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A341853
Number of triangulations of a fixed pentagon with n internal nodes.
5
5, 21, 105, 595, 3675, 24150, 166257, 1186680, 8717940, 65572325, 502957455, 3922142574, 31021294850, 248377859100, 2010068042625, 16421073515280, 135277629836412, 1122788441510820, 9381874768828100, 78871575753345375, 666727830129370275
OFFSET
0,1
COMMENTS
These may be called rooted [n,2] triangulations.
LINKS
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.
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
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
KEYWORD
nonn
AUTHOR
Andrew Howroyd, Feb 21 2021
STATUS
approved