%I
%S 11,20,23,35,39,48,83,96,131,231,303,375,384,519,771,848,1400,1983,
%T 2280,2640,2715,3359,6144,7736,7911,11079,13224,16664,24263,36168,
%U 130439,406583
%N Numbers n such that the Phi_n(2) is the product of exactly two primes and is divisible by 2n+1.
%C Here Phi_n is the nth cyclotomic polynomial.
%C Is this sequence infinite?
%C Phi_n(2)/(2n+1) is only a probable prime for n > 16664.
%C a(33) > 2000000.
%C Subsequence of A005097 (2 * a(n) + 1 are all primes)
%C Subsequence of A081858.
%C 2 * a(n) + 1 are in A115591.
%C Primes in this sequence are listed in A239638.
%C A085021(a(n)) = 2.
%C All a(n) are congruent to 0 or 3 (mod 4). (A014601)
%C All a(n) are congruent to 0 or 2 (mod 3). (A007494)
%C Except the term 20, all even numbers in this sequence are divisible by 8.
%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 Will Edgington, <a href="https://web.archive.org/web/20111107151843/http://www.garlic.com/~wedgingt/factoredM.txt">Factorization of completely factored Phi_n(2)</a> [from Internet Archive Wayback Machine]
%H Henri Lifchitz and Renaud Lifchitz, <a href="http://www.primenumbers.net/prptop/searchform.php?form=%282%5Ea1%29%2Fb&action=Search">PRP records. Search for (2^a1)/b</a>
%H Samuel Wagstaff, <a href="http://homes.cerias.purdue.edu/~ssw/cun/index.html">The Cunningham project</a>
%e Phi_11(2) = 23 * 89 and 23 = 2 * 11 + 1, so 11 is in this sequence.
%e Phi_35(2) = 71 * 122921 and 71 = 2 * 35 + 1, so 35 is in this sequence.
%e Phi_48(2) = 97 * 673 and 97 = 2 * 48 + 1, so 48 is in this sequence.
%t Select[Range[10000], PrimeQ[2*# + 1] && PowerMod[2, #, 2*# + 1] == 1 &&
%t PrimeQ[Cyclotomic[#, 2]/(2*#+1)] &]
%o (PARI) isok(n) = if (((x=polcyclo(n, 2)) % (2*n+1) == 0) && (omega(x) == 2), print1(n, ", ")); \\ _Michel Marcus_, Mar 13 2015
%Y Cf. A239638, A085724, A072226, A005384, A005097, A081858, A014601, A014664, A001917, A115591.
%K nonn,more,hard
%O 1,1
%A _Eric Chen_, Mar 13 2015
