A341853 Number of triangulations of a fixed pentagon with n internal nodes.

5, 21, 105, 595, 3675, 24150, 166257, 1186680, 8717940, 65572325, 502957455, 3922142574, 31021294850, 248377859100, 2010068042625, 16421073515280, 135277629836412, 1122788441510820, 9381874768828100, 78871575753345375, 666727830129370275, 5664526697522326590

Offset

0,1

Comments

These may be called rooted [n,2] triangulations.

Links

Andrew Howroyd, Table of n, a(n) for n = 0..500

K. A. Penson, K. Górska, A. Horzela, and G. H. E. Duchamp, Hausdorff moment problem for combinatorial numbers of Brown and Tutte: exact solution, arXiv:2209.06574 [math.CO], 2022-2023.

Formula

a(n) = 210*binomial(4*n+5, n)/((3*n+6)*(3*n+7)).

D-finite with recurrence 3*n*(3*n+5)*(3*n+7)*(n+2)*a(n) - 8*(4*n+5)*(2*n+1)*(4*n+3)*(n+1)*a(n-1) = 0. - R. J. Mathar, Jul 31 2024

a(n) ~ 35 * 2^(8*n+23/2) / (3^(3*n+13/2) * n^(5/2) * sqrt(Pi)). - Amiram Eldar, Oct 24 2025

Example

The a(0) = 5 triangulations correspond with the dissections of a pentagon by nonintersecting diagonals into 3 triangles. Although there is only one essentially different dissection, each rotation is counted separately here.

Mathematica

Array[210 Binomial[4 # + 5, #]/((3 # + 6)*(3 # + 7)) &, 21, 0] (* Michael De Vlieger, Feb 22 2021 *)

Prog

(PARI) a(n) = {210*binomial(4*n+5, n)/((3*n+6)*(3*n+7))}

Crossrefs

Column k=2 of A146305.

Sequence in context: A097175 A100284 A337168 * A260845 A325157 A218299

Adjacent sequences: A341850 A341851 A341852 * A341854 A341855 A341856

Keyword

nonn,easy

Author

Andrew Howroyd, Feb 21 2021

Status

approved