OFFSET
1,1
COMMENTS
Original name was: Primes of the form 3*x^2 + 2*x*y - 4*y^2 (as well as of the form 3*x^2 + 8*x*y + y^2).
Discriminant = 52. Class = 1. Binary quadratic forms a*x^2+b*x*y+c*y^2 have discriminant d = b^2 - 4ac and gcd(a,b,c) = 1.
From Tito Piezas III, Dec 28 2008: (Start)
Primes of the form u^2 - 13v^2. Using the transformation {u,v} = {4x+y,x} yields the second quadratic form in the title.
This is probably identical to A038883.
(End)
From Jianing Song, Sep 20 2018: (Start)
Yes, this is a duplicate of A038883. For primes p congruent to {1, 3, 4, 9, 10, 12} mod 13, they split in the quadratic field Q(sqrt(13)). Since Z[(1+sqrt(13))/2], the ring of integers of Q(sqrt(13)), is a UFD, they are reducible in Z[(1+sqrt(13))/2], so we have p = +-((u' + v'*sqrt(13))/2)*((u' - v'*sqrt(13))/2) = +-(u'^2 - 13*v'^2)/4, where u', v' have the same parity. Since there are elements in Z[(1+sqrt(13))/2] with norm -1 (for example, (3+sqrt(13))/2), we can suppose that p = (u'^2 - 13*v'^2)/4. If u', v' are even, then pick u = u'/2, v = v'/2 and we have p = u^2 - 13*v^2; if u', v' are odd, pick u = (11*u'+-39*v')/4, v = (3*u'+-11*v')/4 by choosing signs appropriately so that u, v are integers.
On the other hand, u^2 - 13*v^2 == 0, 1, 3, 4, 9, 10, 12 (mod 13). So these two sequences are the same.
Also primes of the form x^2 - x*y - 3*y^2 (discriminant 13) with 0 <= x <= y (or x^2 + x*y - 3*y^2 with x, y nonnegative). (End) [Comment partly rewritten by Jianing Song, Oct 11 2022]
EXAMPLE
a(9) = 79 because we can write 79 = 3*5^2 + 2*5*2 - 4*2^2 (or 79 = 3*3^2 + 8*3*2 + 2^2).
MATHEMATICA
Reap[For[p = 2, p < 1000, p = NextPrime[p], If[FindInstance[p == 3*x^2 + 2*x*y - 4*y^2, {x, y}, Integers, 1] =!= {}, Print[p]; Sow[p]]]][[2, 1]] (* Jean-François Alcover, Oct 25 2016 *)
CROSSREFS
KEYWORD
dead
AUTHOR
Laura Caballero Fernandez, Lourdes Calvo Moguer, Maria Josefa Cano Marquez, Oscar Jesus Falcon Ganfornina and Sergio Garrido Morales (sergarmor(AT)yahoo.es), Jun 12 2008
STATUS
approved