march 2014. patterns vs invariants 142857 2 × 142857 = 285714
TRANSCRIPT
![Page 1: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/1.jpg)
March 2014
![Page 2: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/2.jpg)
PATTERNSvs
INVARIANTS
![Page 3: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/3.jpg)
142857
![Page 4: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/4.jpg)
142857
2 × 142857 = 285714
![Page 5: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/5.jpg)
142857
2 × 142857 = 285714
3 × 142857 = 428571
![Page 6: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/6.jpg)
142857
2 × 142857 = 285714
3 × 142857 = 428571
4 × 142857 = 571428
![Page 7: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/7.jpg)
142857
2 × 142857 = 285714
3 × 142857 = 428571
4 × 142857 = 571428
5 × 142857 = 714285
![Page 8: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/8.jpg)
142857
2 × 142857 = 285714
3 × 142857 = 428571
4 × 142857 = 571428
5 × 142857 = 714285
6 × 142857 = 857142
![Page 9: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/9.jpg)
142857
2 × 142857 = 285714
3 × 142857 = 428571
4 × 142857 = 571428
5 × 142857 = 714285
6 × 142857 = 857142
7 × 142857 = 999999
![Page 10: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/10.jpg)
0.142857142857...17
![Page 11: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/11.jpg)
Black Holes
Give me any whole number
(# of even digits)(# of odd digits)(total number of digits)
Repeat process
For Example: 19500723
![Page 12: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/12.jpg)
19500723 # of even digits = 3
# of odd digits = 5
Total # of digits = 8
358 → 123
22221111 → 448 → 303 → 123
22222222 → 808 → 303 → 123
![Page 13: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/13.jpg)
WHY ?
![Page 14: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/14.jpg)
ooe –––> 123
oeo –––> 123
eoo –––> 123
eee –––> 303 –––> 123
![Page 15: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/15.jpg)
201320142015
→ 7512
→ 134
→ 123
![Page 16: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/16.jpg)
Take any 3-digit number.
Rearrange it: Largest – Smallest
674: 764 – 467 = 303
303: 330 – 033 = 297
297: 972 – 279 = 693
693: 963 – 369 = 594
594: 954 – 459 = 495
![Page 17: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/17.jpg)
738 873 – 378 = 495
954 – 459 = 495
215 512 – 125 = 396
963 – 369 = 594
954 – 459 = 495
123 321 – 123 = 198
981 – 189 = 792
972 – 279 = 693
963 – 369 = 594
954 – 459 = 495
![Page 18: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/18.jpg)
1. What is so special about 495?
2. Is there any 3–digit number does not fall into this black hole?
3. Is there an upper limit of number of steps to get to this black hole?
4. Is there any type of numbers require more steps than other types?
5. Is there a black hole for 2–digit numbers?
![Page 19: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/19.jpg)
What about 4-digit numbers?
What about 5-digit numbers?
What about 6-digit numbers?
Is there a pattern?
![Page 20: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/20.jpg)
Take 2–digit numbers.
87: 87 – 78 = 9
90: 90 – 09 = 81
81 – 18 = 63
63 – 36 = 27
72 – 27 = 45
54 – 45 = 9
80: 80 – 08 = 72
72 – 27 = 45
54 – 45 = 9
![Page 21: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/21.jpg)
Is “9” a Black Hole?
Is there any other Black Hole for 2–digit numbers?
Which 2–digit number takes the longest to reach the Black Hole?
![Page 22: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/22.jpg)
What about 4–digit numbers?
2341: 4321 – 1234 = 3087
8730 – 0378 = 8352
8532 – 2358 = 6174
7641 – 1467 = 6174
Is 6174 the Black Hole for 4–digit numbers?
![Page 23: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/23.jpg)
What about 5–digit numbers?
12345: 54321 – 12345 = 41976
97641 – 14679 = 82962
98622 – 22689 = 75933
97533 – 33579 = 63954
96543 – 34569 = 61974
97641 – 14679 = 82962
Is 82962 the Black Hole for 5–digit numbers?
![Page 24: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/24.jpg)
Try 22220.
22220: 22220 – 02222 = 19998
99981 – 18999 = 80982
98820 – 02889 = 95931
99531 – 13599 = 85932
98532 – 23589 = 74943
97443 – 34479 = 62964
96642 – 24669 = 71973
97731 – 13779 = 83952
What happened? No Black Hole for 5–digit numbers?
