Timeline for Is there any number other than 109 whose reciprocal contains the Fibonacci sequence? [closed]
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Feb 23, 2018 at 10:04 | history | closed |
Gro-Tsen Peter Heinig user6976 Mikhail Katz Stefan Kohl♦ |
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Feb 23, 2018 at 1:36 | comment | added | Gerry Myerson | What are you talking about, Esdras? $${1\over89}=.0+.01+.001+.0002+.00003+.000005+.0000008+.00000013+\cdots$$ The whole Fibonacci sequence, out to infinity, is there. | |
Feb 22, 2018 at 23:17 | comment | added | Esdras E E Dansha | #Gerry Myerson 89 is not a candidate... it does not give full fibonacci sequence | |
Feb 22, 2018 at 23:15 | answer | added | Aaron Meyerowitz | timeline score: 5 | |
Feb 22, 2018 at 22:16 | comment | added | Taneli Huuskonen | The factors of numbers of the form $10^{2n}\pm 10^{n}-1$ give rise to Fibonacci-like sequences not starting from 1. For instance, $999899=179\cdot 5581$, and the digit triples of $1/5581$ start like this: 000, 179, 179, 358, 537, ... | |
Feb 22, 2018 at 21:51 | comment | added | Taneli Huuskonen | You get the $n$-digit versions with $10^{2n}\pm 10^{n}-1$, with the direction of the Fibonacci sequence depending on the sign. | |
Feb 22, 2018 at 20:57 | comment | added | Gerry Myerson | Look at $1/89$. | |
Feb 22, 2018 at 19:42 | review | Close votes | |||
Feb 23, 2018 at 10:04 | |||||
Feb 22, 2018 at 19:02 | history | edited | Peter Heinig | CC BY-SA 3.0 |
Mainly stylistic corrections ('ending sequence' replaced by something more informative).
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Feb 22, 2018 at 18:47 | history | edited | Peter Heinig | CC BY-SA 3.0 |
Purely grammatical corrections in title and OP text. Style and content respected.
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Feb 22, 2018 at 17:41 | comment | added | Gerhard Paseman | There probably is. Check Joe Roberts text Elementary Number Theory. I imagine one can find n-digit versions (e.g. ending in 0003000200010001). This post is more suited for math.stackexchange. Gerhard "There Are Also Generating Functions" Paseman, 2018.02.22. | |
Feb 22, 2018 at 17:26 | history | asked | Esdras E E Dansha | CC BY-SA 3.0 |