3 ms·

My ten year old found Fibonacci numbers in the reciprocal of 89.

DavidSJ 4y ago
That's excellent! I think it's pretty interesting that this sequence can be hiding within the 44 repeating digits of the decimal expansion of 1/89: 0.1123595505617977528089887640449438202247191011235… [1] A proof can be found here: https://www.cantorsparadise.com/why-does-1-89-represent-the-fibonacci-sequence-7e09873533c3 https://www.cantorsparadise.com/why-does-1-89-represent-the-... [1] I couldn't easily find this many digits online, but here's a Python program to calculate them: num = int(input('numerator: ')) denom = int(input('denom: ')) sig = int(input('significant digits: ')) div = num // denom rem = num % denom print(f'{div}.{(rem*(10**sig))//denom}')
- 1letterunixname 4y ago
  0.01123595505617977528089887640449438202247191011235955056179775280898876404494382022471910112359550561797752808988764044943820224719101123595505617977528089887640449438202247191011235955056179775280898876404494382022471910112359550561797752808988764044943820... Off by an OoM.
  - DavidSJ 4y ago
    Oops yeah, I should have used some left zero padding. ;)
- dylan604 4y ago
  okay, so i'm confused 1+1 = 2 1+2 = 3 2+3 = 5 3+5 = 9???
  - sugarkjube 4y ago
    3+5=8 8+5=13 I assume the '1' of the '13' carries into the 8 => 8+1=9 Same for the numbers after that.
    - dylan604 4y ago
      Ha! In four lines, you described much more simply than whatever the sibling commenter decided to do ;) Yet again, foiled by forgetting to carry the one!!!! Those are the kinds of "finds" that I'm way too out of practice in searching for patterns to have noticed the answer. Or I'm just too lazy and out of practice and called it quits too quickly. Now, show me a syntax error of missing }, ), ], etc, and I'll find that pattern with/without an IDE!
  - raymondh 4y ago
    There is a carry from the 5+8=13. Add these up and you get closer and closer to 1 / 89: 0.0 0.01 0.001 0.0002 0.00003 0.000005 0.0000008 0.00000013 0.000000021 0.0000000034 0.00000000055 0.000000000089 0.0000000000144 0.00000000000233 0.000000000000377 0.0000000000000610 0.00000000000000987 0.000000000000001597 0.0000000000000002584 0.00000000000000004181 0.000000000000000006765 0.0000000000000000010946 0.00000000000000000017711 0.000000000000000000028657 0.0000000000000000000046368 0.00000000000000000000075025 0.000000000000000000000121393 0.0000000000000000000000196418 0.00000000000000000000000317811 0.000000000000000000000000514229 0.0000000000000000000000000832040 0.00000000000000000000000001346269 0.000000000000000000000000002178309 0.0000000000000000000000000003524578 0.00000000000000000000000000005702887 0.000000000000000000000000000009227465 0.0000000000000000000000000000014930352 0.00000000000000000000000000000024157817 0.000000000000000000000000000000039088169 0.0000000000000000000000000000000063245986 0.00000000000000000000000000000000102334155 0.000000000000000000000000000000000165580141 0.0000000000000000000000000000000000267914296 