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http://list.seqfan.eu/pipermail/seqfan/2012-September/010196.html

is the link to "The SeqFan Archives" thread "Recursions in decimal expansions"

This thread discusses whether there is a generalization, which would allow to formulate the definitive rule for finding "Positive integers n such that the initial part of the decimal expansion of 1/n reveals a recursive sequence".

So this is the question, for which it would be interesting to find the answer.

P.S. The subject of interest for submitting this seq was to find such (and only such) positive integers, which decimal expansion INITIALLY reveals some algorithmically defined (and already known, thus easily visually recognizable) sequence (recursive or not) but THEN LATER (after some number of terms being revealed) such resemblance gets distorted (scrambled) and eventually goes away (fades, vanishes).

The follow up question, which was intriguing me was whether such resemblance is another incarnation of the Law of Small Numbers or whether such effect has deterministic origin.

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  • $\begingroup$ I don't think there is any way to make this rigorous. Any finite number of terms can be defined recursively (in the sense used by computability theorists). $\endgroup$
    – Alex Becker
    Sep 25, 2012 at 23:41
  • $\begingroup$ I think that OEIS doesn't require it to be strongly rigorous $\endgroup$
    – Alex
    Sep 25, 2012 at 23:46
  • $\begingroup$ So why isn't say 1/1171 on the list? It starts off following the sequence $0,0,0,8,5,3,9,0,0,0,8,5,3,9,\ldots$ (which can be defined recursively) but begins to differ from it at the 8th term. $\endgroup$
    – Alex Becker
    Sep 26, 2012 at 19:26
  • $\begingroup$ My point is the same logic applies to any $n$. I picked $1171$ off the top of my head and made a sequence to fit. $\endgroup$
    – Alex Becker
    Sep 27, 2012 at 19:24
  • $\begingroup$ Not "any" n - terminating 1/n fractions will not fit (say n=2 for example). Periodic ones should also be excluded . Perhaps "recursive" is not the best definition word. I thought I explained what I meant in one of the comments above more clearly but may be more clarifications are needed. $\endgroup$
    – Alex
    Sep 27, 2012 at 19:34

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An easy to find such is $\frac 13+\frac 1{10^{100}}$ for example. It will start with $99 \ 3$'s, then a $4$. If you don't want just one perturbation, you can use $\frac 13+\frac 1{10^{100}-7}$

Another question in the vein is this one regarding $\frac 1{243}$

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