

A111931


Smallest prime p such that 1/2, 2/3, 3/4, ..., (m1)/m are nth power nonresidues modulo p for maximum possible m (=A000236(n)).


1




OFFSET

2,1


COMMENTS

A000236(n) is the maximum length of a run of consecutive residues modulo prime p, starting with 1, where no two adjacent elements belong to the same nth power residue class (in other words, there is no nth power residue modulo p in the sequence of ratios 1/2, 2/3, ..., (A000236(n)1)/A000236(n)). a(n) equals the smallest p admitting a run of maximum length A000236(n).


LINKS

Table of n, a(n) for n=2..6.


EXAMPLE

a(2)=11 since A000236(2)=3 and 1/2=6, 2/3=8 are nonsquares modulo 11, and there is no smaller prime modulo which 1/2 and 2/3 are nonsquares.


CROSSREFS

Cf. A000236, A000445.
Sequence in context: A120792 A228032 A145833 * A066433 A038741 A292490
Adjacent sequences: A111928 A111929 A111930 * A111932 A111933 A111934


KEYWORD

hard,nonn,more


AUTHOR

Max Alekseyev, Aug 21 2005


EXTENSIONS

a(6) from Don Reble, added by Max Alekseyev, Sep 03 2017


STATUS

approved



