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%I #21 May 07 2020 02:55:29
%S 1,4,1,1,1,3,3,10,17,8,83,92,525,1888,20
%N a(n) is the smallest k such that exactly one of k*2^(2^n) - 2*k + 1 and k*2^(2^n) + 2*k - 1 is a prime.
%C Conjecture: a(n) > 1 for all n > 4.
%C a(n) = 1, n > 1, is equivalent to F(n) = 2^(2^n) + 1 being a Fermat prime, because the Mersenne number M(2^n) = 2^(2^n) - 1 is prime only for n = 1 (since divisible by 3 for all n >= 1), where F(1) is also prime. - The two considered numbers can also be written 2*k*A077585(n) + 1 resp. 2*k*A000051(A000225(n)) - 1. - _M. F. Hasler_, Mar 05 2020
%e a(0) = 1 because 1*2^(2^0) - 2*1 + 1 = 1 is a nonprime and 1*2^(2^0) + 2*1 - 1 = 3 is a prime.
%e a(1) = 4 because 4*2^(2^1) - 2*4 + 1 = 9 is a composite and 4*2^(2^1) + 2*4 - 1 = 23 is a prime.
%e a(2) = 1 because 1*2^(2^2) - 2*1 + 1 = 15 is a composite and 1*2^(2^2) + 2*1 - 1 = 17 is a prime.
%e a(3) = 1 because 1*2^(2^3) - 2*1 + 1 = 255 is a composite and 1*2^(2^3) + 2*1 - 1 = 257 is a prime.
%e a(4) = 1 because 1*2^(2^4) - 2*1 + 1 = 65535 is a composite and 1*2^(2^4) + 2*1 - 1 = 65537 is a prime.
%o (PARI) a(n) = {my(k=1, m=2^2^n); while(ispseudoprime(k*m-2*k+1)-ispseudoprime(k*m+2*k-1)==0, k++); k; } \\ _Jinyuan Wang_, Feb 26 2020
%Y Cf. A019434, A331487.
%Y Cf. A000215 (Fermat numbers 2^2^n + 1), A000225 (Mersenne numbers 2^n - 1).
%K nonn,more
%O 0,2
%A _Juri-Stepan Gerasimov_, Feb 26 2020
%E Offset changed to 0 and a(11)-a(14) from _Jinyuan Wang_, Feb 26 2020
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