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%I #16 Jun 06 2021 20:11:41
%S 5,11,13,29,43,53,283,557,563,613,691,2731,5147,5323,9533,10771,
%T 221891,235099,305867,311027,333227,792061,1347781,1669219,1882787,
%U 2305781
%N Primes p such that (2^p - 2^((p+1)/2) + 1)/5 is prime.
%C PrimePi[ a(n) ] = {3, 5, 6, 10, 14, 16, 61, 102, 103, 112, 125, 399, 686, 705, 1180, 1312, 19768, 20843, 26482, 26882, 28656, ...}. (2^p - 2^((p+1)/2) + 1) is the Aurifeuillan cofactor of 4^p + 1, where p is odd prime. All a(n) belong to A124112(n) = {5, 7, 9, 11, 13, 17, 29, 43, 53, 89, 283, 557, 563, 613, 691, 1223, 2731, ...} Numbers n such that ((1+I)^n+1)/(2+I) is a Gaussian prime. 5 largest currently known terms found by Jean Penne in Nov 2006: {221891, 235099, 305867, 311027, 333227}.
%H Henri Lifchitz and Renaud Lifchitz: <a href="http://www.primenumbers.net/prptop/prptop.php">PRP Records. Probable Primes Top 10000</a>
%t Do[p=Prime[n];f=(2^p-2^((p+1)/2)+1)/5;If[PrimeQ[f],Print[{PrimePi[p],p}]],{n,1,28656}]
%o (PARI) is(p)=isprime(p)&&ispseudoprime((2^p - 2^((p+1)/2) + 1)/5) \\ _Charles R Greathouse IV_, May 15 2013
%Y Cf. A124165 (primes p such that (2^p + 2^((p+1)/2) + 1)/5 is prime).
%Y Cf. A124112 (numbers n such that ((1+i)^n+1)/(2+i) is a Gaussian prime).
%K hard,more,nonn
%O 1,1
%A _Alexander Adamchuk_, Dec 04 2006
%E a(23-25) = 1347781, 1669219, 1882787 were found by Borys Jaworski between 2008 and 2012 (see the PRP Records link). - _Alexander Adamchuk_, Nov 27 2008
%E a(22) = 792061 was found out-of-sequence by Thomas Ritschel in March of 2014 (see the PRP Records link). - _Serge Batalov_, Mar 31 2014
%E a(26) = 2305781 from _Serge Batalov_, Mar 31 2014
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