%I #15 Sep 21 2022 15:16:57
%S 5,21,105,595,3675,24150,166257,1186680,8717940,65572325,502957455,
%T 3922142574,31021294850,248377859100,2010068042625,16421073515280,
%U 135277629836412,1122788441510820,9381874768828100,78871575753345375,666727830129370275
%N Number of triangulations of a fixed pentagon with n internal nodes.
%C These may be called rooted [n,2] triangulations.
%H Andrew Howroyd, <a href="/A341853/b341853.txt">Table of n, a(n) for n = 0..500</a>
%H K. A. Penson, K. Górska, A. Horzela, and G. H. E. Duchamp, <a href="https://arxiv.org/abs/2209.06574">Hausdorff moment problem for combinatorial numbers of Brown and Tutte: exact solution</a>, arXiv:2209.06574 [math.CO], 2022.
%F a(n) = 210*binomial(4*n+5, n)/((3*n+6)*(3*n+7)).
%e 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.
%t Array[210 Binomial[4 # + 5, #]/((3 # + 6)*(3 # + 7)) &, 21, 0] (* _Michael De Vlieger_, Feb 22 2021 *)
%o (PARI) a(n) = {210*binomial(4*n+5, n)/((3*n+6)*(3*n+7))}
%Y Column k=2 of A146305.
%K nonn
%O 0,1
%A _Andrew Howroyd_, Feb 21 2021
