

A118915


Primes p such that (p^2 + 5)/6 is prime.


16



5, 13, 19, 23, 37, 41, 89, 113, 127, 131, 139, 149, 167, 197, 229, 233, 239, 251, 271, 359, 373, 401, 433, 449, 463, 503, 523, 541, 607, 631, 643, 653, 701, 719, 743, 769, 811, 827, 877, 881, 887, 919, 967, 971, 1009, 1013, 1021, 1093, 1097, 1283, 1301
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OFFSET

1,1


COMMENTS

For all primes q>3, we have q=6k+1 for some k, which makes it easy to show that 6 divides q^2+5.
(n^2+5)/6 is an integer for all primes except 2 and 3.  Michael B. Porter, Apr 14 2010


LINKS



PROG

(PARI) isA118915(n)=if(n^2%6==1, isprime(n)&&isprime((n^2+5)/6), 0) \\ Michael B. Porter, Apr 14 2010


CROSSREFS

Cf. A109953 (primes p such that (p^2+1)/3 is prime), A118918 (primes p such that (p^2+11)/12 is prime).


KEYWORD

nonn


AUTHOR



STATUS

approved



