cp's OEIS Frontend

Terms are odd: by Morehead's theorem, 3*2^(2*n) + 1 can never divide a Fermat number.

No other terms below 7516000.

Is this sequence the same as "Numbers k such that 3*2^k + 1 is a factor of a Fermat number 2^(2^m) + 1 for some m"? - Arkadiusz Wesolowski, Nov 13 2018

The last sentence of Morehead's paper is: "It is easy to show that composite numbers of the forms 2^kappa * 3 + 1, 2^kappa * 5 + 1 can not be factors of Fermat's numbers." [a proof is needed]. - Jeppe Stig Nielsen, Jul 23 2019

Any factor of a Fermat number 2^(2^m) + 1 of the form 3*2^k + 1 is prime if k < 2*m + 6. - Arkadiusz Wesolowski, Jun 12 2021

If, for any m >= 0, F(m) = 2^(2^m) + 1 has a prime factor p of the form 3*2^k + 1, then F(m)/p is congruent to 11 mod 30. - Arkadiusz Wesolowski, Jun 13 2021

A number k belongs to this sequence if and only if the order of 2 modulo p is not divisible by 3, where p is a prime of the form 3*2^k + 1 (see Golomb paper). - Arkadiusz Wesolowski, Jun 14 2021

No other terms below 5330000.

The reason all terms are odd is that if k is even, then 5*2^k + 1 == (-1)*(-1)^k + 1 = (-1)*1 + 1 = 0 (mod 3). So if k is even, then 3 divides 5*2^k + 1, and since 3 divides no other Fermat number than F_0=3 itself, we do not have a Fermat factor. - Jeppe Stig Nielsen, Jul 21 2019

18233956 belongs to this sequence, but its position is currently unknown. - Jeppe Stig Nielsen, Oct 05 2020

Fernando (Remark 5.2) shows that all terms are odd. - Jeppe Stig Nielsen, Jan 02 2025