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Numbers k such that 3*153479820268467961^2*2^k + 1 is prime.
2

%I #12 Apr 22 2023 19:30:10

%S 600,810,1074,7974,22290,43086

%N Numbers k such that 3*153479820268467961^2*2^k + 1 is prime.

%C Let p be a prime number of the form 3*153479820268467961^2*2^k + 1 with k > 0, then the multiplicative order of 2 modulo p is not of the form 2^(m+1), m >= 0. Hence, p does not divide any Fermat number F(m) = 2^(2^m) + 1.

%t Select[Range[2, 10^4, 2], PrimeQ[3*153479820268467961^2*2^# + 1] &]

%Y Cf. A000215, A229852, A351332, A361898, A361899.

%K nonn,more

%O 1,1

%A _Arkadiusz Wesolowski_, Mar 28 2023