cp's OEIS Frontend

From Jianing Song, Oct 13 2022: (Start)

Rational primes that decompose in the field Q(sqrt(15)).

Primes p such that kronecker(60,p) = 1.

Primes congruent to 1, 7, 11, 17, 43, 49, 53, 59 modulo 60. (End)

The number of (x, y) pairs in the rows are: 1, 2, 2, 1, 2, 2, 2, 2, 2, 2, 4, 2, 2, 2, 2, 2, 2, 2, 4, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 4, ...

For details on the general proper representations of a negative integer k by forms with discriminant Disc = 60 = 4*15 see A378710, with references. For the Pell case x^2 - 15*y^2 only a subset of these k values is permitted, namely those that have representative reduced primitive forms (rpapfs) Fpa(-k, j) (see A378710) equivalent to the principal reduced form CR(1) = [1, 6, -6], which is in turn equivalent to the Pell form FPell = [1, 0, -15].

Some rules for the represented -A378710(n) values are: the negative of the prime factors 2, 3 and 5 of 15 are not represented, they are equivalent to some of the other three 2-cycles forms. Powers of these three primes can never occur because they cannot be lifted (see the Apostol reference, Theorem 5.20, pp. 121-122). The products -2*3 and -3*5 are represented but not -2*5 (the rpapf [-10, 10, -1] is equivalent to [-1, 6, 6], a member of the 2-cycle called CRhat in A378710). -2*3*5 is also not represented ([-30, 30, -7] is equivalent to [2, 6, -3] from the cycle Chat).

The reservoir for the odd primes >= 7 is given by Legendre(15, p) = +1 (A097956 or A038887(n), n >= 4). These primes can be lifted uniquely, but one has to find out for each case, also for the product of powers of these primes (together with 2, 3, and 5 factors) if the rpapfs reach the fundamental cycle CR.

The number of infinite families of proper solutions for k = -A378710(n) with positive y is determined by 2^P(n), where P(n) is the number of primes >= 7 in A378710(n). These numbers 2^P(n) are given in the table below, and in the first comment.

The proper family of solutions {(x(n,i), y(n,i))}_{i = -infinity ... +infinity} are found from the rpapf(-A378710(n), j) = [-A378710(n), 2*j, (15 - j^2)/A378710(n)] with the help of the formula (x(n, i), y(n, i))^T = (+ or - B15)*(-Auto15)^i*Rtvalues(n,j)^(-1)*(1, 0)^T, for the solutions of j^2 - 15 == 0 (mod(A378710(n)), for j from 0, 1,.., A378710(n) - 1, (T for transpose) where B15 = R(0)*R(3) = -Matrix([1, 3], [0, 1]), Auto15 = R(-1)*R(6) = - Matrix([1, 6], [1, 7]). For the R(t)-transformation matrix see A378710(n). Rtvalues(n,j) is the product of R(t) matrices with the t-values leading from the rpapf(-A378710(n), j) to the form CR(1). The sign of B15 is chosen such that no negative values for y appear.

The powers (- Auto15)^i = Matrix([S(i, 8) - 7*S(i-1, 8), 6*S(i-1, 8)], [S(-i, 8), S(i, 8) - S(i-1, 8)]), with the Chebyshev polynomial S(i, 8) given, for i >= -1, in A001090(i) and S_(-i, 8) = -S(i-2, 8), for i >= 2.

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).