A248576 - OEIS

%I #8 Oct 09 2014 03:58:12

%S 2,5,3,17,2,5,2,17,3,5,2,17,2,5,3,769,2,5,2,17,3,5,2,17,2,13,3,17,2,5,

%T 2,193,3,17,2,17,2,5,3,17,2,5,2,17,3,5,2,97,2,5,3,17,2,5,11,17,3,5,2,

%U 17,2,5,3,257,2,5,2,17,3,5,2,17,2,37,3,17,2,13,2,769,3,5,2

%N Least prime p such that m^n+1 is divisible by p^2, where m = min{ b>0 | b^n+1 not squarefree} = A248214(n).

%C See the main entry A248214 for all further information.

%o (PARI) a(n,bound=b->n*b*20)=for(b=1,9e9,forprime(p=1,bound(b),Mod(b,p^2)^n+1||return(p))) \\ The given default bound is experimental. You may use, e.g., a(n,b->10^5) for a fixed bound. Especially for n = 2^k >= 32, there might be a larger p leading to a smaller b, than the one found with this bound.

%Y Cf. A248214.

%K nonn

%O 1,1

%A _M. F. Hasler_, Oct 08 2014