A141158 Duplicate of A038872.

5, 11, 19, 29, 31, 41, 59, 61, 71, 79, 89, 101, 109, 131, 139, 149, 151, 179, 181, 191, 199, 211, 229, 239, 241, 251, 269, 271, 281, 311, 331, 349, 359, 379, 389, 401, 409, 419, 421, 431, 439, 449, 461, 479, 491, 499, 509, 521, 541, 569, 571, 599, 601, 619

Offset

1,1

Comments

Original name was: Primes of the form x^2 + 4*x*y - y^2.

Discriminant = 20. Class number = 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 (primitive).

Values of the quadratic form are {0, 1, 4} mod 5, so this is a subsequence of A038872. - R. J. Mathar, Jul 30 2008

Is this the same sequence as A038872? [Yes. See a comment in A038872, and the comment by Jianing Song below. - Wolfdieter Lang, Jun 19 2019]

Also primes of the form u^2 - 5v^2. The transformation {u,v}={x+2y,y} transforms it into the one in the title. - Tito Piezas III, Dec 28 2008

From Jianing Song, Sep 20 2018: (Start)

Yes, this is a duplicate of A038872. For primes p congruent to {1, 4} mod 5, they split in the ring Z[(1+sqrt(5))/2]. Since Z[(1+sqrt(5))/2] is a UFD, they are reducible in Z[(1+sqrt(5))/2], so we have p = e*((a + b*sqrt(5))/2)*((a - b*sqrt(5))/2), where a and b have the same parity and e = +-1. WLOG we can suppose e = 1, otherwise substitute a, b by (a+5*b)/2 and (a+b)/2. Now we show that there exists integer u, v such that p = (u + v*sqrt(5))*(u - v*sqrt(5)) = u^2 - 5*v^2.

(i) If u, v are both even, then choose u = a/2, v = b/2.

(ii) If u, v are both odd, 4 | (a-b), then choose u = (3*a+5*b)/4, v = (3*b+a)/4.

(iii) If u, v are both odd, 4 | (a+b), then choose u = (3*a-5*b)/4, v = (3*b-a)/4.

Hence every prime congruent to {1, 4} mod 5 is of the form u^2 - 5*v^2. On the other hand, u^2 - 5*v^2 == 0, 1, 4 (mod 5). So these two sequences are the same.

Also primes of the form x^2 - x*y - y^2 (discriminant 5) with 0 <= x <= y (or x^2 + x*y - y^2 with x, y nonnegative). (End) [Comment revised by Jianing Song, Feb 24 2021]

Links

Table of n, a(n) for n=1..54.

Example

a(3) = 19 because we can write 19 = 2^2 + 4*2*5 - 5^2.

Mathematica

lim = 25; Select[Union[Flatten[Table[x^2 + 4 x y - y^2, {x, 0, lim}, {y, 0, lim}]]], # > 0 && # < lim^2 && PrimeQ[#] &] (* T. D. Noe, Aug 31 2012 *)

Crossrefs

Sequence in context: A089270 A275068 A038872 * A239732 A130828 A244241

Adjacent sequences: A141155 A141156 A141157 * A141159 A141160 A141161

Keyword

dead

Author

Laura Caballero Fernandez, Lourdes Calvo Moguer, Maria Josefa Cano Marquez, Oscar Jesus Falcon Ganfornina and Sergio Garrido Morales (oscfalgan(AT)yahoo.es), Jun 12 2008

Status

approved