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A126960
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Primes p such that (3p)^2 + 2 is prime.
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0
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3, 5, 7, 11, 13, 19, 37, 41, 73, 79, 83, 101, 103, 107, 139, 149, 151, 167, 191, 227, 233, 251, 269, 311, 337, 443, 457, 479, 499, 503, 521, 541, 601, 613, 647, 673, 761, 811, 829, 863, 877, 883, 887, 907, 919, 941, 983, 997
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,1
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COMMENTS
| A generalization of this would be primes p such that (kp)^2+2 is prime. Then k=3 is the only solution. This follows from the fact that k of the form 3m+/-1 will give 9m^2+6m+1+2 a multiple of 3.
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REFERENCES
| Garath A. Jones and Mary Jones, Elementary Number Theory, Springer - Verlag London, 1998; p. 35, Exercise 2.17.
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MATHEMATICA
| lst={}; Do[p=Prime@n; If[PrimeQ@((3*p)^2+2), AppendTo[lst, p]], {n, 6!}]; lst [From Vladimir Orlovsky (4vladimir(AT)gmail.com), Aug 11 2009]
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PROG
| (PARI) g(n) = forprime(x=0, n, y=9*x^2+2; if(isprime(y), print1(x", ")))
(MAGMA) [ p: p in PrimesUpTo(1000) | IsPrime((3*p)^2+2)] [From Vincenzo Librandi, Jan 29 2011]
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CROSSREFS
| Sequence in context: A040993 A078425 A020587 * A119753 A169969 A174350
Adjacent sequences: A126957 A126958 A126959 * A126961 A126962 A126963
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KEYWORD
| easy,nonn
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AUTHOR
| Cino Hilliard (hillcino368(AT)hotmail.com), Mar 19 2007
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EXTENSIONS
| Entries confirmed by Zak Seidov, Mar 19 2007
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