![Page 25: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/25.jpg)
22220: 22220 – 02222 = 19998
99981 – 18999 = 80982
98820 – 02889 = 95931
99531 – 13599 = 85932
98532 – 23589 = 74943
97443 – 34479 = 62964
96642 – 24669 = 71973
97731 – 13779 = 83952
98532 – 23589 = 74943
However, 74943 is different from 82962. Could it be that 5–digit numbers have more than one Black Holes?
![Page 26: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/26.jpg)
What about 6–digit, 7–digit, …, n–digit numbers?
![Page 27: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/27.jpg)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18…
![Page 28: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/28.jpg)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18…
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 …
![Page 29: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/29.jpg)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 …
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 …
1 4 9 16 25 36 49 64 81…
![Page 30: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/30.jpg)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 …
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 …
1 4 9 16 25 36 49 64 81…
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 …
1 3 7 12 19 27 37 48 61 75 91…
1 8 27 64 125 216…
![Page 31: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/31.jpg)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 …
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 …
1 4 9 16 25 36 49 64 81…
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 …
1 3 712 19 27 37 48 61 75 91…
1 8 27 64 125 216…
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 …
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 …
2 3 6 11 17 24 33 43 54 67 81…
1 4 15 32 65 108 175 256…
3 16 81 256…
![Page 32: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/32.jpg)
Can you generalize this process and extend it? Is it same true for 5th power or 6th power or nth power?
![Page 33: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/33.jpg)
Give me any number n.
Use the following process:
n = odd 3 n +1
n = even n /2
![Page 34: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/34.jpg)
Give me any number n.
Use the following process:
n = odd 3n+1
n = even n/2
n=13…40, 20, 10, 5, 16, 8, 4, 2, 1
n=37…112, 56, 28, 14, 7, 22, 11, 34, 17, 52, 26, 13,…,1
n=101…304, 152, 76, 38, 19, 58, 29, 88, 44, 22, 11,…,1
![Page 35: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/35.jpg)
1. Does it always go to 1?
2. In how many steps?
3. What kind of numbers require more step?
4. Is there a maximum number of steps?
5. Is there a pattern to the number of steps required?
![Page 36: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/36.jpg)
1 = 0
2 = 1
3 = 7
4 = 2
5 = 5
6 = 8
7 = 16
8 = 3
9 = 19
10 = 6
![Page 37: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/37.jpg)
1 = 0 11 = 14 21 = 6
2 = 1 12 = 9 22 = 15
3 = 7 13 = 9 23 = ?
4 = 2 14 = 17 24 = 10
5 = 5 15 = ? 25 = ?
6 = 8 16 = 4 26 = 10
7 = 16 17 = 12 27 = ?
8 = 3 18 = 20 28 = 18
9 = 19 19 = ? 29 = 17
10 = 6 20 = 7 30 = ?
![Page 38: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/38.jpg)
199 is a prime
Rearrange 199 into 919 (a prime).
Rearrange 199 into 991 (a prime).
199 is called a “Permutable” or “Absolute” prime
![Page 39: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/39.jpg)
How many permutable primes?
1-digit:
2-digit:
3-digit:
![Page 40: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/40.jpg)
1-digit: 4 (2, 3, 5, 7)
2-digit: 9 (11, 13, 17, 31, 37, 71, 73, 79, 97)
5 (11, 13, 17, 37, 79)
3-digit: 8 (113, 131, 199, 311, 337, 733, 919, 991)
3 (113, 199, 337)
19-digit: 1 (R19 = 11111…111)
23-digit: 1 (R23)
317-digit: 1 (R317)
1031-digit: 1 (R1031)
![Page 41: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/41.jpg)
Facts about Permutable Primes:
1. Must compose from digits 1, 3, 7, and 9 if it is not 1-digit.
2. There are no permutable primes with number of
digits3 < n < 6×10175 except those that are composed all 1’s. (Not Proven)
3. There are no permutable primes with only digit 1
other than what are listed in the previous slide.(Not Proven)
![Page 42: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/42.jpg)
Facts about Permutable Primes:
4. No permutable prime exists which contains 3 different numbers of the 4 digits 1, 3, 7, 9.
(For example: 731 = 17 × 43)
5. No permutable primes that are composed of 2 or more of each of the 4 digits 1, 3, 7, 9.
![Page 43: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/43.jpg)
“Unprimable” Numbers
= A composite number that changes
any digit and still composite.
