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A118915
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Primes p such that (p^2+5)/6 is prime.
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6
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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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,1
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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. [From Michael B. Porter (michael_b_porter(AT)yahoo.com), Apr 14 2010]
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PROG
| (PARI) isA118915(n)=if(n^2%6==1, isprime(n)&&isprime((n^2+5)/6), 0) [From Michael B. Porter (michael_b_porter(AT)yahoo.com), Apr 14 2010]
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CROSSREFS
| Cf. A109953 (primes p such that (p^2+1)/3 is prime), A118918 (primes p such that (p^2+11)/12 is prime).
Sequence in context: A156111 A162174 A171603 * A084442 A023270 A103804
Adjacent sequences: A118912 A118913 A118914 * A118916 A118917 A118918
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KEYWORD
| nonn
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AUTHOR
| T. D. Noe (noe(AT)sspectra.com), May 05 2006
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