%I #57 Apr 06 2024 10:03:11
%S 1,2,5,15,50,176,638,2354,8789,33099,125477,478193,1830271,7030571,
%T 27088871,104647631,405187826,1571990936,6109558586,23782190486,
%U 92705454896,361834392116,1413883873976,5530599237776,21654401079326,84859704298202,332818970772254
%N a(n) = (1/2)*(1 + Sum_{k=0..n} binomial(2*k, k)).
%C Total number of leaves in all rooted ordered trees with at most n edges. - _Michael Somos_, Feb 14 2006
%C 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
%C Hankel transform is A010892. - _Paul Barry_, Apr 28 2009
%C Binomial transform of A005773. - _Philippe Deléham_, Dec 13 2009
%C 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
%H Guo-Niu Han, <a href="/A196265/a196265.pdf">Enumeration of Standard Puzzles</a>, 2011. [Cached copy]
%H Guo-Niu Han, <a href="https://arxiv.org/abs/2006.14070">Enumeration of Standard Puzzles</a>, arXiv:2006.14070 [math.CO], 2020.
%H Sherry H. F. Yan, <a href="https://arxiv.org/abs/0805.2465">Schröder Paths and Pattern Avoiding Partitions</a>, arXiv:0805.2465 [math.CO], 2008-2009.
%F a(n) = A079309(n) + 1.
%F G.f.: 1/((1 - x)*(2 - C)), where C = g.f. for the Catalan numbers A000108. - _N. J. A. Sloane_, Aug 30 2002
%F Given g.f. A(x), then x * A(x - x^2) is the g.f. of A024494. - _Michael Somos_, Feb 14 2006
%F G.f.: (1 + 1 / sqrt(1 - 4*x)) / (2 - 2*x). - _Michael Somos_, Feb 14 2006
%F 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
%F 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
%F 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
%F 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
%F a(n) = [x^n] 1/((1-x+x^2) * (1-x)^n). - _Seiichi Manyama_, Apr 06 2024
%e G.f. = 1 + 2*x + 5*x^2 + 15*x^3 + 50*x^4 + 176*x^5 + 638*x^6 + ...
%p a := n -> ((1-I/sqrt(3))/2-binomial(2*n+1,n)*hypergeom([n+3/2,1],[n+2],4));
%p seq(simplify(a(n)), n=0..24); # _Peter Luschny_, May 18 2018
%t Table[Sum[Binomial[2k-1,k-1],{k,0,n}],{n,0,100}] (* _Emanuele Munarini_, May 18 2018 *)
%o (PARI) a(n) = (1 + sum(k=0, n, binomial(2*k, k)))/2; \\ _Michel Marcus_, May 18 2018
%Y Partial sums of A088218.
%Y Bisection of A086905.
%Y Second column of triangle A102541.
%Y Cf. A000108, A006134, A024494, A079309.
%K nonn
%O 0,2
%A _Clark Kimberling_
%E Name edited by _Petros Hadjicostas_, Aug 04 2020