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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: 0.11235955

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

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