A141188 Duplicate of A038883.

3, 13, 17, 23, 29, 43, 53, 61, 79, 101, 103, 107, 113, 127, 131, 139, 157, 173, 179, 181, 191, 199, 211, 233, 251, 257, 263, 269, 277, 283, 311, 313, 337, 347, 367, 373, 389, 419, 433, 439, 443, 467, 491, 503, 521, 523, 547, 563, 569, 571, 599, 601, 607, 641

Offset: 1

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

Keywords

• dead

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]

Examples

			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).
		

Programs

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 *)