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A328896 Primes p such that p divides at least one of the integers Fibonacci(2k) for 2k <= p-1.

%I A328896 #33 Apr 12 2020 14:29:45
%S A328896 11,19,29,31,41,47,59,61,71,79,89,101,107,109,113,131,139,149,151,179,
%T A328896 181,191,199,211,229,233,239,241,251,263,269,271,281,307,311,331,347,
%U A328896 349,353,359,379,389,401,409,419,421,431,439,449,461,479,491,499,509
%N A328896 Primes p such that p divides at least one of the integers Fibonacci(2k) for 2k <= p-1.
%C A328896 Is the sequence infinite?
%C A328896 Yes, it contains all primes p == 1 or 4 (mod 5), because such p divide Fibonacci(p-1). - _Robert Israel_, Nov 05 2019
%H A328896 Robert Israel, <a href="/A328896/b328896.txt">Table of n, a(n) for n = 1..10000</a>
%e A328896 There are two integers k with 2*k <= 29-1 such that 29 divides Fibonacci(2*k), namely k = 7 and 14, so 29 is a term of the sequence.
%p A328896 filter:= proc(p) local f,k,a,b,t;
%p A328896 a:= -1; b:= 0;
%p A328896 for k from 1 to (p-1)/2 do
%p A328896    t:= a+2*b mod p;
%p A328896    a:= a+b mod p; b:= t;
%p A328896    if t = 0 then return true fi;
%p A328896 od;
%p A328896 false
%p A328896 end proc:
%p A328896 select(filter, [seq(ithprime(i),i=2..100)]); # _Robert Israel_, Nov 05 2019
%o A328896 (PARI) forprime(p=1, 100, for(k=1, (p-1)/2, if(Mod(fibonacci(2*k), p)==0, print1(p, ", "); break)))
%o A328896 (Sage)
%o A328896 def isA328896(p):
%o A328896     return any(p.divides(fibonacci(2*k)) for k in (1..(p-1)//2))
%o A328896 print([p for p in primes(1,510) if isA328896(p)]) # _Peter Luschny_, Nov 01 2019
%Y A328896 Cf. A000045, A000928, A120337, A321217.
%K A328896 nonn
%O A328896 1,1
%A A328896 _Felix Fröhlich_, Oct 30 2019
%E A328896 Definition corrected by _Robert Israel_, Nov 05 2019