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A352597 a(n) is the smallest k > 1 such that k^n + 1 has all prime divisors p == 1 (mod n).

%I A352597 #21 May 05 2022 04:41:45
%S A352597 2,2,6,2,10,6,28,2,18,10,22,6,52,14,60,2,102,36,190,20,756,66,46,18,
%T A352597 2550,26,2970,28,58,120,310,2,330,170,11550,6,148,570,156,140,82,2184,
%U A352597 172,88,3040020,184,282,42,7252,110,7548,312,106,1440,41800,42,11172
%N A352597 a(n) is the smallest k > 1 such that k^n + 1 has all prime divisors p == 1 (mod n).
%C A352597 Equivalently, a(n) is the smallest k > 1 such that for all divisors d of k^n + 1, d == 1 (mod n).
%C A352597 A298299 is a subsequence.
%C A352597 All terms in this sequence are even since for odd k the expression k^n + 1 is divisible by 2 which is not congruent to 1 (mod n) for any n > 1.
%C A352597 If n is odd, a(n)^n + 1 is divisible by a(n) + 1. Therefore, a(n) + 1 == 1 (mod n) and so n | a(n) for odd n.
%C A352597 Theorem: a(n) = 2 if and only if n is a power of 2.
%F A352597 a(2n) = A298299(n).
%e A352597 a(3) = 6 since 6^3 + 1 = 217 = 7 * 31 and both factors are congruent to 1 (mod 3).
%o A352597 (PARI) isok(k,n) = my(f=factor(k^n+1)); for (i=1, #f~, if (Mod(f[i,1], n) != 1, return(0))); return(1);
%o A352597 a(n) = my(k=2); while (!isok(k, n), k+=2); k; \\ _Michel Marcus_, Mar 22 2022
%Y A352597 Cf. A298076, A298299 (bisection), A298310, A298398.
%K A352597 nonn,hard
%O A352597 1,1
%A A352597 _Kevin P. Thompson_, Mar 21 2022