and 11986100 = 1199060 and it holds In fact if

119899 ⩽ 1199060 then we have ℓ119899 = 119899 and

119861119899119899 = 1199060 (

1199060

1

+ 119903) sdot sdot sdot (

1199060

119899

+ 119903) ⩾ 1199060(1 + 119903)119899 (5)

When 119899 ⩾ 1199060 we have (119899119903+1199060)(119903+1) ⩽ (119899119903+119899)(119903+1) ⩽119899 and ℓ119899 = lfloor(119899119903 + 1199060)(119903 + 1)rfloor It remains to derive the resultfor the case 119899 + 1 from that of 119899 ⩾ 1199060

It is obvious that ℓ119899 ⩽ ℓ119899+1 ⩽ ℓ119899 + 1 and 119906119899(119903 + 1) minus 1 ⩽ℓ119899 ⩽ 119906119899(119903 + 1)

If ℓ119899 = ℓ119899+1 then (119899 + 1)119903 + 1199060 ⩽ (ℓ119899 + 1)(119903 + 1) and119906119899minusℓ119899

= 1199060 + (119899 minus ℓ119899)119903 ⩽ ℓ119899 + 1Thus

119906119899+1

119906119899minusℓ119899

=

119906119899minusℓ119899+ (ℓ119899 + 1) 119903

119906119899minusℓ119899

= 1 +

ℓ119899 + 1

119906119899minusℓ119899

119903 ⩾ 1 + 119903 (6)

Hence

119861119899+1ℓ119899+1=

119906119899+1

119906119899minusℓ119899

sdot 119861119899ℓ119899⩾ 1199060(119903 + 1)

119899+1 (7)

If ℓ119899+1 = ℓ119899 + 1 then

119861119899+1ℓ119899+1=

119906119899+1

ℓ119899 + 1

sdot 119861119899ℓ119899⩾ 1199060(119903 + 1)

119899+1 (8)

In either case the principle of mathematical inductionassures the result

We can complete our proof now Suppose that 119899 ⩾ 119903(119903+1)Then

ℓ119899 = lfloor

119906119899

119903 + 1

rfloor ⩾ lfloor1199032+

1199060

119903 + 1

rfloor ⩾ 1199032 (9)

By considering the first 119903multiples of 119903 we have 119903119903+1 | ℓ119899Since (119903 1199060) = 1 we deduce that for all 119896 isin 0 119899 (119903 119906119896) = 1Writing the result of Lemma 3 for ℓ = ℓ119899 as

ℓ119899 sdot 119871119899ℓ119899= 119860119899ℓ119899

sdot 119906119899minusℓ119899sdot sdot sdot 119906119899 (10)

we conclude that 119903119903+1 | 119860119899ℓ119899 Using the Lemma 4 we obtain

119871119899 ⩾ 119871119899ℓ119899= 119860119899ℓ119899

119861119899ℓ119899⩾ 1199060119903119903+1(119903 + 1)

119899 (11)

which is our conclusionConsider the case 119899 = 1199032 + 119903 minus 1 If 1199060 gt 119903 then we still

have ℓ119899 ⩾ 1199032

Supposing now that 119903 ⩾ 2 and 1199060 ⩽ min2 119903minus1 we shallprove that for 1198990 = 119903

2+ 119903 minus 1 it is still possible to choose

ℓ1015840

1198990= 1199032 and 1198990 minus ℓ

1015840

1198990= 119903 minus 1 so that 1198611198990 ℓ10158401198990 ⩾ 1199060(119903 + 1)

1198990 On the one hand when 119899 = 1199032 minus 119903 minus 1 we have

ℓ119899 = lfloor

1199060 + 119903 (1199032minus 119903 minus 1)

119903 + 1

rfloor

= lfloor

1199032(119903 + 1) minus 2119903 (119903 + 1) + 119903 + 1199060

119903 + 1

rfloor

= 1199032minus 2119903 + lfloor

119903 + 1199060

119903 + 1

rfloor

= 1199032minus 2119903 + 1

(12)

