A024718 a(n) = (1/2)*(1 + Sum_{k=0..n} binomial(2*k, k)).

1, 2, 5, 15, 50, 176, 638, 2354, 8789, 33099, 125477, 478193, 1830271, 7030571, 27088871, 104647631, 405187826, 1571990936, 6109558586, 23782190486, 92705454896, 361834392116, 1413883873976, 5530599237776, 21654401079326, 84859704298202, 332818970772254

Offset: 0

Author

Clark Kimberling

Keywords

• nonn

Comments

Total number of leaves in all rooted ordered trees with at most n edges. - Michael Somos, Feb 14 2006

Also: Number of UH-free Schroeder paths of semilength n with horizontal steps only at level less than two [see Yan]. - R. J. Mathar, May 24 2008

Hankel transform is A010892. - Paul Barry, Apr 28 2009

Binomial transform of A005773. - Philippe Deléham, Dec 13 2009

Number of vertices all of whose children are leaves in all ordered trees with n+1 edges. Example: a(3) = 15; for an explanation see David Callan's comment in A001519. - Emeric Deutsch, Feb 12 2015

Examples

			G.f. = 1 + 2*x + 5*x^2 + 15*x^3 + 50*x^4 + 176*x^5 + 638*x^6 + ...
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 0..1664
• Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
• Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
• Sherry H. F. Yan, Schröder Paths and Pattern Avoiding Partitions, arXiv:0805.2465 [math.CO], 2008-2009.

Crossrefs

Partial sums of A088218.

Bisection of A086905.

Second column of triangle A102541.

Cf. A000108, A006134, A024494, A079309.

Programs

Maple

A024718 := n -> (binomial(2*n, n)*hypergeom([1, -n], [1/2 - n], 1/4) + 1)/2:
seq(simplify(A024718(n)), n = 0..26); # Peter Luschny, Dec 15 2024

Mathematica

Table[Sum[Binomial[2k-1,k-1],{k,0,n}],{n,0,100}] (* Emanuele Munarini, May 18 2018 *)

PARI

a(n) = (1 + sum(k=0, n, binomial(2*k, k)))/2; \\ Michel Marcus, May 18 2018

Formula

a(n) = A079309(n) + 1.

G.f.: 1/((1 - x)*(2 - C)), where C = g.f. for the Catalan numbers A000108. - N. J. A. Sloane, Aug 30 2002

Given g.f. A(x), then x * A(x - x^2) is the g.f. of A024494. - Michael Somos, Feb 14 2006

G.f.: (1 + 1 / sqrt(1 - 4*x)) / (2 - 2*x). - Michael Somos, Feb 14 2006

D-finite with recurrence: n*a(n) - (5*n-2)*a(n-1) + 2*(2*n-1)*a(n-2) = 0. - R. J. Mathar, Dec 02 2012

Remark: The above recurrence is true (it can be easily proved by differentiating the generating function). Notice that it is the same recurrence satisfied by the partial sums of the central binomial coefficients (A006134). - Emanuele Munarini, May 18 2018

0 = a(n)*(16*a(n+1) - 22*a(n+2) + 6*a(n+3)) + a(n+1)*(-18*a(n+1) + 27*a(n+2) - 7*a(n+3)) + a(n+2)*(-3*a(n+2) + a(n+3)) for all n in Z if a(n) = 1/2 for n < 0. - Michael Somos, Apr 23 2014

a(n) = ((1 - I/sqrt(3))/2 - binomial(2*n+1, n)*hypergeom([n+3/2, 1], [n+2], 4)). - Peter Luschny, May 18 2018

a(n) = [x^n] 1/((1-x+x^2) * (1-x)^n). - Seiichi Manyama, Apr 06 2024

Extensions

Name edited by Petros Hadjicostas, Aug 04 2020