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A088218
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Total number of leaves in all rooted ordered trees with n edges.
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75
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1, 1, 3, 10, 35, 126, 462, 1716, 6435, 24310, 92378, 352716, 1352078, 5200300, 20058300, 77558760, 300540195, 1166803110, 4537567650, 17672631900, 68923264410, 269128937220, 1052049481860, 4116715363800, 16123801841550, 63205303218876, 247959266474052
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OFFSET
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0,3
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COMMENTS
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Essentially the same as A001700, which has more information.
Note that the unique rooted tree with no edges has no leaves, so a(0)=1 is by convention. - Michael Somos, Jul 30 2011
Number of ordered partitions of n into n parts, allowing zeros (cf. A097070) is binomial(2*n-1,n) = a(n) = essentially A001700. - Vladeta Jovovic, Sep 15 2004
Hankel transform is A000027; example: Det([1,1,3,10;1,3,10,35;3,10,35,126; 10,35,126,462]) = 4. - Philippe Deléham, Apr 13 2007
a(n) is the number of functions f:[n]->[n] such that for all x,y in [n] if x<y then f(x)<=f(y). So 2*a(n)-n = A045992(n). - Geoffrey Critzer, Apr 02 2009
Hankel transform of the aeration of this sequence is A000027 doubled: 1,1,2,2,3,3,... - Paul Barry, Sep 26 2009
The Fi1 and Fi2 triangle sums of A039599 are given by the terms of this sequence. For the definitions of these triangle sums see A180662. - Johannes W. Meijer, Apr 20 2011
Alternating row sums of Riordan triangle A094527. See the Philippe Deléham formula. - Wolfdieter Lang, Nov 22 2012
(-2)*a(n) is the Z-sequence for the Riordan triangle A110162. For the notion of Z- and A-sequences for Riordan arrays see the W. Lang link under A006232 with details and references. - Wolfdieter Lang, Nov 22 2012
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REFERENCES
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L. W. Shapiro and C. J. Wang, Generating identities via 2 X 2 matrices, Congressus Numerantium, 205 (2010), 33-46.
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LINKS
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Vincenzo Librandi, Table of n, a(n) for n = 0..200
Paul Barry, On Integer-Sequence-Based Constructions of Generalized Pascal Triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4.
B. Dasarathy and C. Yang, A transformation on ordered trees, Comput. J. 23 (1980) 161-164. - Nachum Dershowitz, Sep 01 2016
T. Mansour, M. Shattuck, A statistic on n-color compositions and related sequences, Proc. Indian Acad. Sci. (Math. Sci.) Vol. 124, No. 2, May 2014, pp. 127-140.
Mircea Merca, A Note on Cosine Power Sums, J. Integer Sequences, Vol. 15 (2012), Article 12.5.3.
A. Vogt, Resummation of small-x double logarithms in QCD: semi-inclusive electron-positron annihilation, arXiv preprint arXiv:1108.2993 [hep-ph], 2011.
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FORMULA
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a(n) = (0^n + C(2n, n))/2. - Paul Barry, May 21 2004
a(n) is the coefficient of x^n in 1 / (1 - x)^n and also the sum of the first n coefficients of 1 / (1 - x)^n. Given B(x) with the property that the coefficient of x^n in B(x)^n equals the sum of the first n coefficients of B(x)^n, then B(x) = B(0) / (1 - x).
G.f.: 1 / (2 - C(x)) = (1 - x*C(x))/sqrt(1-4*x) where C(x) is g.f. for Catalan numbers A000108. Second equation added by Wolfdieter Lang, Nov 22 2012.
G.f.: (1 + 1 / sqrt(1 - 4*x)) / 2. a(n) = binomial( 2*n - 1, n).
