[seqfan] Re: Conway's subprime Fibonacci sequences

Tanya Khovanova mathoflove-seqfan at yahoo.com
Fri Jul 27 20:49:25 CEST 2012


Another possibility is to list the cycles by the largest prime. The largest number in a cycle is prime. Also, if there is another cycle the largest prime will be bigger than the known ones. One problem: we didn't prove that two different cycles can't have the same largest number. Given that if other cycles exist, there are very rare, we can expect that this is not a problem.



________________________________
 From: "franktaw at netscape.net" <franktaw at netscape.net>
To: seqfan at list.seqfan.eu 
Sent: Friday, July 27, 2012 11:06 AM
Subject: [seqfan] Re: Conway's subprime Fibonacci sequences
 
So submit the cycle lengths ordered by the first starting pair that generates that cycle (in anti-diagonal order). And add the starting values as two more sequences. (This assumes that you can determine whether a sequence will cycle; is that true?)

Franklin T. Adams-Watters

-----Original Message-----
From: Hans Havermann <gladhobo at teksavvy.com>

Wouter Meeussen:

> I submit the cycle lengths (1), 10, 11, 18, 19, 56, 136 as
> 'potentially finite'...

Tanya Khovanova:

> We do not have a proof that these are the only cycle lengths, there
> could be cycle lengths in between. Also, we do not have a proof that
> there couldn't be two cycles of the same length.

A way around this might be to submit the smallest n such that the pair
{1, n} evolves into a distinct non-trivial cycle. For example, n = {1,
4, 36, 62, 6946, 27857, ...} evolving into the cycles of length {18,
136, 56, 19, 11, 10, ...}. (Unfortunately, if there are other cycles,
there is no proof that there exists an n such that {1, n} necessarily
evolves into it.)

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