0.00000000000000000000000000000000000433494437 0.000000000000000000000000000000000000701408733 0.0000000000000000000000000000000000001134903170 0.00000000000000000000000000000000000001836311903 0.000000000000000000000000000000000000002971215073 0.0000000000000000000000000000000000000004807526976 0.00000000000000000000000000000000000000007778742049 0.000000000000000000000000000000000000000012586269025 0.0000000000000000000000000000000000000000020365011074 0.00000000000000000000000000000000000000000032951280099 0.000000000000000000000000000000000000000000053316291173 0.0000000000000000000000000000000000000000000086267571272 0.00000000000000000000000000000000000000000000139583862445 0.000000000000000000000000000000000000000000000225851433717 0.0000000000000000000000000000000000000000000000365435296162 0.00000000000000000000000000000000000000000000000591286729879 + 0.000000000000000000000000000000000000000000000000956722026041 -------------------------------------------------------------- 0.011235955056179775280898876404494382022471910112174867308031 ≈ 1 / 89 ============================================================== For those interested in running this with more iterations, here is a Python script: from decimal import * n = 60 getcontext().prec = n t = Decimal(0) p, q = 0, 1 for i in range(1, n+1): s = f'0.{p:0{i}}' print(s) t += Decimal(s) p, q = q, p+q print('-' * (n + 2)) print(t, '≈ 1 / 89') print('=' * (n + 2)) print() print(1 / t)
- sugarkjube 4y ago
  > couldn't easily find this many digits I still find 'bc' usefull when needing many digits $ echo 'scale=100;1/89'|bc
orlp 4y ago
That's because the infinite sum of F(n) x^n, also known as its generating function, is x/(1 - x - x^2). If you substitute x = 10^-1 you get 10/89, but also F(1) 10^-1 + F(2) 10^-2 + F(3) 10^-3 + F(4) 10^4 + F(5) 10^5 + ... 1 * 0.1 + 1 * 0.01 + 2 * 0.001 + 3 * 0.0001 + 5 * 0.00001 + ... So it's no surprise that the first digits of 1/89 contain the Fibonacci series as it's just 10/89 shifted by one decimal place. The pattern breaks down eventually because you overflow a single digit. But you can delay how long until this occurs by substituting in smaller powers of 10. For example if we substitute x = 10^-3 we get 1000/998999 or 0.0010010020030050080130210340550891442333776109885995881... I have also used this neat fact to make this golfed loopless/recursionless exact (no floating-point arithmetic) Python implementation of Fibonacci, by substituting in binary powers instead: F=lambda n:(4<<n*(3+n))//((4<<2*n)-(2<<n)-1)&~-(2<<n) If someone is not aware of generating functions it's essentially impossible to understand why this generates Fibonacci.
- mbrudd 4y ago
  > So it's no surprise that the first digits of 1/89 contain the Fibonacci series as it's just 10/89 shifted by one decimal place. >If someone is not aware of generating functions it's essentially impossible to understand why this generates Fibonacci. ??
  - orlp 4y ago
    Maybe "it's no surprise" is a poor choice of idiom in this context in hindsight.

