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%I #9 Feb 07 2021 22:53:50
%S 113,499,2081,2287,5807,6151,7823,9203,9629,11069,11497,13463,16987,
%T 17891,18049,19889,24091,26981,27259,27953,28319,28597,31219,35899,
%U 39047,41381,41603,43403,44839,45343,49529,50753,50857,55079,60793,62219,66721,72679,76771
%N Primes p such that (p^128 + 1)/2 is prime.
%C Expressions of the form m^j + 1 can be factored (e.g., m^3 + 1 = (m + 1)*(m^2 - m + 1)) for any positive integer j except when j is a power of 2, so (p^j + 1)/2 for prime p cannot be prime unless j is a power of 2. A005383, A048161, A176116, A340480, A341210, A341224, A341229, and this sequence list primes of the form (p^j + 1)/2 for j=2^0=1, j=2^1=2, ..., j=2^7=128, respectively.
%H Jon E. Schoenfield, <a href="/A341230/b341230.txt">Table of n, a(n) for n = 1..1000</a>
%e (3^128 + 1)/2 = 5895092288869291585760436430706259332839105796137920554548481 = 257*275201*138424618868737*3913786281514524929*153849834853910661121, so 3 is not a term.
%e (113^128 + 1)/2 = 3111793506...0421698561 (a 263-digit number) is prime, so 113 is a term. Since 113 is the smallest prime p such that (p^128 + 1)/2 is prime, it is a(1) and is also A341211(7).
%o (PARI) isok(p) = (p>2) && isprime(p) && ispseudoprime((p^128 + 1)/2); \\ _Michel Marcus_, Feb 07 2021
%Y Primes p such that (p^(2^k) + 1)/2 is prime: A005383 (k=0), A048161 (k=1), A176116 (k=2), A340480 (k=3), A341210 (k=4), A341224 (k=5), A341229 (k=6), (this sequence) (k=7).
%Y Cf. A341211 (Smallest prime p such that (p^(2^n) + 1)/2 is prime).
%K nonn
%O 1,1
%A _Jon E. Schoenfield_, Feb 07 2021
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