A101406 a(n) = least k such that k^n*(k^n-1)-1 is prime.

3, 2, 3, 2, 2, 3, 3, 19, 2, 2, 45, 7, 15, 7, 5, 5, 44, 2, 4, 3, 84, 62, 128, 5, 4, 90, 16, 37, 15, 11, 311, 15, 295, 72, 3, 3, 242, 2, 126, 64, 152, 11, 78, 26, 2, 13, 14, 26, 140, 2, 24, 16, 157, 4, 49, 13, 2, 123, 64, 16, 61, 206, 6, 76, 412, 31, 84, 23, 24, 9, 471, 26, 422, 227, 8

Offset: 1

Author

Pierre CAMI, Jan 24 2005

Keywords

• nonn

Comments

Under the Bunyakovsky conjecture, a(n) exists for every n. [Charles R Greathouse IV, Dec 27 2011]

Examples

			2^5*(2^5-1) - 1 = 32*31 - 1 = 991 (prime) so for n=5 a(5)=2.
		

Links

• Pierre CAMI, Table of n, a(n) for n = 1..700

Crossrefs

Cf. A101446.

Programs

Mathematica

a = {}; Do[ k = 1; While[c = k^n; t = c*(c - 1) - 1; ! PrimeQ[t], k++ ]; AppendTo[a, k];, {n, 75}]; a (* Ray Chandler, Jan 27 2005 *)