Simple(?) Fib problem

[Ralf Stephan]

> This is an interesting problem. If m is a multiple of 8, then the Fibs do 
> not have every residue mod m since no Fib is congruent to 4 mod 8.
> Otherwise, I need to think about it further.

 From the numbers so far I'd conjecture the numbers contain
1. all powers of three
2. the numbers 1,2,3,4,6,7,14
3. if x is in the sequence, then 5x is, too.


ralf