A292186 Number of rooted unlabeled connected four-regular maps on a compact closed oriented surface with n vertices (and thus 2*n edges), with a(0) = 1.

1, 3, 24, 297, 4896, 100278, 2450304, 69533397, 2247492096, 81528066378, 3280382613504, 145009234904922, 6986546222800896, 364418301804218028, 20459842995693256704, 1230262900677124568397, 78884016707711348637696, 5372823210133041283250178, 387394283866652086938107904

Offset: 0

Author

Sasha Kolpakov, Sep 11 2017

Keywords

• nonn
• easy

Comments

Equivalently, the number of rooted quadrangulations of oriented surfaces with n quadrangles (and thus 2*n edges) for n > 0.

Equivalently, the number of pairs (alpha,sigma) of permutations on a pointed set of size 4*n up to simultaneous conjugacy such that alpha (resp. sigma) has only cycles of length 2 (resp. 4) and the subgroup generated by them acts transitively, for n > 0.

This is also the S(4, -6, 1) sequence of Martin and Kearney, if the offset is set to 1.

This sequence is not D-finite (or holonomic).

Links

• Sasha Kolpakov, Table of n, a(n) for n = 0..119
• Laura Ciobanu and Alexander Kolpakov, Free subgroups of free products and combinatorial hypermaps, Discrete Mathematics, 342 (2019), 1415-1433; arXiv:1708.03842 [math.CO], 2017-2019.
• R. J. Martin and M. J. Kearney, An exactly solvable self-convolutive recurrence, Aequat. Math., 80 (2010), 291-318; arXiv:1103.4936 [math.CO], 2011.
• A. N. Stokes, Continued fraction solutions of the Riccati equation, Bull. Austral. Math. Soc. Vol. 25 (1982), 207-214.

Crossrefs

Column 4 of A380622.

Cf. A172455, A292187, A128709.

Programs

Python

from sympy.core.cache import cacheit
@cacheit
def a(n): return 1 if n == 0 else (4*n - 2)*a(n - 1) + sum([a(k)*a(n - k - 1) for k in range(n)])
[a(n) for n in range(21)]

Formula

a(0)=1, a(1)=3, a(n) = 4*n*a(n-1) + Sum_{k=1..n-2} a(k)*a(n-k-1) for n>=2.

From Peter Bala, Aug 22 2023: (Start)

The o.g.f. A(x) = 1 + 3*x + 24*x^2 + 297*x^3 + 4896*x^4 + 100278*x^5 + 2450304*x^6 + ... satisfies the Riccati differential equation (4*x^2)*A'(x) = -1 + (1 - 2*x)*A(x) - x*A(x)^2 with A(0) = 1.

O.g.f. as a continued fraction of Stieltjes type: 1/(1 - 3*x/(1 - 5*x/(1 - 7*x/(1 - 9*x/(1 - ... - (2*n+1)*x/(1 - ... )))))).

Also A(x) = 1/(1 + 2*x - 5*x/(1 - 3*x/(1 - 9*x/(1 - 7*x/(1 - ... - (4*n+1)*x/(1 - (4*n-1)*x/(1 - ... ))))))). (End)

Extensions

Edited by Andrey Zabolotskiy, Jan 23 2025