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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^(m-2) divides (m-1)! and m divides (m-1)! + 1. - Thomas Ordowski, Nov 25 2014
From Jaroslav Krizek, Mar 06 2016: (Start)
Also primes p such that sigma(p-1) = 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(x-1)=floor((3x-2)/2). - Benoit Cloitre, Mar 02 2018
For n > 1, if 2^n + 1 divides 3^(2^(n-1)) + 1, then 2^n + 1 is a prime. - Jinyuan Wang, Oct 13 2018
The prime numbers occurring in A003401. Also, the prime numbers dividing at least one term of A003401. - Jeppe Stig Nielsen, Jul 24 2019
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