cp's OEIS Frontend

The "1/k" heuristic predicts that primes of the form k*2^m + 1 with k odd and m > 0 have almost a 1/k chance of being Fermat divisors (Dubner and Keller). This sequence yields a correction to the "1/k" heuristic, because it generates special values of k.

If:

1) k is of the form 3*a^2, where a is an odd positive integer not divisible by 3,

2) k is not a Sierpiński number,

3) for all odd positive integers m the numbers k*2^m + 1 are composite,

then the probability that a Fermat number is divisible by a prime of the form k*2^m + 1 equal to 0.

Every term meets the first and third condition. For any n, at least one of the primes from A361898 (except 3) divides every integer in the sequence a(n)*2^m + 1 with m odd.

What is the smallest odd integer k such that every prime of the form k*2^m + 1 (m > 0) does not divide any Fermat number?

Supersequence of A351332. Thus every prime congruent to 1 mod 3 that divides a Fermat number is in this sequence.

Every Fermat number that is a semiprime has a prime of this form as a factor.

Let p be a prime number of the form 3*153479820268467961^2*2^k + 1 with k > 0, then the multiplicative order of 2 modulo p is not of the form 2^(m+1), m >= 0. Hence, p does not divide any Fermat number F(m) = 2^(2^m) + 1.