cp's OEIS Frontend

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 for this 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 where 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, Borys Jaworski (discoverer of the presently largest known GM) also discovered 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, Jun 20 2007

The term 12503723 is also in the sequence but its position is unknown. - Serge Batalov, Jul 17 2020

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}.

PrimePi[ a(n) ] = {3, 11, 16, 38, 97, 178,...}.

PrimePi[ a(n) ] = {5, 12, 15, 25, 39, 151, 307, 341, ...}.