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a(n) is the smallest prime p such that p^2 - 1 has 2*n divisors, or -1 if no such prime exists.
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%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