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The row length of this irregular triangle is 2*A307372(n).

The indefinite binary quadratic Pell form is F = [1, 0, -D(n)], with D(n) = A000037(n) (D not a square). This form is not reduced (see the Buell or Scholz-Schoeneberg references, and the W. Lang link in A225953 for the definition).

The first reduced form, obtained after two equivalence transformations, is FR(n) = [1, 2*s(n), -(D(n) - s(n)^2)] where s(n) = A000194(n) = D(n) - n, for n >= 1. Hence FR(n) = [1, 2*s(n), -(n - s(n)*(s(n)-1))]. For the two transformations invoving R(t) = matrix([0, -1], [1, t]), first with t = 0 then with t = s(n) see a comment in A000194, and the proposition in the W. Lang link given below. FR(n) is the principal form F_p(n) of discriminant 4*D(n).

Each reduced form FR(n) leads to a cycle of reduced forms with (primitive) period P(n) = 2*p(n) = 2*A307372(n). The sequence of R-transformations is given by the parameter tuple (t_1(n), ..., t_{2*p(n)}(n)) with alternating signs which give the row entries T(n, k) = t_k(n). See also Table 2 of the W. Lang link.

The automorphic transformation is obtained by the matrix Auto(n) = R(t_1(n))*R(t_2(n))*...*R(t_{2*p(n)}(n)). Together with the matrix B(n) := R(0)*R(s(n)) = Matrix([-1, s(n)], [0, -1]) one finds all solutions of the Pell equation x^2 - D(n)*y^2 = +1. For each n >= 1 there is one family (also called class) of proper solutions. The general solution is (x(n;j), y(n;j))^T = B(n)*(Auto(n))^j*(1,0)^T, for integer j (T for transposed). One can always choose x >= 1 by an overall sign flip in x and y.

For the general Pell equation x^2 - D(n)*y^2 = N, for integer N, the parallel forms equivalent to FR(n) become important. For details see the W. Lang link given below, section 3.

The corresponding y values are given in A339882.

The Diophantine equation x^2 - p*y^2 = -2, of discriminant Disc = 4*p > 0 (indefinite binary quadratic form), with prime p can have proper solutions (gcd(x, y) = 1) only for primes p = 2 and p == 3 (mod 8) by parity arguments.

There are no improper solutions (with g >= 2, g^2 does not divide 2).

The prime p = 2 has just one infinite family of proper solutions with nonnegative x values. The fundamental proper solutions for p = 2 is (0, 1).

If a prime p congruent to 3 modulo 8, (p(n) = A007520(n)) has a solution then it can have only one infinite family of (proper) solutions with positive x value.

This family is self-conjugate (also called ambiguous, having with each solution (x, y) also (x, -y) as solution). This follows from the fact that there is only one representative parallel primitive form (rpapf), namely F_{pa(n)} = [-2, 2, -(p(n) - 1)/2].

The reduced principal form of Disc(n) = 4*p(n) is F_{p(n)} = [1, 2*s(n), -(p(n) - s(n)^2)], with s(n) = A000194(n'(n)), if p(n) = A000037(n'(n)). The corresponding (reduced) principal cycle has length L(n) = 2*A307372(n'(n)).

The number of all cycles, the class number, for Disc(n) is h(n'(n)) = A324252(n'(n)), Note that in the Buell reference, Table 2B in Appendix 2, p. 241, all Disc(n) <= 4*1051 = 4204 have class number 2, except for p = 443, 499, 659 (Disc = 1772, 1996, 26).

See the W. Lang link Table 1 for some principal reduced forms F_{p(n)} (there for p(n) in the D-column, and F_p is called FR(n)) with their t-tuples, giving the automporphic matrix Auto(n) = R(t_1) R(t_1) ... R(t_{L(n)}), where R(t) := Matrix([[0, -1], [1, t]]), and the length of the principal cycle L(n) given above, and in Table 2 for CR(n).

To prove the existence of a solution one would have to show that the rpapf F{pa(n)} is properly equivalent to the principal form F_{pa(n)}.

The length of row n is 2*A307236(n). This is the number of primitive reduced binary quadratic forms of discriminant 4*D(n), with D(n) = A000037(n).

The number of cycles in row n is A307359(n), the class number h(n) of binary quadratic forms of discriminant 4*D(n).

The principal cycle starts with F_p(n) = [1, 2*s(n), -(D(n) -s(n))^2], with s(n) = A000194(n). Its period is A307372(n). This is the only cycle (the class number is 1) for n = 1, 3, 10, 13, 24, ...

For class number h(n) >= 2 the cycles come mostly in pairs of cycles which can be transformed into each other by a sign flip operation on the outer entries of the forms of the cycle (called outer sign flip). Exceptions occur if cycles are identical with their outer sign flipped ones. This happens, e.g., for n = 7 with two cycles: one of length 2 (the principal cycle CR(2)) and one of length 6. This 6-cycle is also identical to the outer sign flipped one. See the example below.

See the Buell and Scholz-Schoeneberg references for cycles and class number, and also the W. Lang link given in A324251, with Table 2.