A171749 Odd primes of the form (1+n)*(2+2*n)+n*(3+2*n) = 4*n^2+7*n+2.

13, 59, 137, 389, 563, 769, 1277, 1579, 1913, 2677, 5147, 5737, 6359, 7013, 7699, 9949, 12487, 13397, 15313, 16319, 18427, 20663, 23027, 26813, 32309, 36767, 38317, 41513, 43159, 51869, 61379, 63377, 65407, 73847, 78259, 80513, 82799, 89849

Offset: 1

Author

Vladimir Joseph Stephan Orlovsky, Dec 17 2009

Keywords

• nonn
• easy

Comments

This sequence is infinite under the Bunyakovsky conjecture. - Charles R Greathouse IV, Apr 04 2012

Also primes of the form 16*m^2-2*m-1, by the substitution n=2*m-1. [Note that n is odd because otherwise 4n^2+7n+2 is even]. - Bruno Berselli, Jul 03 2012

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A171748.

Programs

Mathematica

f[n_] := (1+n)(2+2*n)+n*(3+2*n); lst={}; Do[If[PrimeQ[f[n]], AppendTo[lst, f[n]]], {n, 6!}]; lst
Select[Table[4*n^2+7*n+2,{n, 1000}],PrimeQ] (* Vincenzo Librandi, Aug 01 2012 *)