A292187 - OEIS

%I #49 Jan 23 2025 21:57:11

%S 1,2,12,112,1392,21472,394752,8421632,204525312,5572091392,

%T 168331164672,5585571889152,201973854584832,7905697598963712,

%U 333049899230625792,15025907115679875072,722841343143300759552,36935846945562562527232,1997902532753538016346112,114050521905958855289864192,6852141240070150728132329472

%N Number of rooted unlabeled bipartite cubic maps on a compact closed oriented surface with 2*n vertices (and thus 3*n edges), with a(0) = 1.

%C Equivalently, the number of rooted bicolored triangulations with 2*n triangles (and thus 3*n edges) for n > 0.

%C Equivalently, the number of pairs of permutations (alpha,sigma) up to simultaneous conjugacy on a pointed set of size 3*n with alpha^3=sigma^3=1, acting transitively and without fixed points, for n > 0.

%C This is also the S(3, -5, 1) sequence of Martin and Kearney, if the offset is set to 1.

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

%H Sasha Kolpakov, <a href="/A292187/b292187.txt">Table of n, a(n) for n = 0..119</a>

%H Laura Ciobanu and Alexander Kolpakov, <a href="https://doi.org/10.1016/j.disc.2019.01.014">Free subgroups of free products and combinatorial hypermaps</a>, Discrete Mathematics, 342 (2019), 1415-1433; arXiv:<a href="https://arxiv.org/abs/1708.03842">1708.03842</a> [math.CO], 2017-2019.

%H R. J. Martin and M. J. Kearney, <a href="https://doi.org/10.1007/s00010-010-0051-0">An exactly solvable self-convolutive recurrence</a>, Aequat. Math., 80 (2010), 291-318; arXiv:<a href="https://arxiv.org/abs/1103.4936">1103.4936</a> [math.CO], 2011.

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

%F From _Peter Bala_, Sep 01 2023: (Start)

%F The o.g.f. A(x) = 1 + 2*x + 12*x^2 + 112*x^3 + 1392*x^4 + 21472*x^5 + 394752*x^6 + ... satisfies the Riccati differential equation (3*x^2)*A'(x) = -1 + (1 - x)*A(x) - x*A(x)^2 with A(0) = 1.

%F O.g.f. as a continued fraction of Stieltjes type: A(x) = 1/(1 - 2*x/(1 - 4*x/(1 - 5*x/(1 - 7*x/(1 - 8*x/(1 - 10*x/(1 - ... ))))))).

%F Also A(x) = 1/(1 + 2*x - 4*x/(1 - 2*x/(1 - 7*x/(1 - 5*x/(1 - 10*x/(1 - 8*x/(1 - ... ))))))). (End)

%o (Python)

%o from sympy.core.cache import cacheit

%o @cacheit

%o def a(n): return 1 if n == 0 else (3*n - 1)*a(n - 1) + sum([a(k)*a(n - k - 1) for k in range(1, n)])

%o [a(n) for n in range(21)]

%Y Cf. A000698, A062980, A292186.

%K nonn,easy

%O 0,2

%A _Sasha Kolpakov_, Sep 11 2017

%E Edited by _Andrey Zabolotskiy_, Jan 23 2025