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
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
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
Comments