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A197271 a(n) = (10 / ((3*n+1)*(3*n+2))) * binomial(4*n, n). 6
5, 2, 5, 20, 100, 570, 3542, 23400, 161820, 1159400, 8544965, 64448228, 495508780, 3872033900, 30680401500, 246041115600, 1993987498284, 16310419381080, 134519771966180, 1117653277802000, 9347742311507600, 78652006531467930, 665393840873409150, 5657273782416664200, 48318619683648190500 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,1
COMMENTS
A combinatorial interpretation for this sequence in terms of a family of plane trees is given in [Schaeffer, Corollary 2 with k = 4].
For n>=1, the number of rooted strict triangulations of a square with n-1 internal vertices, where a triangulation is "strict" if no two distinct edges have the same pair of ends. See equation (1) in [Tutte 1980] (who references [Brown 1964]) for the number of rooted strict near-triangulations of type (n,m), with m=1. - Noam Zeilberger, Jan 04 2023
LINKS
William G. Brown, Enumeration of Triangulations of the Disk, Proc. Lond. Math. Soc. s3-14 (1964) 746-768.
W. G. Brown, Enumeration of Triangulations of the Disk, Proc. Lond. Math. Soc. s3-14 (1964) 746-768. [Annotated scanned copy]
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.
William T. Tutte, On the enumeration of convex polyhedra, J. Combin. Theory Ser. B 28 (1980), 105-126.
FORMULA
a(n) = 10/((3*n+1)*(3*n+2))*binomial(4*n,n).
a(n) = A000260(n) * 5*(n+1)/(4*n+1). - Noam Zeilberger, May 20 2019
a(n) ~ c*(256/27)^n / n^(5/2), where c = (10/9)*sqrt(2/(3*Pi)) = 0.511843.... - Peter Luschny, Jan 05 2023
MATHEMATICA
Table[10/((3n+1)(3n+2)) Binomial[4n, n], {n, 0, 30}] (* Harvey P. Dale, Jan 27 2015 *)
CROSSREFS
Column m=1 of A146305.
Sequence in context: A071546 A154649 A100040 * A248260 A253382 A253384
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Oct 12 2011
STATUS
approved

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Last modified May 12 08:55 EDT 2024. Contains 372432 sequences. (Running on oeis4.)