[seqfan] Re: Conway's subprime Fibonacci sequences

[franktaw at netscape.net]

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.

Franklin T. Adams-Watters