That's excellent! I think it's pretty interesting that this sequence can be hiding within the 44 repeating digits of the decimal expansion of 1/89: 0.1123595505617977528089887640449438202247191011235… [1] A proof can be found here: https://www.cantorsparadise.com/why-does-1-89-represent-the-fibonacci-sequence-7e09873533c3 https://www.cantorsparadise.com/why-does-1-89-represent-the-... [1] I couldn't easily find this many digits online, but here's a Python program to calculate them: num = int(input('numerator: ')) denom = int(input('denom: ')) sig = int(input('significant digits: ')) div = num // denom rem = num % denom print(f'{div}.{(rem*(10**sig))//denom}')

1letterunixname 4y ago
0.01123595505617977528089887640449438202247191011235955056179775280898876404494382022471910112359550561797752808988764044943820224719101123595505617977528089887640449438202247191011235955056179775280898876404494382022471910112359550561797752808988764044943820... Off by an OoM.
- DavidSJ 4y ago
  Oops yeah, I should have used some left zero padding. ;)
dylan604 4y ago
okay, so i'm confused 1+1 = 2 1+2 = 3 2+3 = 5 3+5 = 9???
- sugarkjube 4y ago
  3+5=8 8+5=13 I assume the '1' of the '13' carries into the 8 => 8+1=9 Same for the numbers after that.
  - dylan604 4y ago
    Ha! In four lines, you described much more simply than whatever the sibling commenter decided to do ;) Yet again, foiled by forgetting to carry the one!!!! Those are the kinds of "finds" that I'm way too out of practice in searching for patterns to have noticed the answer. Or I'm just too lazy and out of practice and called it quits too quickly. Now, show me a syntax error of missing }, ), ], etc, and I'll find that pattern with/without an IDE!
- raymondh 4y ago
  There is a carry from the 5+8=13. Add these up and you get closer and closer to 1 / 89: 0.0 0.01 0.001 0.0002 0.00003 0.000005 0.0000008 0.00000013 0.000000021 0.0000000034 0.00000000055 0.000000000089 0.0000000000144 0.00000000000233 0.000000000000377 0.0000000000000610 0.00000000000000987 0.000000000000001597 0.0000000000000002584 0.00000000000000004181 0.000000000000000006765 0.0000000000000000010946 0.00000000000000000017711 0.000000000000000000028657 0.0000000000000000000046368 0.00000000000000000000075025 0.000000000000000000000121393 0.0000000000000000000000196418 0.00000000000000000000000317811 0.000000000000000000000000514229 0.0000000000000000000000000832040 0.00000000000000000000000001346269 0.000000000000000000000000002178309 0.0000000000000000000000000003524578 0.00000000000000000000000000005702887 0.000000000000000000000000000009227465 0.0000000000000000000000000000014930352 0.00000000000000000000000000000024157817 0.000000000000000000000000000000039088169 0.0000000000000000000000000000000063245986 0.00000000000000000000000000000000102334155 0.000000000000000000000000000000000165580141 0.0000000000000000000000000000000000267914296 0.00000000000000000000000000000000000433494437 0.000000000000000000000000000000000000701408733 0.0000000000000000000000000000000000001134903170 0.00000000000000000000000000000000000001836311903 0.000000000000000000000000000000000000002971215073 0.0000000000000000000000000000000000000004807526976 0.00000000000000000000000000000000000000007778742049 0.000000000000000000000000000000000000000012586269025 0.0000000000000000000000000000000000000000020365011074 0.00000000000000000000000000000000000000000032951280099 0.000000000000000000000000000000000000000000053316291173 0.0000000000000000000000000000000000000000000086267571272 0.00000000000000000000000000000000000000000000139583862445 0.000000000000000000000000000000000000000000000225851433717 0.0000000000000000000000000000000000000000000000365435296162 0.00000000000000000000000000000000000000000000000591286729879 + 0.000000000000000000000000000000000000000000000000956722026041 -------------------------------------------------------------- 0.011235955056179775280898876404494382022471910112174867308031 ≈ 1 / 89 ============================================================== For those interested in running this with more iterations, here is a Python script: from decimal import * n = 60 getcontext().prec = n t = Decimal(0) p, q = 0, 1 for i in range(1, n+1): s = f'0.{p:0{i}}' print(s) t += Decimal(s) p, q = q, p+q print('-' * (n + 2)) print(t, '≈ 1 / 89') print('=' * (n + 2)) print() print(1 / t)
sugarkjube 4y ago
> couldn't easily find this many digits I still find 'bc' usefull when needing many digits $ echo 'scale=100;1/89'|bc

orlp 4y ago

That's because the infinite sum of F(n) x^n, also known as its generating function, is x/(1 - x - x^2). If you substitute x = 10^-1 you get 10/89, but also F(1) 10^-1 + F(2) 10^-2 + F(3) 10^-3 + F(4) 10^4 + F(5) 10^5 + ... 1 * 0.1 + 1 * 0.01 + 2 * 0.001 + 3 * 0.0001 + 5 * 0.00001 + ... So it's no surprise that the first digits of 1/89 contain the Fibonacci series as it's just 10/89 shifted by one decimal place. The pattern breaks down eventually because you overflow a single digit. But you can delay how long until this occurs by substituting in smaller powers of 10. For example if we substitute x = 10^-3 we get 1000/998999 or 0.0010010020030050080130210340550891442333776109885995881... I have also used this neat fact to make this golfed loopless/recursionless exact (no floating-point arithmetic) Python implementation of Fibonacci, by substituting in binary powers instead: F=lambda n:(4<<n*(3+n))//((4<<2*n)-(2<<n)-1)&~-(2<<n) If someone is not aware of generating functions it's essentially impossible to understand why this generates Fibonacci.

mbrudd 4y ago
> So it's no surprise that the first digits of 1/89 contain the Fibonacci series as it's just 10/89 shifted by one decimal place. >If someone is not aware of generating functions it's essentially impossible to understand why this generates Fibonacci. ??
- orlp 4y ago
  Maybe "it's no surprise" is a poor choice of idiom in this context in hindsight.