A109953 Primes p such that p^2+2 is a semiprime.

2, 7, 11, 17, 29, 37, 43, 53, 73, 79, 83, 97, 137, 191, 233, 251, 263, 269, 271, 277, 281, 359, 379, 389, 433, 461, 479, 521, 541, 577, 601, 631, 647, 677, 691, 719, 739, 827, 829, 863, 881, 929, 947, 983, 997, 1033, 1063, 1087, 1109, 1187, 1223

Offset: 1

Author

Zak Seidov, Jul 06 2005

Keywords

• nonn

Comments

Cf. A048161 Primes p such that p^2+1 is a semiprime.

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

Examples

			a(2) = 7 is o.k. because 7^2+2=51=3*17 (semiprime), and 17 = A289135(2).
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..2500

Crossrefs

Cf. A048161, A289135.

Cf. A118915 (primes p such that (p^2+5)/6 is prime).

Programs

Mathematica

A109953=Select[Prime[Range[200]], Plus@@Last/@FactorInteger[ #^2+2]==2&]
Select[Prime[Range[200]],PrimeOmega[#^2+2]==2&] (* Harvey P. Dale, Nov 19 2011 *)

Formula

a(n) = sqrt(3*A289135(n) - 2). See the T. D. Noe comment above. - Wolfdieter Lang, Jul 19 2017