1,1

These numbers have been proved prime only up to exponent a[25] = 12583.

With the only exception of a[3] = 9, it is easy to prove that ((1+I)^a[n]+1)/(2+I) prime => a[n] prime. Following an idea of Harsh Aggarwal, many of these numbers have been discovered as by-products of the search for prime Gaussian-Mersenne norms. The reason of that is the Aurifeuillan factorization of M(k) = 2^(2k) + 1 with k odd. These numbers can be written as : M(k) = GM(k)*GQ(k)*5 GM(k) is the norm of the Gaussian-Mersenne (1+I)^k-1 while GQ(k) is the norm of ((1+I)^a[n]+1)/(2+I) This allowed us to write a program which can simultaneously prove the primality of GM(k) and, without extra cost, the probable primality of GQ(k). Using this program, Boris Jaworski, (discoverer of the presently largest known GM) discovered also an outlier of this sequence : a[?] = 1127239

The terms 1127239 and 1148729 were found by Borys Jaworski in 2006-2007 (see PRP Records link). These two terms also belong to A124165(n) = Primes p such that (2^p + 2^((p+1)/2) + 1)/5 is prime. a(n) is a union of the only composite term a(3) = 9 and two prime sequences: A124165(n) and A125742(n) = Primes p such that (2^p - 2^((p+1)/2) + 1)/5 is prime. - Alexander Adamchuk (alex(AT)kolmogorov.com), Jun 20 2007

Henri Lifchitz & Renaud Lifchitz : PRP Records. Probable Primes Top 10000.

For n = 27 : ((1+I)^36479+1)/(2+I) is a probable Gaussian prime because its norm : (2^36479+2^18240+1)/5 is a Fermat PRP

Cf. A124165 = Primes p such that (2^p + 2^((p+1)/2) + 1)/5 is prime. Cf. A125742 = Primes p such that (2^p - 2^((p+1)/2) + 1)/5 is prime.

Sequence in context: A081001 A075025 A075394 this_sequence A049758 A064077 A026282

Adjacent sequences: A124109 A124110 A124111 this_sequence A124113 A124114 A124115

more,nonn

David J. Broadhurst and Jean Penne (jpenne(AT)wanadoo.fr), Nov 27 2006