What is the smallest “unprimable” number?
Example: 200
![Page 44: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/44.jpg)
For many generations, prime numbers have always had mystical and magical appeals to mathematicians. Dating back to Euclid’s days, he had already proved that there are infinitely many prime numbers. However, from what we had seen before, it is very difficult to prove even some of the simple properties about prime numbers.
![Page 45: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/45.jpg)
1. There are infinitely many prime numbers (Proved by Euclid about 2000 years ago).
2. There are infinitely many twin prime numbers (still not yet proved).
![Page 46: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/46.jpg)
3. There is always at least one prime number exists between n and 2n where n is a natural number larger than 1. (Proved in 1850 by Chebyshev but he used very difficult and deep mathematics in his proof. However, Erdos, the second most prolific mathematicians in history, used a very simply proof to prove this result when he was a freshman in college.)
![Page 47: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/47.jpg)
There are some numbers that are comparable to prime numbers. The great late Indian mathematician Ramanujan had considered some numbers called “highly composite” numbers.
![Page 48: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/48.jpg)
Prime numbers have the minimum number of factors…only two…1 and itself.
Highly composite numbers have the maximum number of factors.
![Page 49: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/49.jpg)
Prime numbers have the minimum number of factors…only two…1 and itself.
Highly composite numbers have the maximum number of factors.
A composite number is Highly Composite if it has more distinct factors than any composite numbers before it.
![Page 50: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/50.jpg)
4 has 3 distinct factors.
6 has 4 distinct factors.
8 has 4 distinct factors.
9 has 3 distinct factors.
10 has 4 distinct factors.
![Page 51: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/51.jpg)
4 has 3 distinct factors…. Yes
6 has 4 distinct factors…. Yes
8 has 4 distinct factors…. No
9 has 3 distinct factors…. No
10 has 4 distinct factors…. No
![Page 52: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/52.jpg)
What about 12?
![Page 53: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/53.jpg)
12 has 6 distinct factors …. Yes
What is the next highly composite number?
![Page 54: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/54.jpg)
24 has 8 distinct factors …. Yes
What is the next highly composite number?
![Page 55: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/55.jpg)
36 has 9 distinct factors …. Yes
Ramanujan made a list of all highly composite numbers up to 6,746,328,388,800. He examined the highly composite numbers expressed as product of primes.
![Page 56: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/56.jpg)
For instance:
4 = 22
6 = 21 x 31
12 = 22 x 31
24 = 23 x 31
36 = 22 x 32
![Page 57: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/57.jpg)
4 = 22
6 = 21 x 31
12 = 22 x 31
24 = 23 x 31
36 = 22 x 32
He discovered that the final exponent of all highly composite numbers,
except for 4 and 36, is 1. Also, in the prime factorization of any highly
composite number, the first exponent always equals or exceeds the
second exponent and the second always equals or exceeds the third, and
so on.
![Page 58: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/58.jpg)
For example:
332,640 = 25 x 33 x 51 x 71 x 111
43,243,200 = 26 x 33 x 52 x 71 x 111 x 131
2,248,776,129,600 = 26 x 33 x 52 x 72 x 111
x 131 x 171 x 191 x 231
He wrote his thesis on the proof of this to get his
college degree from Cambridge.
![Page 59: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/59.jpg)
Pythagoras had considered a category of numbers with very interesting properties that are either very difficult to prove or not yet proven….“friendly numbers”.
![Page 60: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/60.jpg)
220 and 284
![Page 61: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/61.jpg)
Factors of 220 = 1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110 with a sum of 284.
Factors of 284 = 1, 2, 4, 71, 142 with a sum of 220.
![Page 62: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/62.jpg)
This was discovered about 2000 years ago by Greeks. The next pair did not get discovered until 1636 by Fermat.
17,296 and 18,416.
![Page 63: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/63.jpg)
The second smallest pair of friendly numbers was not discovered until 1866 by a 16 years old Italian schoolboy.
1184 and 1210.
