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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).
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.
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