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A001761
a(n) = (2*n)!/(n+1)!.
(Formerly M3635 N1478)
40
1, 1, 4, 30, 336, 5040, 95040, 2162160, 57657600, 1764322560, 60949324800, 2346549004800, 99638080819200, 4626053752320000, 233153109116928000, 12677700308232960000, 739781100339240960000, 46113021921146019840000
OFFSET
0,3
COMMENTS
According to the Beineke and Pippert paper, the number of dissections of a disk is given by D(n)=R(n)/(n-2)!, where R(n)=A001761(n-2) is the number of labeled planar 2-trees having n vertices and rooted at a given exterior edge. [Clarified by M. F. Hasler, Feb 22 2012]
a(n+1) is the number of labeled incomplete ternary trees on n vertices in which each left and middle child have a larger label than their parent. - Brian Drake, Jul 28 2008
For n>0: a(n) = A173333(2*n,n+1); cf. A006963, A001813. - Reinhard Zumkeller, Feb 19 2010
REFERENCES
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
L. W. Beineke and R. E. Pippert, Enumerating labeled k-dimensional trees and ball dissections, pp. 12-26 of Proceedings of Second Chapel Hill Conference on Combinatorial Mathematics and Its Applications, University of North Carolina, Chapel Hill, 1970. Reprinted in Math. Annalen, Vol. 191 (1971), pp. 87-98.
Allan Bickle, A Survey of Maximal k-degenerate Graphs and k-Trees, Theory and Applications of Graphs 0 1 (2024) Article 5.
Peter J. Cameron, Some treelike objects, Quart. J. Math. Oxford, Vol. 38, No. 2 (1987), pp. 155-183. See p. 166. - N. J. A. Sloane, Apr 18 2014
Ali Chouria, Vlad-Florin Drǎgoi, and Jean-Gabriel Luque, On recursively defined combinatorial classes and labelled trees, arXiv:2004.04203 [math.CO], 2020.
K. A. Penson and J.-M. Sixdeniers, Integral Representations of Catalan and Related Numbers, J. Integer Sequences, Vol. 4 (2001), Article 01.2.5.
Karol A. Penson and Allan I. Solomon, Coherent states from combinatorial sequences, arXiv:quant-ph/0111151, 2001.
FORMULA
a(n) = n!*Catalan(n) =n!* A000108(n). - N. J. A. Sloane, Apr 18 2014
a(n+2) = sum(A038455(n, m), m=1..n), n >= 1. - Wolfdieter Lang
E.g.f. for this sequence = o.g.f. for A000108. - Len Smiley, Dec 07 2001
Integral representation as the moment of a positive function on the positive half-axis: in Maple notation, a(n)=int(x^n*(-1/2+exp(-x/4)/sqrt(Pi*x)+erf(sqrt(x)/2)/2), x=0..infinity), n=0, 1... This representation is unique. - Karol A. Penson, Aug 21 2001
G.f.: If G_N(x)=1+sum('(2*k)!*(x^k)/(k+1)!', 'k'=1..N), G_N(x)=1+2*x/(G(0)-2*x); G(k)=4*x*(k^2)+6*k*x+k+2*x+2-2*x*(2*k+3)*((k+2)^2)/G(k+1) ; (continued fraction). - Sergei N. Gladkovskii, Nov 24 2011
a(n) = Sum_{k=0..n} (-1)^(n-k) * (n+1)^(k-1) * Stirling1(n,k). - Paul D. Hanna, Nov 09 2012
G.f.: Q(0) where Q(k) = 1 + x*(2*k+1)*(4*k+1)/(k+1 - 4*x*(k+1)^2*(4*k+3)/(4*x*(k+1)*(4*k+3) + (2*k+3)/Q(k+1) )); (continued fraction ). - Sergei N. Gladkovskii, Apr 05 2013
G.f.: G(0)/2, where G(k)= 1 + 1/(1 - x/(x + (k+2)/(2*k+2)/(2*k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 03 2013
Let A(x) = sum(k>=0, a(k)*x^k /(2*k)! ) = ( exp(x)-1)/x, then A(x) = 1/Q(0), where Q(k) = 1 - x/( 1 + (2*k+1)/(1 - x/( 1 + 2*(k+1)/Q(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Nov 24 2013
From Ilya Gutkovskiy, Jan 21 2017: (Start)
a(n) ~ sqrt(2)*4^n*n^(n-1)/exp(n).
Sum_{n>=0} 1/a(n) = (7*exp(1/4)*sqrt(Pi)*erf(1/2) + 10)/8 = 2.2865189388213215..., where erf() is the error function. (End)
D-finite with recurrence: (n+1)*a(n) -2*n*(2*n-1)*a(n-1)=0. - R. J. Mathar, Feb 16 2020
Sum_{n>=0} (-1)^n/a(n) = 3/4 - 5*sqrt(Pi)*erfi(1/2)/(8*exp(1/4)), where erfi() is the imaginary error function. - Amiram Eldar, Apr 03 2022
MAPLE
seq(mul((n+k), k=2..n), n=0..17); # Zerinvary Lajos, Feb 15 2008
MATHEMATICA
Table[(2*n)!/(n+1)!, {n, 0, 20}] (* Vincenzo Librandi, Feb 23 2012 *)
PROG
(MuPAD) combinat::catalan(n)*n! $ n = 0..17; // Zerinvary Lajos, Feb 15 2007
(Sage) [binomial(2*n, n)/(1+n)*factorial(n) for n in range(0, 18)] # Zerinvary Lajos, Dec 03 2009
(PARI) A001761(n)=binomial(2*n, n+1)*(n-1)! \\ M. F. Hasler, Feb 23 2012
(PARI) {Stirling1(n, k)=n!*polcoeff(binomial(x, n), k)}
{a(n)=sum(k=0, n, (-1)^(n-k)*(n+1)^(k-1)*Stirling1(n, k))} \\ Paul D. Hanna, Nov 09 2012
CROSSREFS
Main diagonal of A255982, A256061.
Sequence in context: A006149 A207833 A121413 * A292220 A099712 A370931
KEYWORD
nonn
STATUS
approved