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%I #11 Jul 19 2017 18:00:00
%S 2,7,11,17,29,37,43,53,73,79,83,97,137,191,233,251,263,269,271,277,
%T 281,359,379,389,433,461,479,521,541,577,601,631,647,677,691,719,739,
%U 827,829,863,881,929,947,983,997,1033,1063,1087,1109,1187,1223
%N Primes p such that p^2+2 is a semiprime.
%C Cf. A048161 Primes p such that p^2+1 is a semiprime.
%C Primes p such that (p^2+2)/3 is prime. For all primes q>3, we have q=6k+-1 for some k, which makes it easy to show that 3 divides q^2+2. Hence if q^2+2 is a semiprime then (q^2+2)/3 must be prime. - _T. D. Noe_, May 05 2006
%H Harvey P. Dale, <a href="/A109953/b109953.txt">Table of n, a(n) for n = 1..2500</a>
%F a(n) = sqrt(3*A289135(n) - 2). See the T. D. Noe comment above. - _Wolfdieter Lang_, Jul 19 2017
%e a(2) = 7 is o.k. because 7^2+2=51=3*17 (semiprime), and 17 = A289135(2).
%t A109953=Select[Prime[Range[200]], Plus@@Last/@FactorInteger[ #^2+2]==2&]
%t Select[Prime[Range[200]],PrimeOmega[#^2+2]==2&] (* _Harvey P. Dale_, Nov 19 2011 *)
%Y Cf. A048161, A289135.
%Y Cf. A118915 (primes p such that (p^2+5)/6 is prime).
%K nonn
%O 1,1
%A _Zak Seidov_, Jul 06 2005
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