Thus 119899 minus ℓ119899 = 119903 minus 2 and119906119903minus2 sdot sdot sdot 1199061199032minus119903minus1

(1199032minus 2119903 + 1)

⩾ 1199060(119903 + 1)1199032minus119903minus1 (13)

On the other hand we write

1198611198990 ℓ1015840

1198990

=

119906119903minus1 sdot sdot sdot 1199061199032+119903minus1

1199032

=

119906119903minus2 sdot sdot sdot 1199061199032minus119903minus1

(1199032minus 2119903 + 1)

sdot

1199061199032minus1199031199061199032

119906119903minus2 (1199032minus 119903 + 1)

sdot

119903minus1

prod

119904=1

1199061199032minus119903+1199041199061199032+119903minus119904

(1199032minus 2119903 + 119904 + 1) (119903


(14)



(1199032minus 2119903 + 119904 + 1) (119903


⩾ (119903 + 1)2 (15)

1199061199032minus1199031199061199032

119906119903minus2 (1199032minus 119903 + 1)

⩾ (119903 + 1)2 (16)



2minus (119904 minus 1)) (119903 + 1)

2

(17)


119891119904 (119903) = 211990601199033minus (4119904 minus 1) 119903

2+ 2119904 (119904 minus 1) 119903 + 119904

2+ 1199062

0minus 1 ⩾ 0

(18)


119891119904 (119904 + 1) = 2 (1199060 minus 1) 1199043+ 6 (1199060 minus 1) 119904

2

+ 2 (31199060 minus 2) 119904 + 1199062

0+ 21199060 ⩾ 0

1198911015840

119904(119903) ⩾ 6119903


⩾ 2 (119903 minus 119904) (3119903 minus 119904 + 1) ⩾ 0

(19)


(1199060 + (1199032minus 119903) 119903) (1199060 + 119903

3)

⩾ (1199060 + (119903 minus 2) 119903) (1199032minus 119903 + 1) (119903 + 1)

2

(20)

or

(2 minus 1199060) 1199034+ (1199060 minus 1) 119903

3+ (1 minus 1199060) 119903

2

+ (2 minus 1199060) 119903 + 1199062

0minus 1199060 ⩾ 0

(21)



1198611198990 ℓ1015840

1198990

⩾ 1199060(119903 + 1)1198990 (22)





1198711015840

119898119899= [119898

2+ 119888 (119898 + 1)

2+ 119888 119899

2+ 119888] (23)

Then

1198711015840

119898119899⩾

prod119899

119896=119898radic1198962+ 119888

(119899 minus 119898)

(24)


variable

int

1

0

119909119898minus1+radic119888120484

(1 minus 119909)119899minus119898119889119909 (25)


int

1

0

119909119898minus1+radic119888120484

(1 minus 119909)119899minus119898119889119909

=

119909119898+radic119888120484

(1 minus 119909)119899minus119898

119898 + radic119888120484

1003816100381610038161003816100381610038161003816100381610038161003816

1

0

+

119899 minus 119898

119898 + radic119888120484

int

1

0

119909119898+radic119888120484


119889119909

=

(119899 minus 119898)

prod119899minus1

119896=119898(119896 + radic119888120484)

int

1

0

119909119899minus1+radic119888120484

119889119909

=

(119899 minus 119898)

prod119899

119896=119898(119896 + radic119888120484)

(26)


int

1

0

119909119898minus1+radic119888120484

(1 minus 119909)119899minus119898119889119909

= int

1

0

119899minus119898

sum

119896=0

(minus1)119896(

119899 minus 119898

119896)119909119898minus1+119896+radic119888120484

119889119909

=

119899minus119898

sum

119896=0

(minus1)119896(119899minus119898

119896)

119898 + 119896 + radic119888120484

(27)




1198711015840

119898119899⩾

prod119899

119896=119898radic1198962+ 119888

(119899 minus 119898)

(28)


1198711015840

119898119899⩾

119899

(119898 minus 1) (119899 minus 119898)

