2 together with the Fermat primes A019434.
Obviously if 2^n + 1 is a prime then either n = 0 or n is a power of 2.  N. J. A. Sloane, Apr 07 2004
Numbers m > 1 such that 2^(m2) divides (m1)! and m divides (m1)! + 1.  Thomas Ordowski, Nov 25 2014
From Jaroslav Krizek, Mar 06 2016: (Start)
Also primes p such that sigma(p1) = 2p  3.
Also primes of the form 2^n + 3*(1)^n  2 for n >= 0 because for odd n, 2^n  5 is divisible by 3.
Also primes of the form 2^n + 6*(1)^n  5 for n >= 0 because for odd n, 2^n  11 is divisible by 3.
Also primes of the form 2^n + 15*(1)^n  14 for n >= 0 because for odd n, 2^n  29 is divisible by 3. (End)
Exactly the set of primes p such that any number congruent to a primitive root (mod p) must have at least one prime divisor that is also congruent to a primitive root (mod p). See the links for a proof.  Rafay A. Ashary, Oct 13 2016
Conjecture: these are the only solutions to the equation A000010(x)+A000010(x1)=floor((3x2)/2).  Benoit Cloitre, Mar 02 2018
For n > 1, if 2^n + 1 divides 3^(2^(n1)) + 1, then 2^n + 1 is a prime.  Jinyuan Wang, Oct 13 2018
