1,1

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

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

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

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 *)

Cf. A101446.

Sequence in context: A056564 A082844 A279124 * A245219 A097509 A095206

Adjacent sequences: A101403 A101404 A101405 * A101407 A101408 A101409

nonn

Pierre CAMI, Jan 24 2005

approved