A248576 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).

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

Offset: 1

Author

M. F. Hasler, Oct 08 2014

Keywords

• nonn

Comments

See the main entry A248214 for all further information.

Crossrefs

Cf. A248214.

Programs

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.