cp's OEIS Frontend

From Klaus Purath, Feb 17 2024: (Start)

Positive numbers of the form 15x^2 - y^2. The reduced form is -x^2 + 6xy + 6y^2.

Even powers of terms as well as products of an even number of terms belong to A243188. This can be proved with respect to the forms [a,0,-c] and [a, 0, +c] by the following identities: (au^2 - cv^2)(ax^2 - cy^2) = (aux + cvy)^2 - ac(uy + vx)^2 and (au^2 + cv^2)(ax^2 + cy^2) = (aux - cvy)^2 + ac(uy + vx)^2 for all a, c, u, v, x, y in R. This can be verified by expanding both sides of the equations. Generalization (conjecture): This multiplication rule applies to all sequences represented by any binary quadratic form [a, b, c].

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 forms [a,0,-c] and [a, 0, +c] by the following identities: (as^2 - ct^2)(au^2 - cv^2)(ax^2 - cy^2) = a[s(aux + cvy) + ct(uy + vx)]^2 - c[as(uy + vx) + t(aux + cvy)]^2 and (as^2 + ct^2)(au^2 + cv^2)(ax^2 + cy^2) = a[s(aux - cvy) - ct(uy + vx)]^2 + c[as(uy + vx) + t(aux - cvy)]^2 for all a, c, s, t, u, v, x, y in R. This can be verified by expanding both sides of the equations. Generalization (conjecture): This multiplication rule applies to all sequences represented by any binary quadratic form [a, b, c].

If we denote any term of this sequence by B and correspondingly of A243189 by C and of A243190 by D, then B*C = D, C*D = B and B*D = C. This can be proved by the following identities, where the sequence (B) is represented by [kn, 0, -1], (C) by [n, 0, -k] and (D) by [k, 0, -n].

Proof of B*C = D: (knu^2 - v^2)(nx^2 - ky^2) = k(nux + vy)^2 - n(kuy + vx)^2 for k, n, u, v, x, y in R.

Proof of C*D = B: (nu^2 - kv^2)(kx^2 - ny^2) = kn(ux + vy)^2 - (nuy + kvx)^2 for k, n, u, v, x, y in R.

Proof of B*D = C: (knu^2 - v^2)(kx^2 - ny^2) = n(kux + vy)^2 - k(nuy + vx)^2 for k, n, u, v, x, y in R. This can be verified by expanding both sides of the equations.

Generalization (conjecture): If there are three sequences of a given positive discriminant that are represented by the forms [a1, b1, c1], [a2, b2, c2] and [a1*a2, b3, c3] for a1, a2 != 1, then the BCD rules apply to these sequences. (End)

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).