= 119898(

119899

119898) (29)


1198981015840 119899⩾

1198981015840(119899

1198981015840 ) and 119899 = 21198981015840 or 21198981015840 minus 1


(

119899

119890

)

119899

radic2120587119899 ⩽ 119899 ⩽ (

119899

119890

)

119899

radic21205871198991198901(12119899) (30)


allows us to obtain

(

2119899

119899) ⩾

22119899

radic119899120587

119890minus1(6119899)

(31)

If 119899 = 21198981015840 then

1198981015840(

21198981015840

1198981015840 ) ⩾ 2

21198981015840

radic1198981015840

120587

119890minus1(6119898

1015840)

(32)


1198981015840(

21198981015840

1198981015840 ) ⩾ 2

21198981015840

(33)


1198981015840(

21198981015840minus 1

1198981015840 ) ⩾ (2119898

1015840minus 1)(

2 (1198981015840minus 1)

1198981015840minus 1

)

⩾ 221198981015840minus11198981015840minus 12

radic1198981015840120587

119890minus1(6119898

1015840)

⩾ 221198981015840minus1

(34)


1198711015840

119898119899⩾ 2119899 (35)



[16 + 119888 25 + 119888 36 + 119888] ⩾ [25 + 119888 36 + 119888]

=

(25 + 119888) (36 + 119888)

gcd (25 + 119888 36 + 119888)

⩾

25 times 36

11

⩾ 64 = 26

(36)


[9 + 119888 16 + 119888 25 + 119888] ⩾ [16 + 119888 25 + 119888]

⩾

16 times 25

25 minus 16

⩾ 25

(37)


Acknowledgment


References





















Volume 2014




Journal of











Journal of


Function Spaces






Algebra











(1199032minus 2119903 + 119904 + 1) (119903


⩾ (119903 + 1)2 (15)

1199061199032minus1199031199061199032

119906119903minus2 (1199032minus 119903 + 1)

⩾ (119903 + 1)2 (16)



2minus (119904 minus 1)) (119903 + 1)

2

(17)


119891119904 (119903) = 211990601199033minus (4119904 minus 1) 119903

2+ 2119904 (119904 minus 1) 119903 + 119904

2+ 1199062

0minus 1 ⩾ 0

(18)


119891119904 (119904 + 1) = 2 (1199060 minus 1) 1199043+ 6 (1199060 minus 1) 119904

2

+ 2 (31199060 minus 2) 119904 + 1199062

0+ 21199060 ⩾ 0

1198911015840

119904(119903) ⩾ 6119903


⩾ 2 (119903 minus 119904) (3119903 minus 119904 + 1) ⩾ 0

(19)


(1199060 + (1199032minus 119903) 119903) (1199060 + 119903

3)

⩾ (1199060 + (119903 minus 2) 119903) (1199032minus 119903 + 1) (119903 + 1)

2

(20)

or

(2 minus 1199060) 1199034+ (1199060 minus 1) 119903

3+ (1 minus 1199060) 119903

2

+ (2 minus 1199060) 119903 + 1199062

0minus 1199060 ⩾ 0

(21)



1198611198990 ℓ1015840

1198990

⩾ 1199060(119903 + 1)1198990 (22)





1198711015840

119898119899= [119898

2+ 119888 (119898 + 1)

2+ 119888 119899

2+ 119888] (23)

Then

1198711015840

119898119899⩾

prod119899

119896=119898radic1198962+ 119888

(119899 minus 119898)

(24)


variable

int

1

0

119909119898minus1+radic119888120484

(1 minus 119909)119899minus119898119889119909 (25)


int

1

0

119909119898minus1+radic119888120484

(1 minus 119909)119899minus119898119889119909

=

119909119898+radic119888120484

(1 minus 119909)119899minus119898

119898 + radic119888120484

1003816100381610038161003816100381610038161003816100381610038161003816

1

0

+

119899 minus 119898

119898 + radic119888120484

int

1

0

119909119898+radic119888120484


119889119909

=

(119899 minus 119898)

prod119899minus1

119896=119898(119896 + radic119888120484)

int

1

0

119909119899minus1+radic119888120484

119889119909

=

(119899 minus 119898)

prod119899

119896=119898(119896 + radic119888120484)

