%I #5 Dec 05 2022 08:36:42
%S 2,3,-1,5,7,-1,-1,11,17,23,-1,19,-1,31,73,29,-1,383,-1,41,97,-1,-1,79,
%T -1,-1,127,223,-1,71,-1,109,-1,-1,2593,197,-1,-1,-1,281,-1,1439,-1,
%U 34303,199,-1,-1,181,-1,647,-1,6143,-1,7057,-1,929,-1,-1,-1,521,-1
%N a(n) is the smallest prime p such that p^2 - 1 has 2*n divisors, or -1 if no such prime exists.
%e For p = 11, p^2 - 1 = 121 - 1 = 120 = 2^3 * 3 * 5 has 16 divisors. 11 is the smallest prime p such that p^2 - 1 has 16 = 2*8 divisors, so a(8) = 11.
%e There does not exist any prime p such that p^2 - 1 has 6 = 2*3 divisors, so a(3) = -1.
%Y Cf. A000005, A000040, A341655, A341658, A341660, A350780.
%K sign
%O 1,1
%A _Jon E. Schoenfield_, Dec 04 2022