A186283 Least number k such that k*n+1 is a prime dividing 2^n-1.

1, 2, 1, 6, 1, 18, 2, 8, 1, 2, 1, 630, 3, 2, 1, 7710, 1, 27594, 2, 6, 1, 2, 10, 24, 105, 9728, 1, 8, 1, 69273666, 8, 18166, 1285, 2, 1, 6, 4599, 2, 1, 326, 1, 10, 2, 14, 1, 50, 2, 90462791808, 5, 2, 1, 120, 1615, 16, 2, 568, 1, 3050, 1, 37800705069076950, 11545611, 2, 4, 126, 1, 2891160, 2, 145690999102, 1

Offset

2,2

Comments

The smallest prime factor of 2^n-1 of the form k*n+1 is A186522(n).

By Fermat's little theorem, a(n) = 1 if and only if n+1 is an odd prime. Further, for prime p, a(p) = 2 if and only if p is in A002515. - Thomas Ordowski, Sep 03 2017

References

Kenneth H. Rosen, Elementary Number Theory and Its Applications, 3rd Ed, Theorem 6.12, p. 225

Links

Max Alekseyev, Table of n, a(n) for n = 2..1236 (terms to a(300) from David A. Corneth)

Will Edgington, Mersenne Page [from Internet Archive Wayback Machine]

Eric W. Weisstein, MathWorld: Mersenne Number

Eric W. Weisstein, MathWorld: Mersenne Prime

Formula

a(n) = (A186522(n)-1)/n.

Example

For n=8, 2^n-1 = 255 = 3 * 5 * 17.  The smallest prime factor of the form k*n+1 is 17 = 2*8+1. Hence, a(8) = 2.

Mathematica

Table[p=First/@FactorInteger[2^n-1]; (Select[p, Mod[#1, n] == 1 &, 1][[1]] - 1)/n, {n, 2, 70}]

Prog

(PARI) a(n) = {if(isprime(n+1), return(1)); my(f = factor(2^n - 1)[, 1]); for(i=1, #f, if(f[i]%n == 1, return((f[i]-1) / n)))} \\ David A. Corneth, Sep 03 2017

Crossrefs

Cf. A000225, A049479, A079324.

Sequence in context: A225769 A280736 A279095 * A307374 A173279 A277440

Adjacent sequences: A186280 A186281 A186282 * A186284 A186285 A186286

Keyword

nonn

Author

Bill McEachen, Feb 16 2011

Status

approved