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Discriminant = 60. Class number = 4. Binary quadratic forms a*x^2+b*x*y+c*y^2 have discriminant d=b^2-4ac and gcd(a,b,c)=1 if they are primitive.

The Pell form X^2 - 15*Y^2 represents the negative primes -a(n), for n >= 1. - Wolfdieter Lang, Nov 28 2024

Nonnegative numbers of the form 8x^2 - y^2. - Jon E. Schoenfield, Jun 03 2022

Positive integers n such that x^2 - 99 y^2 + n = 0 has integer solutions. - Robert Israel, Oct 22 2024

Discriminant 60.

Nonnegative numbers of the form 5x^2 - 3y^2. - Jon E. Schoenfield, Jun 03 2022

From Klaus Purath, Jul 26 2023: (Start)

Nonnegative integers k such that 3x^2 - 5y^2 + k = 0 has integer solutions.

Also nonnegative integers of the form 2x^2 + (4m+2)xy + (2m^2+2m-7)y^2 for integers m. This includes the form in the name with m = 1.

Also nonnegative integers of the form 5x^2 + 10mxy + (5m^2-3)y^2 for integers m. This includes the form from Jon E. Schoenfield above with m = 0.

There are no squares in this sequence. Even powers of terms as well as products of an even number of terms belong to A243188.

Odd powers of terms as well as products of an odd number of terms belong to the sequence. This can be proved with respect to the form 5x^2 - 3y^2 by the following identity: (na^2 - kb^2)(nc^2 - kd^2)(ne^2 - kf^2) = n[a(nce + kdf) + bk(cf + de)]^2 - k[na(cf + de) + b(nce + kdf)]^2 for all a, b, c, d, e, f, k, n in R. This can be verified by expanding both sides of the equation.

(End)

Discriminant 60.

Also: nonnegative 3x^2-5y^2 since 3y^2+6xy-2x^2 = 3(y+x)^2-5x^2. - R. J. Mathar, Jun 10 2020

This is a subsequence of A237606. There the uninteresting numbers that have improper representations are also recorded.

The primes in the sequence are given in A141302.

A primitive indefinite form F(a,b,c;x,y) = a*x^2 + b*x*y + c*y^2, or [a, b, c] with gcd(a, b, c) = 1 and even discriminant Disc = b^2 - 4*a*c = 60 = 4*D, D = 15, has class number A307359(12) = 4. The four reduced 2-cycle forms are the principal cycle CR = {[1, 6, -6], [6, 6,-1]}, CRhat = {[-1, 6, 6], [-6, 6,1]}, (outer signs flipped), Cy ={[2, 6, -3], [-3, 6, 2]} and Cyhat ={[-2, 6, 3], [3, 6, -2]}.

A proper representation of an integer k (not 0) by such a form F is determined by the rpapfs (representative parallel primitive forms) FPa(k, j) = [k, 2*j, (j^2 - 15)/k], where j from {0, 1, ...,|k|-1} is determined by the congruence j^2 - 15 = = 0 (mod |k|).

The equivalence transformations R(t) of a form F = [a, b, c] is [c , -b +2*c*t, 1 - b*t + c*t^2]. This corresponds to R(t) = Matrix([0, -1], [1, t]). Half-reduced R-transformations use the choice t = ceiling((8 + b)/(2*c) - 1), if c > 0, and t = floor(1 - (8 + b)/(2*|c|)) if c < 0. (c = 0 is not considered because Disc becomes a square).

Because any form F of Disc = 60 represents a negative integer -k if it is equivalent to one of the rpapfs FPa(-k, j), the allowed values are

k = 2^{e_2}*3^{e_3}*5^{e_5}*Product_{i=1..P} p_i^{e_j}, where p_i is an odd prime >= 7 from the sequence A097956 or A038887(n), n >= 4, the p with Legendre(15, p) = +1. The exponents for 2, 3, and 5 are from {0, 1} (these primes are not liftable to powers) and e_i >= 0 (p_i is uniquely liftable to powers, see the Apostol reference), but not all exponents should be 0, because -1 is not represented. The number of infinite families of proper solutions (x, y), with positive values y, is 2^(P).

The present sequence is a proper subset of these generally allowed k values. One has to check if the rpapfs Fpa(-k, j) reach the principal cycle CR, then if so k is a member of the present sequence. This is because the Pell form FPell = [1, 0, -15] reaches (taking t to be first 0 then 3) the cycle member CR(1) = [1, 6, -6], the reduced principal form.

For details see the W. Lang paper in the links.

For the fundamental proper positive solutions of the infinite families for - a(n) see A378711. Note that -a(n) may also have improper solutions besides the proper ones whenever even powers of primes satisfying Legendre(15, p) = +1 appear, e.g., the first instance being -294 = -2*3*7^2).