A307378 Irregular triangle T(n, k) read by rows: row n gives the periods of the cycles of binary quadratic forms of discriminant 4*D(n), with D(n) = A000037(n).

2, 2, 2, 2, 2, 2, 4, 4, 2, 2, 2, 6, 2, 2, 2, 2, 10, 4, 4, 2, 2, 2, 2, 2, 2, 2, 6, 6, 2, 2, 6, 6, 6, 6, 4, 4, 2, 2, 4, 4, 2, 6, 2, 2, 4, 4, 10, 2, 2, 4, 4, 8, 8, 4, 4, 4, 4, 4, 4, 6, 6, 2, 2, 2, 2, 2, 6, 6, 2, 2, 2, 2, 6, 6, 2, 2, 4, 4, 6, 2, 2, 2, 2, 10, 10, 8, 8, 6, 6, 12, 12, 4, 4, 2, 2, 2, 2, 2, 6, 2, 2, 6, 6, 6, 6

Offset: 1

Author

Wolfdieter Lang, Apr 21 2019

Keywords

• nonn
• tabf

Comments

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.

Examples

			The irregular triangle T(n, k) begins:
n,  D(n) \k   1  2  3  4 ...              2*A307236
---------------------------------------------------
1,   2:       2                              2
2,   3:       2  2                           4
3,   5:       2                              2
4,   6:       2  2                           4
5,   7:       4  4                           8
6,   8:       2  2                           4
7,  10:       2  6                           8
8,  11:       2  2                           4
9,  12:       2  2                           4
10, 13:      10                             10
11, 14:       4  4                           8
12, 15:       2  2  2  2                     8
13, 17:       2                              2
14, 18:       2  2                           4
15, 19:       6  6                          12
16, 20:       2  2                           4
17, 21:       6  6                          12
18, 22:       6  6                          12
19, 23:       4  4                           8
20, 24:       2  2  4  4                    12
...
---------------------------------------------------
n = 1, D(1) = 2: the only cycle is the principal 2-cycle [[1, 2, -1],[-1, 2, 1]] with discriminant 8.
n = 2, D(2) = 3: besides the principal 2-cycle [[1, 2, -2], [-2, 2, 1]] there is another 2-cycle with sign flips in the outer form entries [[2, 2, -1], [-1, 2, 2]], all with discriminant 12.
n = 7, D(7) = 10: the principal 2-cycle CR(7) is ([1, 6, -1], [-1, 6, 1]). The other 6-cyle is ([3, 4, -2], [-2, 4, 3], [3, 2, -3], [-3, 4, 2], [2, 4, -3], [-3, 2, 3]). Both cycles are invariant under outer entries sign flips.
		

References

• D. A. Buell, Binary Quadratic Forms, Springer, 1989.
• A. Scholz and B. Schoeneberg, Einführung in die Zahlentheorie, 5. Aufl., de Gruyter, Berlin, New York, 1973.

Crossrefs

Cf. A000037, A000194, A307236, A307359, A307372, A324251.

Formula

T(n, k) = length of k-th cycle of reduced forms of discriminant 4*D(n), with D(n) = A000037(n).