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A368134
Characteristic numbers of Markov triples in the binary tree A368546.
1
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
0,1
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).
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
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).
EXAMPLE
The initial rows of the binary tree are
2
5 12
13 75 179 70
34 507 2923 1120 2673 15571 6089 408
PROG
(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
CROSSREFS
Cf. A368546.
Sequence in context: A324601 A305310 A353244 * A039586 A352785 A114217
KEYWORD
nonn,tabf
AUTHOR
William P. Orrick, Jan 11 2024
STATUS
approved