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

Author

Bill McEachen, Feb 16 2011

Keywords

• nonn

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

Examples

			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.
		

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

Crossrefs

Cf. A000225, A049479, A079324.

Programs

Mathematica

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

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

Formula

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