![Page 64: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/64.jpg)
Now, with the help of some powerful computers, hundreds of pair of friendly numbers had been discovered.
Question: Are there infinitely many pairs of friendly numbers existed? Not yet proven.
![Page 65: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/65.jpg)
More Patterns.
What is so important about n2+n+17 ?
![Page 66: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/66.jpg)
n2+n+17
n=1 12+1+17 = 19 a prime
n=2 22+2+17 = 23 a prime
n=3 32+3+17 = 29 a prime
n=4 42+4+17 = 37 a prime
![Page 67: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/67.jpg)
n=5 52+5+17 = 47 a prime
n=6 62+6+17 = 59 a prime
n=7 72+7+17 = 73 a prime
n=8 82+8+17 = 89 a prime
n=9 92+9+17 = 107 a prime
n=10 102+10+17 = 127 a prime
n=11 112+11+17 = 149 a prime
n=12 122+12+17 = 173 a prime
![Page 68: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/68.jpg)
You may be tempted to think that you have found a formula that produces only prime numbers.
n=13 132+13+17 = 199 a prime
n=14 142+14+17 = 227 a prime
n=15 152+15+17 = 257 a prime
n=16 162+16+17 = 289 a prime???
n=17 172+17+17 = 323 a prime???
![Page 69: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/69.jpg)
Of course, if you use logic plus patterns, you will see that
n2+n+17 = (17m)2+(17m)+17 =
= 17(17m2+m+1)
not a prime
when n is a multiple of 17
such as 17m. So 323 = (17)(19)
![Page 70: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/70.jpg)
Euler came up with another formula n2+n+41. It produces prime numbers from n=1 to 39 but it fails at n = 40 and 41. However, people proved that this formula is actually quite good generating prime numbers under 10 million a respectable 47.5% of the time.
![Page 71: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/71.jpg)
28 ÷ 7 = ?
![Page 72: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/72.jpg)
28 ÷ 7 = 13
![Page 73: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/73.jpg)
13
7 | 28
7 21
21
![Page 74: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/74.jpg)
Check:
13
x 7
21
7 28
![Page 75: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/75.jpg)
Another check by adding:
13
13
13
13
13
13
+ 13
![Page 76: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/76.jpg)
Peasant’s Multiplication
37 × 28 = 1036
37 28
38 56
9 112
4 224
2 448
1 896
37 × 28 = 28 + 112 + 896 = 1036
![Page 77: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/77.jpg)
Why does it work?
28 = 111002 = 1×24 + 1×23 + 1×22 + 0×21 + 0×20
37 = 1001012 = 1×25 + 0×24 + 0×23 + 1×22 + 0×21 + 1×20
= 32 + 4 + 1
37 28 20 × 28
38 56
9 112 22 × 28
4 224
2 448
1 896 25 × 28
37 × 28 = (32 + 4 + 1) × 28 = (25 + 22 + 20) × 28 = 896 + 112 + 28) = 1036
![Page 78: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/78.jpg)
What is the sum of all three angles of any triangle?
![Page 79: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/79.jpg)
What is the sum of all three angles of any triangle?
180°
![Page 80: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/80.jpg)
What is the sum of all four angles of any quadralateral?
![Page 81: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/81.jpg)
What is the sum of all four angles of any quadralateral?
360°
![Page 82: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/82.jpg)
What is the sum of all five angles of any pentagon?
![Page 83: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/83.jpg)
What is the sum of all five angles of any pentagon?
540°
![Page 84: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/84.jpg)
In general, what is the sum of all n angles of any
n-polygon?
![Page 85: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/85.jpg)
In general, what is the sum of all n angles of any
n-polygon?
![Page 86: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/86.jpg)
In general, what is the sum of all n angles of any
n-polygon?
180(n–2)°
![Page 87: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/87.jpg)
How do you prove that?
180(n–2)°
![Page 88: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/88.jpg)
Interior angle
vs
Exterior angle
![Page 89: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/89.jpg)
What is the sum of all three exteror angles of any triangle?
![Page 90: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/90.jpg)
What is the sum of all n exterior angles of any
n-polygon?
![Page 91: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/91.jpg)
New Project:
Best way to build a high speed rail train to go from San Francisco to New York.
![Page 92: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/92.jpg)
It takes about 42 minutes to go from anywhere to anywhere.
