OFFSET
1,1
COMMENTS
Under the Bunyakovsky conjecture, a(n) exists for every n. [Charles R Greathouse IV, Dec 27 2011]
LINKS
Pierre CAMI, Table of n, a(n) for n = 1..700
EXAMPLE
2^5*(2^5-1) - 1 = 32*31 - 1 = 991 (prime) so for n=5 a(5)=2.
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 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Pierre CAMI, Jan 24 2005
STATUS
approved