A226366 Numbers k such that 5*2^k + 1 is a prime factor of a Fermat number 2^(2^m) + 1 for some m.

7, 25, 39, 75, 127, 1947, 3313, 23473, 125413

Offset: 1

Author

Arkadiusz Wesolowski, Jun 05 2013

Keywords

• nonn
• hard
• more

Comments

No other terms below 5330000.

The reason all terms are odd is that if k is even, then 5*2^k + 1 == (-1)*(-1)^k + 1 = (-1)*1 + 1 = 0 (mod 3). So if k is even, then 3 divides 5*2^k + 1, and since 3 divides no other Fermat number than F_0=3 itself, we do not have a Fermat factor. - Jeppe Stig Nielsen, Jul 21 2019

Links

• Wilfrid Keller, Fermat factoring status
• J. C. Morehead, Note on the factors of Fermat's numbers, Bull. Amer. Math. Soc., Volume 12, Number 9 (1906), pp. 449-451.
• Eric Weisstein's World of Mathematics, Fermat Number

Crossrefs

Subsequence of A002254.

Cf. A000215, A050526, A057775, A057778, A201364, A204620.

Programs

Mathematica

lst = {}; Do[p = 5*2^n + 1; If[PrimeQ[p] && IntegerQ@Log[2, MultiplicativeOrder[2, p]], AppendTo[lst, n]], {n, 7, 3313, 2}]; lst

PARI

isok(n) = my(p = 5*2^n + 1, z = znorder(Mod(2, p))); isprime(p) && ((z >> valuation(z, 2)) == 1); \\ Michel Marcus, Nov 10 2018