OFFSET
1,1
COMMENTS
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.
REFERENCES
D. A. Buell, Binary Quadratic Forms. Springer-Verlag, NY, 1989, p. 21.
A. Scholz and B. Schoeneberg, Einführung in die Zahlentheorie, 5. Aufl., de Gruyter, Berlin, New York, 1973, p. 112.
LINKS
FORMULA
T(n, k) = t_k(n), the k-th entry of the t-tuple for the R-transformations of the principal cycle for discriminant 4*D(n), with D(n) = A000037(n). See the comments above.
EXAMPLE
The irregular triangle T(n, k) begins:
n, D(n) \k 1 2 3 4 5 6 7 8 9 10 ... 2*A324252(n)
----------------------------------------------------------------------
1, 2: -2 2 2
2, 3: -1 2 2
3, 5: -4 4 2
4, 6: -2 4 2
5, 7: -1 1 -1 4 4
6, 8: -1 4 2
7, 10: -6 6 2
8, 11: -3 6 2
9, 12: -2 6 2
10, 13: -1 1 -1 1 -6 1 -1 1 -1 6 10
11, 14: -1 2 -1 6 4
12, 15: -1 6 2
13, 17: -8 8 2
14, 18: -4 8 2
15, 19: -2 1 -3 1 -2 8 6
16, 20: -2 8 2
17, 21: -1 1 -2 1 -1 8 6
18, 22: -1 2 -4 2 -1 8 6
19, 23: -1 3 -1 8 4
20, 24: -1 8 2
...
--------------------------------------------------------------------
The forms for the cycle CR(5) for D(5) = 7 (discriminant 28) are:
FR(5) = [1, 4, -3], the transformation with R(-1) produces FR1(5) = [-3, 2, 2], from this R(1) leads to FR2(5) = [2, 2, -3], then with R(-1) to FR3(5) = [-3, 4, 1], and with R(4) back to FR(5).
CROSSREFS
KEYWORD
sign,tabf
AUTHOR
Wolfdieter Lang, Apr 19 2019
STATUS
approved