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A092506
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Prime numbers of the form 2^n + 1.
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47
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OFFSET
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1,1
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COMMENTS
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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
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
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
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LINKS
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MATHEMATICA
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PROG
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(GAP) Filtered(List([1..20], n->2^n+1), IsPrime); # Muniru A Asiru, Oct 25 2018
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CROSSREFS
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A019434 is the main entry for these numbers.
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KEYWORD
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nonn,hard
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AUTHOR
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STATUS
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approved
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