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A024718
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a(n) = (1/2)*(1 + Sum_{k=0..n} binomial(2*k, k)).
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29
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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
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OFFSET
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0,2
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COMMENTS
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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
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
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LINKS
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FORMULA
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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
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EXAMPLE
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G.f. = 1 + 2*x + 5*x^2 + 15*x^3 + 50*x^4 + 176*x^5 + 638*x^6 + ...
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MAPLE
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a := n -> ((1-I/sqrt(3))/2-binomial(2*n+1, n)*hypergeom([n+3/2, 1], [n+2], 4));
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MATHEMATICA
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Table[Sum[Binomial[2k-1, k-1], {k, 0, n}], {n, 0, 100}] (* Emanuele Munarini, May 18 2018 *)
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PROG
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(PARI) a(n) = (1 + sum(k=0, n, binomial(2*k, k)))/2; \\ Michel Marcus, May 18 2018
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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