cp's OEIS Frontend

From Giuseppe Coppoletta, May 16 2017: (Start)

All terms after a(52) refer to probabilistic primality tests for 2^a(n) + 1 (see Caldwell's link for the list of the largest certified Wagstaff primes). After a(58), 267017, 269987, 374321, 986191, 4031399 and 4101572 are also terms, but there still remains the remote possibility of some gaps in between. In addition, 13347311 and 13372531 are also terms, but possibly much farther in the numbering (see comments in A000978).

For the relation with Fermat numbers and for the primality of odd terms, see comments in A073936. The terms 9 and 10 give a value of 2^n + 1 which is not squarefree, so they are not in A073936. For the rest, the actually known terms of the two sequences coincide. In order to verify if any other term could be found hereafter that is not in A073936, all we have to do is to examine the terms for which 2^n + 1 is not squarefree. Considering that 3 divides 2^a(n) + 1 for any odd term a(n) and using Zsigmondy's and Mihăilescu-Catalan's theorems (see links), one can verify that any nonsquarefree term greater than 9 has to be of the form a(n) = 2^j * Fj, where Fj is the Fermat prime 2^2^j + 1. So basically we have to see if ((Fj-1)^Fj + 1)/(Fj)^2 is a prime or the power of a prime for any Fermat prime Fj. The case j = 1 gives the term a(n) = 10 because (4^5 + 1)/5^2 = 41 is a prime, while for j = 2, (16^17 + 1)/17^2 = 354689 * 2879347902817 is composite. Similarly (256^257 + 1)/257^2 is neither a prime nor the power of a prime, so there is no contribution from the cases j = 2, 3 (see also comments in A127317).

For j = 4 and for any possible other Fermat prime which could be found later, the question is still open, in the sense that it is not actually known if n = 16 * F4 = 1048592 is a term or not. That seems very unlikely, but in order to decide that question for j = 4, one would have to check if (2^1048592 + 1)/65537^2 is a prime or the power of a prime. As this number has 315649 digits, I wonder if it is possible to handle it with the existing primality tests.

(End)

For primes p = (2^a(n) + 3)/5, n >= 2, there are exactly 5 positive integers m for which the exponents of 2 and p in the prime power factorization of m! are both powers of 2 (for a proof see the Shevelev link (Theorem 2)).

The sequence of such primes p begins 7, 103, 6710887, 7036874417767, etc.

a(11) > 160000. - Michael S. Branicky, Jul 12 2024