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A024718 a(n) = (1/2)*(1 + sum of C(2k,k)) for k = 0,1,2,...,n. 12
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

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

LINKS

Table of n, a(n) for n=0..26.

Guo-Niu Han, Enumeration of Standard Puzzles

Guo-Niu Han, Enumeration of Standard Puzzles [Cached copy]

Sherry H. F. Yan, Schröder Paths and Pattern Avoiding Partitions, arXiv:0805.2465 [math.CO], 2008-2009.

FORMULA

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

Given g.f. A(x), then x * A(x - x^2) is 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

Conjecture: 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 conjectured 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

EXAMPLE

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

MAPLE

a := n -> ((1-I/sqrt(3))/2-binomial(2*n+1, n)*hypergeom([n+3/2, 1], [n+2], 4));

seq(simplify(a(n)), n=0..24); # Peter Luschny, May 18 2018

MATHEMATICA

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

PROG

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

CROSSREFS

Equals A079309(n) + 1. Partial sums of A088218. Bisection of A086905. Second column of triangle A102541. A006134

Sequence in context: A228343 A149949 A149950 * A149951 A157135 A196836

Adjacent sequences:  A024715 A024716 A024717 * A024719 A024720 A024721

KEYWORD

nonn

AUTHOR

Clark Kimberling

STATUS

approved

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Last modified August 21 14:53 EDT 2018. Contains 313954 sequences. (Running on oeis4.)