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Primes p such that (p^2+2)/33 is prime.
3

%I #7 Nov 21 2013 12:49:42

%S 19,47,107,151,173,179,349,487,547,569,641,701,883,971,1009,1097,1213,

%T 1361,1493,1559,1873,1889,1933,2269,2351,2357,2423,2797,2819,2879,

%U 3259,3347,3391,3457,3539,3583,4051,4139,4177,4799,4969,5437,6091,6163,6427

%N Primes p such that (p^2+2)/33 is prime.

%C For (p^2+2)/33 to be an integer, p must be congruent to 8, 14, 19, or 25 (mod 33). Examples: 107 = 8, 47 = 14, 19 = 19, and 487 = 25 (mod 33). [From _Michael B. Porter_, Oct 20 2009]

%H Harvey P. Dale, <a href="/A165672/b165672.txt">Table of n, a(n) for n = 1..1000</a>

%e p=19: (19^2+2)/33=11; p=47: (47^2+2)/33=67; p=107: (107^2+2)/33=347

%t Select[Prime[Range[1000]],PrimeQ[(#^2+2)/33]&] (* _Harvey P. Dale_, Feb 18 2012 *)

%Y Cf. A165671, A165673

%K nonn,less

%O 1,1

%A _Vincenzo Librandi_, Sep 24 2009

%E Corrected and extended by _Charles R Greathouse IV_, Oct 09 2009