(26)


int

1

0

119909119898minus1+radic119888120484

(1 minus 119909)119899minus119898119889119909

= int

1

0

119899minus119898

sum

119896=0

(minus1)119896(

119899 minus 119898

119896)119909119898minus1+119896+radic119888120484

119889119909

=

119899minus119898

sum

119896=0

(minus1)119896(119899minus119898

119896)

119898 + 119896 + radic119888120484

(27)




1198711015840

119898119899⩾

prod119899

119896=119898radic1198962+ 119888

(119899 minus 119898)

(28)


1198711015840

119898119899⩾

119899

(119898 minus 1) (119899 minus 119898)

= 119898(

119899

119898) (29)


1198981015840 119899⩾

1198981015840(119899

1198981015840 ) and 119899 = 21198981015840 or 21198981015840 minus 1


(

119899

119890

)

119899

radic2120587119899 ⩽ 119899 ⩽ (

119899

119890

)

119899

radic21205871198991198901(12119899) (30)


allows us to obtain

(

2119899

119899) ⩾

22119899

radic119899120587

119890minus1(6119899)

(31)

If 119899 = 21198981015840 then

1198981015840(

21198981015840

1198981015840 ) ⩾ 2

21198981015840

radic1198981015840

120587

119890minus1(6119898

1015840)

(32)


1198981015840(

21198981015840

1198981015840 ) ⩾ 2

21198981015840

(33)


1198981015840(

21198981015840minus 1

1198981015840 ) ⩾ (2119898

1015840minus 1)(

2 (1198981015840minus 1)

1198981015840minus 1

)

⩾ 221198981015840minus11198981015840minus 12

radic1198981015840120587

119890minus1(6119898

1015840)

⩾ 221198981015840minus1

(34)


1198711015840

119898119899⩾ 2119899 (35)



[16 + 119888 25 + 119888 36 + 119888] ⩾ [25 + 119888 36 + 119888]

=

(25 + 119888) (36 + 119888)

gcd (25 + 119888 36 + 119888)

⩾

25 times 36

11

⩾ 64 = 26

(36)


[9 + 119888 16 + 119888 25 + 119888] ⩾ [16 + 119888 25 + 119888]

⩾

16 times 25

25 minus 16

⩾ 25

(37)


Acknowledgment


References





















Volume 2014




Journal of











Journal of


Function Spaces






Algebra










allows us to obtain

(

2119899

119899) ⩾

22119899

radic119899120587

119890minus1(6119899)

(31)

If 119899 = 21198981015840 then

1198981015840(

21198981015840

1198981015840 ) ⩾ 2

21198981015840

radic1198981015840

120587

119890minus1(6119898

1015840)

(32)


1198981015840(

21198981015840

1198981015840 ) ⩾ 2

21198981015840

(33)


1198981015840(

21198981015840minus 1

1198981015840 ) ⩾ (2119898

1015840minus 1)(

2 (1198981015840minus 1)

1198981015840minus 1

)

⩾ 221198981015840minus11198981015840minus 12

radic1198981015840120587

119890minus1(6119898

1015840)

⩾ 221198981015840minus1

(34)


1198711015840

119898119899⩾ 2119899 (35)



[16 + 119888 25 + 119888 36 + 119888] ⩾ [25 + 119888 36 + 119888]

=

(25 + 119888) (36 + 119888)

gcd (25 + 119888 36 + 119888)

⩾

25 times 36

11

⩾ 64 = 26

(36)


[9 + 119888 16 + 119888 25 + 119888] ⩾ [16 + 119888 25 + 119888]

⩾

16 times 25

25 minus 16

⩾ 25

(37)


Acknowledgment


References





















Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









note on the lower bound of least common multiple

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