This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A024718 #64 Apr 23 2025 23:45:02
%S A024718 1,2,5,15,50,176,638,2354,8789,33099,125477,478193,1830271,7030571,
%T A024718 27088871,104647631,405187826,1571990936,6109558586,23782190486,
%U A024718 92705454896,361834392116,1413883873976,5530599237776,21654401079326,84859704298202,332818970772254
%N A024718 a(n) = (1/2)*(1 + Sum_{k=0..n} binomial(2*k, k)).
%C A024718 Total number of leaves in all rooted ordered trees with at most n edges. - _Michael Somos_, Feb 14 2006
%C A024718 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 A024718 Hankel transform is A010892. - _Paul Barry_, Apr 28 2009
%C A024718 Binomial transform of A005773. - _Philippe Deléham_, Dec 13 2009
%C A024718 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 A024718 Michael De Vlieger, <a href="/A024718/b024718.txt">Table of n, a(n) for n = 0..1664</a>
%H A024718 Guo-Niu Han, <a href="/A196265/a196265.pdf">Enumeration of Standard Puzzles</a>, 2011. [Cached copy]
%H A024718 Guo-Niu Han, <a href="https://arxiv.org/abs/2006.14070">Enumeration of Standard Puzzles</a>, arXiv:2006.14070 [math.CO], 2020.
%H A024718 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 A024718 a(n) = A079309(n) + 1.
%F A024718 G.f.: 1/((1 - x)*(2 - C)), where C = g.f. for the Catalan numbers A000108. - _N. J. A. Sloane_, Aug 30 2002
%F A024718 Given g.f. A(x), then x * A(x - x^2) is the g.f. of A024494. - _Michael Somos_, Feb 14 2006
%F A024718 G.f.: (1 + 1 / sqrt(1 - 4*x)) / (2 - 2*x). - _Michael Somos_, Feb 14 2006
%F A024718 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 A024718 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 A024718 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 A024718 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 A024718 a(n) = [x^n] 1/((1-x+x^2) * (1-x)^n). - _Seiichi Manyama_, Apr 06 2024
%e A024718 G.f. = 1 + 2*x + 5*x^2 + 15*x^3 + 50*x^4 + 176*x^5 + 638*x^6 + ...
%p A024718 A024718 := n -> (binomial(2*n, n)*hypergeom([1, -n], [1/2 - n], 1/4) + 1)/2:
%p A024718 seq(simplify(A024718(n)), n = 0..26); # _Peter Luschny_, Dec 15 2024
%t A024718 Table[Sum[Binomial[2k-1,k-1],{k,0,n}],{n,0,100}] (* _Emanuele Munarini_, May 18 2018 *)
%o A024718 (PARI) a(n) = (1 + sum(k=0, n, binomial(2*k, k)))/2; \\ _Michel Marcus_, May 18 2018
%Y A024718 Partial sums of A088218.
%Y A024718 Bisection of A086905.
%Y A024718 Second column of triangle A102541.
%Y A024718 Cf. A000108, A006134, A024494, A079309.
%K A024718 nonn
%O A024718 0,2
%A A024718 _Clark Kimberling_
%E A024718 Name edited by _Petros Hadjicostas_, Aug 04 2020