![Page 93: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/93.jpg)
Suppose there are 2 points on a plane. Can you find a
line that separates these two points?
Suppose there are 3 points on a plane. Can you find a
line that has the same number of points on either side?
Suppose there are 10 points.
Suppose there are 11 points.
![Page 94: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/94.jpg)
What if there are
2,000,000 points?
![Page 95: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/95.jpg)
What if there are
2,000,000 points in 3–
space? Can we find a plane
that separates them into 2
sets of 1,000,000 each?
![Page 96: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/96.jpg)
Monte Hall’s
“Let’s Make a Deal”
Problem
![Page 97: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/97.jpg)
Behind three doors. One of them has a car and each of the other two has a goat. You
chose Door #1.
![Page 98: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/98.jpg)
If the host asks you if you want to change
your mind and switch your choice to another
door, would you?
![Page 99: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/99.jpg)
If the host opens the other two doors and shows you that
there is a goat behind each of these two doors and asks you if you want to change your
mind and switch your choice to another door, would you?
![Page 100: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/100.jpg)
If the host opens the one of the other two doors and shows
you that there is a goat behind that door and asks you if you
want to change your mind and switch your choice to another
door, would you?
![Page 101: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/101.jpg)
Not Switch
#1 #2 #3 OutcomeCar Goat Goat WinGoat Car Goat LoseGoat Goat Car LoseProbability of winning = 1/3
![Page 102: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/102.jpg)
Switch
#1 #2 #3 OutcomeCar Goat Goat LoseGoat Car Goat WinGoat Goat Car WinProbability of winning = 2/3
![Page 103: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/103.jpg)
Another problem.
During the old days, when people look up on the sky and see 8 stars lining up on a straight line, they think that special phenomenon must be set up artificially on purpose. The former great astronomer Carl Sagan said that this is really nothing special about it. He pointed out that, actually, if you have looked at larger enough of a sample, that phenomenon of 8 stars lining up happens everywhere. He is actually speaking of a very special topic in mathematics (one which is very popular among math competition problems).
![Page 104: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/104.jpg)
Example.
Pick any two students here. You may find one boy and one girl. As you keep picking out two students many times and suddenly you see both of the students are of the same sex. You are so amazed about this find that you proclaimed that someone must have done this for a purpose.
![Page 105: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/105.jpg)
However, upon examining this, you find that if you widen your pick … picking 3 students at a time, then the phenomenon of having two students of the same gender is nothing special. In fact, every time you do your pick of 3 students, two students of the same gender are among your pick.
![Page 106: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/106.jpg)
So picking 3 students at a time is the smallest number of students you can pick to guarantee this phenomenon happened.
The first person who formalize this theory and process is Frank Ramsey who died in 1930 at the age of 26. We now called this theory the Ramsey Theory in honor of him. It turns out problems from Ramsey Theory are some of the most difficult problems in mathematics.
![Page 107: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/107.jpg)
Here is a classical Ramsey problem.
What is the minimum number of guests that need to be invited so that either at least 3 guests will know each other or at least 3 will be mutual strangers?
![Page 108: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/108.jpg)
Obviously a party of 3 is not enough.
![Page 109: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/109.jpg)
Obviously a party of 3 is not enough.
A – B C
![Page 110: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/110.jpg)
A party of 4 is also not enough.
![Page 111: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/111.jpg)
A party of 4 is also not enough.
A – B – C – D
![Page 112: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/112.jpg)
A party of 5 is not enough.
![Page 113: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/113.jpg)
A party of 5 is not enough.
A – B – D
\ /
C – E
![Page 114: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/114.jpg)
A party of 6 is enough.
Either A knows at least three of the other five or A does not know at least three of the other five.
![Page 115: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/115.jpg)
/ B
Assume A knows B and C and D. A –– C
\
D
If B–C or C–D or B–D, then either ABC or ACD or ABD know each other.
If not, then B, C, and D are total strangers.
![Page 116: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/116.jpg)
Assume A does not know B and C and D.
If B does not know C or C does not know D or B does not know D, then ABC or ACD or ABD are totally strangers.
If B knows C and D and C knows D, BCD know each other.
If B knows C and D but C does not know D, then ACD are totally strangers.
