OFFSET
1,1
COMMENTS
All numbers in this sequence are congruent to 1 mod 4. - Max Alekseyev.
If Fibonacci(n) is divisible by a prime p of the form 4k+3 then n is even. To prove this statement it is enough to show that (1+sqrt(5))/(1-sqrt(5)) is never a square modulo such p (which is a straightforward exercise).
The n-th prime p is an element of this sequence iff A001602(n) is prime and A051694(n)=A000045(A001602(n))>p. - Max Alekseyev
LINKS
Hans Havermann, Table of n, a(n) for n = 1..5000
MATHEMATICA
a = {}; k = {}; Do[If[ !PrimeQ[Fibonacci[Prime[n]]], s = FactorInteger[Fibonacci[Prime[n]]]; c = Length[s]; Do[AppendTo[k, s[[m]][[1]]], {m, 1, c}]], {n, 2, 60}]; Union[k]
CROSSREFS
KEYWORD
nonn
AUTHOR
Artur Jasinski, Dec 08 2007
EXTENSIONS
Edited, corrected and extended by Max Alekseyev, Dec 12 2007
STATUS
approved