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Discriminant = 76. Class = 2. Binary quadratic forms a*x^2 + b*x*y + c*y^2 have discriminant d = b^2 - 4ac.

The number of (x, y) pairs in row n is 1 for n = 1 and 2, and 2^P, with P = P1 + P7, where P1 and P7 are the number of prime factors 1 modulo 8 and 7 modulo 8, respectively, of A057126(n), for n >= 3.

See A057126 for comments concerning its representation by x^2 - 2*y^2.

The numbers A057126 are given by 2^e_2 * Product_{i=1..P1} p_{1,i}^e_{1,i} * Product_{j=1..P7} p_{7,j}^e_{7,j}, with the odd primes p_{1,i} and p_{7,j} congruent to 1 and 7 modulo 8, respectively. See A007519 and A007522 for these odd primes. Together with 2 these primes are given in A038873, and without 2 in A001132. The exponents are e_2 = 0 or 1, and e_{1,i} and e_{7,j} are nonnegative. The a(1) = 1 is obtained if all exponents vanish. For the proof see Lemma 18 of the linked W. Lang paper, pp. 22 - 23.

The general solutions are obtained from each fundamental solution by application of integer powers of the matrix Auto' = Mat([3,4], [2,3]). See the linked paper eq (28), p. 14, and eq. (40), p. 17 for D = 2, and k = A057126(n). For the explicit form of the powers of Auto' in terms of Chebyshev polynomials S(n, 6) = A001109(n+1) see there eq. (38), and Lemma 10, eq. (43), p. 17.

The conversion to the pair of proper solutions (X, Y) of X^2 - 2*Y^2 = A057126(n) is given by (X, Y) = (2*y - x, x - y). This may result in solutions with negative Y values. They are then transformed to the fundamental positive proper solutions via the mentioned matrix Auto'. See the right part of the example below. For this conversion see also the Nov 09 2009 comment in A035251 by Franklin T. Adams-Watters.

Previous name was: Primes of the form 3*x^2 + 5*x*y - 6*y^2 (as well as of the form 6*x^2 + 11*x*y + y^2).

Discriminant = 97. Class = 1. Binary quadratic forms a*x^2 + b*x*y + c*y^2 have discriminant d = b^2-4ac.

Is this the same as A038987? - R. J. Mathar, Oct 12 2013

From Don Reble, Oct 06 2014: (Start)

G. B. Mathews ("Theory of Numbers" by Chelsea publishing) might have an answer to the relation with A038987. In point 59 on page 65 he claims that

- if X is a non-residue of a discriminant of a quadratic form, then X is not representable; and

- if X is a residue of D, then there is a quadratic form of determinant D which represents X.

If all forms of discriminant 97 are equivalent, then that might suffice. (Indeed, either +97 or -97 has class number 1; but I am not sure which sign matters, A003656 vs. A003173.)

(End)

From Jianing Song, Feb 24 2021: (Start)

Also primes of the form u^2 + u*v - 24*v^2. Substitute u, v by u = 9*x+22*y, v = 2*x+5*y gives 3*x^2 + 5*x*y - 6*y^2.

Yes, this is the same as A038987. For primes p being a (coprime) square modulo 97, they split in the ring Z[(1+sqrt(97))/2]. Since Z[(1+sqrt(97))/2] is a UFD, they are reducible in Z[(1+sqrt(97))/2], so we have p = e*(u + v*(1+sqrt(97))/2)*(u + v*(1-sqrt(97))/2) = e*(u^2 + u*v - 24*v^2), e = +-1. WLOG we can suppose e = 1, otherwise substitute u, v by 5035*u+27312*v and 1138*u+6173*v, then p = u^2 + u*v - 24*v^2. On the other hand, if p is a quadratic nonresidue modulo 97, then they remain inert in Z[(1+sqrt(97))/2] and hence cannot be represented as u^2 + u*v - 24*v^2. (End)