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A341853 Number of triangulations of a fixed pentagon with n internal nodes. 5

%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="">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

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