OFFSET
1,1
COMMENTS
Equivalently primes of the form 2n^2 - 2n - 1. a(n)==3 (mod 4).
Equivalently primes p such that 2p+3 is square.
Also 3 followed by primes p of the form 2*n^2+6*n+3 = 2*(n+2)^2-2*(n+2)-1 (see the first comment) such that 2^(p-1)+3 is not prime. - Vincenzo Librandi, Jan 03 2009; M. F. Hasler, Jan 07 2009; R. J. Mathar, Jan 14 2009; Bruno Berselli, Sep 23 2013
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..1000
FORMULA
a(n) = (A109367(n) - 3)/2.
MATHEMATICA
Select[Table[Prime[n], {n, 3500}], IntegerQ[(2# + 3)^(1/2)] &] (* Ray Chandler, Oct 26 2004 *)
PROG
(Magma) [3] cat [ p: p in PrimesUpTo(30100) | exists(t){ n: n in [1..Isqrt(p div 2)] | 2*n^2+6*n+3 eq p } and not IsPrime(2^(p-1)+3) ];
CROSSREFS
KEYWORD
nonn
AUTHOR
Giovanni Teofilatto, Oct 09 2004
EXTENSIONS
Corrected by Ray Chandler, Oct 26 2004
Edited by N. J. A. Sloane, Jan 25 2009
STATUS
approved