%I #9 Mar 21 2021 13:57:34
%S 23,47,71,73,97,167,191,193,239,241,263,311,313,337,359,383,409,431,
%T 433,457,479,503,577,599,601,647,673,719,743,769,839,863,887,911,937,
%U 983,1009,1031,1033,1103,1129,1151,1153,1201,1223,1249,1297,1319,1321,1367
%N Primes p such that p divides 3^((p+1)/2) - 2^((p+1)/2) - 1.
%C Primes = 1 or 23 mod 24. Hence, together with 2, primes such that (2/p) = 1 = (3/p) where (k/p) is the Legendre symbol. - _Charles R Greathouse IV_, Apr 06 2012
%H Charles R Greathouse IV, <a href="/A130063/b130063.txt">Table of n, a(n) for n = 1..10000</a>
%t Select[ Range[2000], PrimeQ[ # ]&&Mod[ PowerMod[3,(#+1)/2,# ] - PowerMod[2,(#+1)/2,# ] - 1, # ]==0&]
%t Select[Prime[Range[250]],Divisible[3^((#+1)/2)-2^((#+1)/2)-1,#]&] (* _Harvey P. Dale_, Mar 21 2021 *)
%o (PARI) is(n)=(n+1)%24<3 && isprime(n) \\ _Charles R Greathouse IV_, Apr 06 2012
%Y Cf. A097934 = Primes p such that p divides 3^((p-1)/2) - 2^((p-1)/2).
%Y Subsequence of A038876.
%Y Cf. A127071, A127072, A127073, A127074. Cf. A130058, A130059, A130060, A130061, A130062.
%K nonn,easy
%O 1,1
%A _Alexander Adamchuk_, May 05 2007