A368134 Characteristic numbers of Markov triples in the binary tree A368546.

2, 5, 12, 13, 75, 179, 70, 34, 507, 2923, 1120, 2673, 15571, 6089, 408, 89, 3468, 51709, 19760, 113922, 1701181, 651838, 16725, 39916, 3472225, 20226717, 1354498, 529673, 3087111, 206855, 2378, 233, 23763, 925943, 353702, 5273811, 205543262, 78545995, 770133

Offset: 0

Author

William P. Orrick, Jan 11 2024

Keywords

• nonn
• tabf

Comments

The characteristic number u of a Markov triple (r, m, s) is the solution in (0, m) of r * x == s (mod m). It satisfies u^2 == -1 (mod m), so that v = (u^2 + 1) / m is also an integer. The other solution in (0,m) of u^2 == -1 (mod m), namely m - u, is always greater than u, so u < m / 2.

The Markov tree may be formulated in terms of a set of Cohn matrices. There is a one-parameter family of such sets, parametrized by an integer c. Given a vertex of the Markov tree with Farey triple (x, y, z) and Markov triple (r, m, s), producing characteristic number u and v = (u^2 + 1) / m, the Cohn matrix C_y(c) with parameter c is

[ c * m + u m ]

[(3 * c - c^2) * m - (2 * c - 3) * u - v (3 - c) * m - u].

Then the vertex is associated with a triple of Cohn matrices, (R, R S, S), where R = C_x(c), RS = C_y(c), and S = C_z(c). See A368546 for a description of Farey and Markov triples. The left child of the vertex is associated with the triple (R, R^2 S, RS) and the right child with (RS, R S^2, S).

Examples

			The initial rows of the binary tree are
                                  2
             5                                       12
   13                 75                    179                70
34    507        2923   1120            2673  15571       6089   408
		

References

• Martin Aigner, Markov's theorem and 100 years of the uniqueness conjecture. A mathematical journey from irrational numbers to perfect matchings. Springer, 2013. x+257 pp. ISBN: 978-3-319-00887-5; 978-3-319-00888-2 MR3098784

Links

• Martin Aigner, Markov's theorem and 100 years of the uniqueness conjecture. A mathematical journey from irrational numbers to perfect matchings, [archive.org copy of the book]

Crossrefs

Cf. A368546.

Programs

SageMath

rowM = [[1,5,2]]
rowU = [[0,2,1]]
a368134 = [2]
for rw in range(1,6):
    prevRowM = rowM
    prevRowU = rowU
    rowM = []
    rowU = []
    for i in range(len(prevRowM)):
        [r,m,s] = prevRowM[i]
        [t,u,v] = prevRowU[i]
        ltM = [r,3*r*m - s,m]
        rtM = [m,3*m*s - r,s]
        ltU = [t,3*r*u - v,u]
        rtU = [u,3*u*s - t,v]
        rowM = rowM + [ltM,rtM]
        rowU = rowU + [ltU,rtU]
        a368134 = a368134 + [ltU[1],rtU[1]]
a368134

Formula

Recurrence: The left child of the Markov triple (r, m, s) is (r, 3rm - s, m); the right child is (m, 3ms - r, s). The corresponding triple of characteristic numbers (t, u, v) has left child (t, 3ru - v, u) and right child (u, 3us - t, v). Initial Markov triple: (1, 5, 2), initial characteristic number triple: (0, 2, 1).