From Paul Barry, Nov 02 2004: (Start)
a(n) = Sum_{k=0..n} binomial(2n, k)cos((n-k)*Pi)};
a(n) = Sum_{k=0..n} binomial(n, (n-k)/2)(1+(-1)^(n-k))cos(k*Pi/2)/2} (with interpolated zeros);
a(n) = Sum_{k=0..floor(n/2)} binomial(n, k)cos((n-2k)Pi/2)} (with interpolated zeros); (End)
a(n) = A110556(n)*(-1)^n, central terms in triangle A110555. - Reinhard Zumkeller, Jul 27 2005
a(n) = Sum_{0<=k<=n} A094527(n,k)*(-1)^k. - Philippe Deléham, Mar 14 2007
From Paul Barry, Mar 29 2010: (Start)
G.f.: 1/(1-x/(1-2x/(1-(1/2)x/(1-(3/2)x/(1-(2/3)x/(1-(4/3)x/(1-(3/4)x/(1-(5/4)x/(1-... (continued fraction);
E.g.f.: (of aerated sequence) (1 + Bessel_I(0, 2*x))/2. (End)
a(n + 1) = A001700(n). a(n) = A024718(n) - A024718(n - 1).
E.g.f.: E(x) = 1+x/(G(0)-2*x) ; G(k) = (k+1)^2+2*x*(2*k+1)-2*x*(2*k+3)*((k+1)^2)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Dec 21 2011
a(n) = Sum_{k=0..n}(-1)^k*binomial(2*n,n+k). - Mircea Merca, Jan 28 2012
a(n) = rf(n,n)/ff(n,n), where rf is the rising factorial and ff the falling factorial. - Peter Luschny, Nov 21 2012
n*a(n) +2*(-2*n+1)*a(n-1) = 0. - R. J. Mathar, Dec 04 2012
a(n) = binomial(2*n-1, n). - Vincenzo Librandi, Aug 07 2014
a(n) = hypergeom([1-n,-n],[1],1). - Peter Luschny, Sep 22 2014
G.f.: 1 + x/W(0), where W(k) = 4*k+1 - (4*k+3)*x/(1 - (4*k+1)*x/(4*k+3 - (4*k+5)*x/(1 - (4*k+3)*x/W(k+1) ))) ; (continued fraction). - Sergei N. Gladkovskii, Nov 13 2014
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EXAMPLE
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G.f. = 1 + x + 3*x^2 + 10*x^3 + 35*x^4 + 126*x^5 + 462*x^6 + 1716*x^7 + ...
The five rooted ordered trees with 3 edges have 10 leaves.
..x........................
..o..x.x..x......x.........
..o...o...o.x..x.o..x.x.x..
..r...r....r....r.....r....
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MAPLE
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seq(binomial(2*n-1, n), n=0..24); # Peter Luschny, Sep 22 2014
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MATHEMATICA
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a[ n_] := SeriesCoefficient[(1 - x)^-n, {x, 0, n}];
c = (1 - (1 - 4 x)^(1/2))/(2 x); CoefficientList[Series[1/(1-(c-1)), {x, 0, 20}], x] (* Geoffrey Critzer, Dec 02 2010 *)
Table[Binomial[2 n - 1, n], {n, 0, 20}] (* Vincenzo Librandi, Aug 07 2014 *)
a[ n_] := If[ n < 0, 0, With[ {m = 2 n}, m! SeriesCoefficient[ (1 + BesselI[0, 2 x]) / 2, {x, 0, m}]]]; (* Michael Somos, Nov 22 2014 *)
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PROG
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(PARI) {a(n) = sum( i=0, n, binomial(n+i-2, i))};
(PARI) {a(n) = if( n<0, 0, polcoeff( (1 + 1 / sqrt(1 - 4*x + x * O(x^n))) / 2, n))};
(PARI) {a(n) = if( n<0, 0, polcoeff( 1 / (1 - x + x * O(x^n))^n, n))};
(PARI) {a(n) = if( n<0, 0, binomial( 2*n - 1, n))};
(PARI) {a(n) = if( n<1, n==0, polcoeff( subst((1 - x) / (1 - 2*x), x, serreverse( x - x^2 + x * O(x^n))), n))};
(Sage)
def A088218(n):
return rising_factorial(n, n)/falling_factorial(n, n)
[A088218(n) for n in (0..24)] # Peter Luschny, Nov 21 2012
(MAGMA) [Binomial(2*n-1, n): n in [0..30]]; // Vincenzo Librandi, Aug 07 2014
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CROSSREFS
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Cf. A001700, A024718, A110556.
Main diagonal of A071919 and of A305161.
Sequence in context: A318117 A110556 A001700 * A300975 A072266 A085282
Adjacent sequences: A088215 A088216 A088217 * A088219 A088220 A088221
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KEYWORD
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nonn
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AUTHOR
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Michael Somos, Sep 24 2003
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STATUS
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approved
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