%I #38 Jul 23 2019 04:22:17
%S 7,25,39,75,127,1947,3313,23473,125413
%N Numbers k such that 5*2^k + 1 is a prime factor of a Fermat number 2^(2^m) + 1 for some m.
%C No other terms below 5330000.
%C 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
%H Wilfrid Keller, <a href="http://www.prothsearch.com/fermat.html">Fermat factoring status</a>
%H J. C. Morehead, <a href="https://doi.org/10.1090/S0002-9904-1906-01371-4">Note on the factors of Fermat's numbers</a>, Bull. Amer. Math. Soc., Volume 12, Number 9 (1906), pp. 449-451.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/FermatNumber.html">Fermat Number</a>
%t lst = {}; Do[p = 5*2^n + 1; If[PrimeQ[p] && IntegerQ@Log[2, MultiplicativeOrder[2, p]], AppendTo[lst, n]], {n, 7, 3313, 2}]; lst
%o (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
%Y Subsequence of A002254.
%Y Cf. A000215, A050526, A057775, A057778, A201364, A204620.
%K nonn,hard,more
%O 1,1
%A _Arkadiusz Wesolowski_, Jun 05 2013