OFFSET
2,1
COMMENTS
a(19) > 155*10^6.
For n == 2 (mod 3), k^n + k + 1 is divisible by k^2 + k + 1. Thus, for n > 2, if n == 2 (mod 3), a(n) = 0.
PROG
(PARI)
a(n)=if(n>2&&n==Mod(2, 3), return(0)); k=1; while(!ispseudoprime(k^n+k+1)||!ispseudoprime(k^n+k-1)||!ispseudoprime(k^n-k+1)||!ispseudoprime(k^n-k-1), k++); k
n=2; while(n<100, print1(a(n), ", "); n++)
CROSSREFS
KEYWORD
nonn
AUTHOR
Derek Orr, Oct 03 2014
EXTENSIONS
a(19)-a(27) from Jon E. Schoenfield, Oct 19 2014
a(28)-a(36) from Jon E. Schoenfield, Oct 22 2014
STATUS
approved