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A204620 Numbers k such that 3*2^k + 1 is a prime factor of a Fermat number 2^(2^m) + 1 for some m. 16

%I

%S 41,209,157169,213321,303093,382449,2145353,2478785

%N Numbers k such that 3*2^k + 1 is a prime factor of a Fermat number 2^(2^m) + 1 for some m.

%C Terms are odd: by Morehead's theorem, 3*2^(2*n) + 1 can never divide a Fermat number.

%C No other terms below 7516000.

%C Is this sequence the same as "Numbers k such that 3*2^k + 1 is a factor of a Fermat number 2^(2^m) + 1 for some m"? - _Arkadiusz Wesolowski_, Nov 13 2018

%C Yes. The last sentence of Morehead's paper is: "It is easy to show that _composite_ numbers of the forms 2^kappa * 3 + 1, 2^kappa * 5 + 1 can not be factors of Fermat's numbers." - _Jeppe Stig Nielsen_, Jul 23 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 = 3*2^n + 1; If[PrimeQ[p] && IntegerQ@Log[2, MultiplicativeOrder[2, p]], AppendTo[lst, n]], {n, 7, 209, 2}]; lst

%o (PARI) isok(n) = my(p = 3*2^n + 1, z = znorder(Mod(2, p))); isprime(p) && ((z >> valuation(z, 2)) == 1); \\ _Michel Marcus_, Nov 10 2018

%Y Subsequence of A002253.

%Y Cf. A000215, A039687, A057775, A057778, A201364, A226366.

%K nonn,hard,more

%O 1,1

%A _Arkadiusz Wesolowski_, Jan 17 2012

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Last modified October 16 21:10 EDT 2019. Contains 328103 sequences. (Running on oeis4.)