![Page 117: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/117.jpg)
Another way to prove this is to list and examine all the combinations of relationships between those 6 people. It turns out these are a total of 32,768 possibilities. This brute force method does not provide any insight.
Now if we increase the number of 3 persons to 4, brute force method is impossible to do because the large number of possibilities.
![Page 118: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/118.jpg)
It has been proved that at least 18 people need to be invited to guarantee at least 4 persons all know each other or all strangers.
What about 5? Nobody knows. People think the least number is between 43 and 49.
What about 6? People think the least number is between 102 and 165.
![Page 119: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/119.jpg)
One of the largest doing with computers to solve math problems occurred in 1993 involving 110 computers running simultaneously to show that the minimum number of guests to invite to a party to guarantee at least 4 mutual friends or at least 5 mutually strangers is 25.
![Page 120: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/120.jpg)
ππ ≈ 3.1415926535897…
![Page 121: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/121.jpg)
ππ ≈ 3.1415926535897…
Pi Day:
March 14, 2015 9:26:53 morning.
![Page 122: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/122.jpg)
Facts About π
1. π was calculated to over 10 trillion digits (1013).
2. Record for most number of digits memorized is over 67,000.
3. π is irrational (1761).
4. π is transcendental (1882).
5. π is NOT normal – all possible sequence of digits appear equally often. That means one can use the digits from π to form random numbers (not proven).
6. A sequence of six consecutive 9’s appeared begins at the 762nd decimal place (Feynman Point).
![Page 123: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/123.jpg)
Facts About πSix 9’s starts at 762.
Next six 9’s starts at 193,034.
Six 8’s starts at 222,299.
Six 0’s starts at 1,699,927.
Six 6’s starts at 45,681,781.
One 9 – 5
Two 9’s – 44
Three 9’s – 762
Four 9’s – 762
Five 9’s – 762
Six 9’s – 762
Seven 9’s – 1,722,776
Eight 9’s – 36,356,642
Nine 9’s – 564,665,206
![Page 124: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/124.jpg)
Joe’s AccomplishmentsJoe lived over 1550 years ago.
Joe calculated 3.1415926 < π < 3.1415927
Joe used two fractions to approximate π
22/7 = 3.142857143… as coarse ratio
355/113 = 3.14159292035… as refined ratio
22/7 was used 2000 years ago by Greek’s Archimedes
Archimedes also calculated
223/72 < π < 22/7
3.140845… < π < 3.142857…
By slight increase to the denominators from 7 and 72 to Joe’s 113, π’s accuracy increases from 2 to 6 decimals
![Page 125: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/125.jpg)
What So Good About 355/113?
1. |31415926/10000000 – π| < 0.00000005
gets 7 decimal accuracy.
31415926/10000000 = 15707963/5000000 gets only 1 more decimal accuracy than 355/113, but 5,000,000 is so much larger than 113.
2. Any fraction with denominator less than 113 cannot get a
better approximation of π.
3. Any fractional approximation of π with denominator less
than 1000 cannot get better estimate than 113
![Page 126: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/126.jpg)
What So Good About 355/113?
4. Even denominator less than 10000 cannot get better estimate than 113
5. In fact, any denominator less than 16604 cannot be better.
What so good about 355/113?
![Page 127: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/127.jpg)
Why 113 Is So Good?Let us try to find a fraction q/p ≠ 355/113 but
gives about the same or better estimate.
0 < 355/113 – π < 0.00000026677
–0.00000026677 < π – q/p < 0.00000026677
Add–0.00000026677 < 355/113 – q/p < 2×0.00000026677
|355/113 – q/p| = |(355p–113q)/113p|
< 2×0.00000026677
![Page 128: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/128.jpg)
Since q/p ≠ 355/113, so |355p – 113q| > 0.
But p & q are integers, so |355p–113q| > 1.
So, 1/113p ≤ |(355p–113q)/113p|
< 2×0.00000026677
p > 1/(113×2×0.00000026677) > 16586.
![Page 129: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/129.jpg)
In fact, 52163/16604 = 3.141592387…
355/113 = 3.1415922035…
π = 3.1415926535897…
Using 16604 gives us 6 decimal accuracy just like 113 but 16604 is more than 100 times larger than 113
![Page 130: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/130.jpg)
How Did Joe Get 355/113?
It is a mystery since Joe did not write it down or publish it on internet.
