This site is supported by donations to The OEIS Foundation.

 Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 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 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 STATUS approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent | More pages
The OEIS Community | Maintained by The OEIS Foundation Inc.

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