A111931 Smallest prime p such that 1/2, 2/3, 3/4, ..., (m-1)/m are n-th power non-residues modulo p for maximum possible m (=A000236(n)).

11, 67, 24077, 29041891, 33699452071

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 n-th power residue class (in other words, there is no n-th 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