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%I #66 Jun 23 2019 17:32:22
%S 2,6,10,14,18,22,30,34,38,42,54,58,66,70,90,102,110,114,126,138,170,
%T 178,242,294,314,326,350,378,462,566,646,726,758,1150,1242,1302,1482,
%U 1558,1638,1710,1770,1970,1994
%N Numbers n such that the right Aurifeuillian primitive part of 2^n+1 is prime.
%C All terms are congruent to 2 modulo 4.
%C Let Phi_n(x) denote the n-th cyclotomic polynomial.
%C Numbers n such that Phi_{2nM(n)}(2) is prime.
%C Let J(n) = 2^n+1, J*(n) = the primitive part of 2^n+1, and this is Phi_{2n}(2).
%C Let M(n) = the Aurifeuillian M-part of 2^n+1, M(n) = 2^(n/2) + 2^((n+2)/4) + 1 for n congruent to 2 (mod 4).
%C Let M*(n) = GCD(M(n), J*(n)), this sequence lists all n such that M*(n) is prime.
%H Eric Chen, Gord Palameta, <a href="https://oeis.org/A250197/a250197_2.txt">Factorization of Phi_n(2) for n up to 1280</a>
%H Samuel Wagstaff, <a href="http://homes.cerias.purdue.edu/~ssw/cun/index.html">The Cunningham project</a>
%H Eric W. Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/AurifeuilleanFactorization.html">Aurifeuillean Factorization</a>.
%e 14 is in this sequence because the right Aurifeuillian primitive part of 2^14+1 is 29, which is prime.
%e 26 is not in this sequence because the right Aurifeuillian primitive part of 2^26+1 is 8321, which equals 53 * 157 and is not prime.
%t Select[Range[2000], Mod[n, 4] == 2 && PrimeQ[GCD[2^(n/2) + 2^((n+2)/4) + 1, Cyclotomic[2*n, 2]]]
%o (PARI) isok(n) = isprime(gcd(2^(n/2) + 2^((n+2)/4) + 1, polcyclo(2*n, 2))); \\ _Michel Marcus_, Jan 27 2015
%Y Cf. A250197, A153443, A019320, A072226, A161508, A085601, A229768, A061443.
%K nonn
%O 1,1
%A _Eric Chen_, Jan 18 2015
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