|
| |
|
|
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
|
|
|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,1
|
|
|
COMMENTS
|
See the main entry A248214 for all further information.
|
|
|
LINKS
|
|
|
|
PROG
|
(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.
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
