OFFSET
1,4
COMMENTS
Note: k must be of the form 2^m (see A058064 for the m values).
Conjecture: For every pair of relatively prime numbers (x, y) there exist at least one number n=2^m and one prime number p such that p = x^n + y^n. This sequence shows one case of this conjecture where y = x + 1. - Tomas Xordan, Jun 02 2007
EXAMPLE
a(101)=16 because 101^16 + 102^16 = 254536435001431070450581794495937.
MATHEMATICA
Do[ k = 0; While[ !PrimeQ[ (n + 1)^(2^k) + n^(2^k) ], k++ ]; Print[ 2^k ], {n, 1, 60} ].
PROG
(PARI) a(n) = my(k=1); while (!isprime(p=(n+1)^k + n^k), k++); k; \\ Michel Marcus, Sep 16 2018
CROSSREFS
KEYWORD
nonn,hard
AUTHOR
Robert G. Wilson v, Nov 14 2000
STATUS
approved