Let us find one possible way that he could have done it.
Remember: We are not trying to get a good estimate for π. We are trying to find a good fraction to estimate π.
![Page 131: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/131.jpg)
π = 3 + 0.141592653… = 3 + a1 = 3 + 1/(1/ a1)
1/a1 = 1/0.141592653 = 7.062513305… = 7 + a2
So,
π = 3 + a1 = 3 + 1/(1/ a1) = 3 + 1/(7+a2) ≈ 3 + 1/7 = 22/7
![Page 132: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/132.jpg)
1/a2 = 1/0.062513305 = 15.99659454…= 15 + a3
So,
π = 3 + 1/(7+(1/(15+a3))) ≈ 3 + 1/(7+1/15)
= 333/106
1/a3 = 1/0.99659454 = 1.003417097…= 1 + a4
So,
π = 3 + 1/(7+(1/(15+1/(1+a4))))
≈ 3 + 1/(7+1/(15+1)) = 355/113
![Page 133: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/133.jpg)
One More Step:
1/a4 = 1/0.003417097 = 292.6460677…= 292 + a5
So,
π = 3 + 1/(7+(1/(15+1/(1+(292+a5)))))
≈ 3 + 1/(7+1/(15+1/(1+1/292)))
= 103993/33102
![Page 134: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/134.jpg)
However, 103993/33102 = 3.141592653011900…
and π = 3.1415926535897…
which is 9 decimals accuracy but 33102 is almost 300 times larger than 113.
Also, 355/113 is easy to remember:
113|355
![Page 135: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/135.jpg)
Note that = 1.414213562…
=
21
11
21
21
22
![Page 136: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/136.jpg)
And the Golden Ratio is:
ᵠ =
= 1.6180339887…
=
11
11
11
11
1
1 5
2
![Page 137: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/137.jpg)
What would be good fractions to approximate
and ᵠ2
![Page 138: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/138.jpg)
Note that = 1.414213562…
=
1, 4/3, 10/7, 24/17, 58/41, 140/99, 338/239, 816/577, …
21
11
21
21
22
![Page 139: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/139.jpg)
1+1 = 2
1 41 1.33333...
1 2 3
1 10
1 1.42857142857...4 713
1 241 1.4117647...
10 1717
1 581 1.4146341463.....
24 41117
![Page 140: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/140.jpg)
ᵠ = 1.6180339887…
=
1, 3/2, 5/3, 8/5, 13/8, 21/13, 34/21, 55/34, 89/55, …
11
11
11
11
1
![Page 141: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/141.jpg)
1+1 = 2
1 31 1.5
2 2
1 51 1.66666...
3 32
1 81 1.6
5 53
1 131 1.625
8 85
![Page 142: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/142.jpg)
1729
![Page 143: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/143.jpg)
Hardy–Ramanujan Number
![Page 144: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/144.jpg)
1729 = 13 + 123
= 93 + 103
![Page 145: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/145.jpg)
(1) 1729 is the smallest natural number that has this property.
(2) 1729 is a “near–miss” for
zn = xn + yn.
1729 = 123 +1 = 93 + 103
![Page 146: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/146.jpg)
Is there a number that can be written as the sum of 4 cubes in at least 10 different ways?
![Page 147: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/147.jpg)
It is hard to find such number but it is relatively easy to show that this number exists.
![Page 148: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/148.jpg)
Note:
There are 1000 cubes that are less than or equal to 1,000,000,000.
Is there a number that can be written as the sum of 4 cubes in at least 10 different ways?
![Page 149: March 2014. PATTERNS vs INVARIANTS 142857 2 × 142857 = 285714](https://reader038.vdocuments.mx/reader038/viewer/2022103122/56649ccf5503460f9499b419/html5/thumbnails/149.jpg)
Order the first thousand cubes as
a1, a2, a3, a4, …, a1000.
There are 1000C4 = 1000×999×998×997/24 > 40×1,000,000,000 ways to pick out any 4 cubes from this sequence.
Any sum of 4 cubes from this sequence cannot exceed 4×1,000,000,000.
So there are at the most 4,000,000,000 values to choose from by these more than 40,000,000,000 sets of sum of 4 cubes. Therefore, there is at least one value (among these 4,000,000,000) that is equal to 10 or more sets of sum of four cubes.