# [seqfan] Re: Conway's subprime Fibonacci sequences

Neil Sloane njasloane at gmail.com
Sun Jul 29 05:11:04 CEST 2012

No one has submitted these two sequences yet - would someone please take
care of it? (So far I think only one sequence
from this whole thread has been submitted, namely Wouter's A214674)

Neil

On Tue, Jul 24, 2012 at 4:49 PM, <franktaw at netscape.net> wrote:

> This suggested a different sequence to me. Start with a(0) = 0 and a(1) =
> 1. Normally a(n) = a(n-1) + a(n-2), but if this is divisible by a prime not
> yet encountered, divide out the smallest such.
>
> If I haven't made any mistakes, this starts:
>
> 0, 1, 1, 1*, 2, 1*, 3, 4, 1*, 1*, 2, 3, 5, 8, 1*, 9, 10, 1*, 1*, 2, 3, 5,
> 8, 13, 21, 2*, 1*, 3, 4, 7, 11, 18, 1*, 19, 20, 39, 1*, 40, 1*, 41, 42, 1*,
> 1*
>
> * marks the entries where a prime has been factored out.
>
> The obvious questions:
>
> 1) Does every prime eventually divide this sequence? I think this is
> almost certainly true. A related question: can one show that any
> Fibonacci-type sequence is divisible by  (infinitely many) primes? If this
> is false (not just unproved), then probably the sequence is divisible by
> only finitely many primes. But I think it must be true, and very likely a
> known result. (Note that "every Fibonacci-type sequence is divisible by
> every prime" is false; e.g. the Lucas numbers mod 5 are 2, 1, 3, 4, 2, 1,
> ...)
>
> We can generate an associated sequence, with the b(n) the index of the
> term divided by Prime(n) in the original sequence. This starts (again, done
> by hand):
>
> 3, 5, 9, 8, 18, 14, 25, 17, 26, 32
>
> 2) Do we ever encounter a term divisible by 2 or more previously unused
> primes? If so, other sequences can be generated by dividing out all unseen
> primes, or the largest unseen prime, instead of the smallest. I suspect
> that the answer to this one is yes, but I have no idea where the first one
> is encountered.
>
> 3) Are there infinitely many 1's in the sequence? If so, do we get 1, 1
> infinitely often? A positive answer to this question would also answer (1)
> in the affirmative.
>
>
>
>
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Dear Friends, I have now retired from AT&T. New coordinates:

Neil J. A. Sloane, President, OEIS Foundation
11 South Adelaide Avenue, Highland Park, NJ 08904, USA