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

11, 19, 29, 31, 41, 47, 59, 61, 71, 79, 89, 101, 107, 109, 113, 131, 139, 149, 151, 179, 181, 191, 199, 211, 229, 233, 239, 241, 251, 263, 269, 271, 281, 307, 311, 331, 347, 349, 353, 359, 379, 389, 401, 409, 419, 421, 431, 439, 449, 461, 479, 491, 499, 509

Offset: 1

Author

Felix Fröhlich, Oct 30 2019

Keywords

• nonn

Comments

Is the sequence infinite?

Yes, it contains all primes p == 1 or 4 (mod 5), because such p divide Fibonacci(p-1). - Robert Israel, Nov 05 2019

Examples

			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.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000045, A000928, A120337, A321217.

Programs

Maple

filter:= proc(p) local f,k,a,b,t;
a:= -1; b:= 0;
for k from 1 to (p-1)/2 do
   t:= a+2*b mod p;
   a:= a+b mod p; b:= t;
   if t = 0 then return true fi;
od;
false
end proc:
select(filter, [seq(ithprime(i),i=2..100)]); # Robert Israel, Nov 05 2019

PARI

forprime(p=1, 100, for(k=1, (p-1)/2, if(Mod(fibonacci(2*k), p)==0, print1(p, ", "); break)))

Sage

def isA328896(p):
    return any(p.divides(fibonacci(2*k)) for k in (1..(p-1)//2))
print([p for p in primes(1,510) if isA328896(p)]) # Peter Luschny, Nov 01 2019

Extensions

Definition corrected by Robert Israel, Nov 05 2019