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A076035
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G.f.: 1/(1-4*x*C) where C = (1/2-1/2*(1-4*x)^(1/2))/x = g.f. for Catalan numbers A000108.
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15
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1, 4, 20, 104, 548, 2904, 15432, 82128, 437444, 2331128, 12426200, 66250672, 353258536, 1883768176, 10045773072, 53573890464, 285714489348, 1523763466296, 8126565627192, 43341046493424, 231149891614008, 1232790669780816, 6574850950474992, 35065749759115104
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OFFSET
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0,2
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COMMENTS
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The Hankel transform of this sequence and that of the aerated sequence with g.f. 1/(1-4x^2*c(x^2)) is 4^n. In general, the expansions of 1/(1-k*x*c(x)) and 1/(1-k*x^2*c(x^2)) have Hankel transform k^n. - Paul Barry, Jan 20 2007
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LINKS
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Vincenzo Librandi, Table of n, a(n) for n = 0..200
S. B. Ekhad, M. Yang, Proofs of Linear Recurrences of Coefficients of Certain Algebraic Formal Power Series Conjectured in the On-Line Encyclopedia Of Integer Sequences, (2017).
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FORMULA
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a(n) = sum{k=0..n, 3^k*C(2n, n-k)(2k+1)/(n+k+1)}. - Paul Barry, Jun 22 2004
a(n) = Sum_{k, 0<=k<=n} A106566(n, k)*4^k. - Philippe Deléham, Sep 01 2005
a(n) = if(n=0,1,sum{k=1..n, C(2n-k-1,n-k)*k*4^k/n}). - Paul Barry, Jan 20 2007
a(n) = Sum{k, 0<=k<=n}A039599(n,k)*3^k. - Philippe Deléham, Sep 08 2007
a(0)=1, a(n)=(16*a(n-1)-4*A000108(n-1))/3. - Philippe Deléham, Nov 27 2007
Conjecture: 3*n*a(n) +2*(9-14*n)*a(n-1) +32*(2*n-3)*a(n-2) = 0. - R. J. Mathar, Nov 14 2011 [proved by Ekhad & Yang, see link]
a(n) ~ 2^(4*n+1) / 3^(n+1). - Vaclav Kotesovec, Feb 13 2014
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MAPLE
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CatalanNumber := n -> binomial(2*n, n)/(n+1):
h := (n, m) -> hypergeom([1+m, m-n], [m+n+2], -3):
a := n -> CatalanNumber(n)*(h(n, 0) + 6*n/(n+2)*h(n, 1)):
seq(simplify(a(n)), n=0..23); # Peter Luschny, Dec 09 2018
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MATHEMATICA
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CoefficientList[Series[1/(1-4*x*(1-Sqrt[1-4*x])/(2*x)), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 13 2014 *)
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CROSSREFS
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Cf. A000108, A000984, A007854, A076036.
Sequence in context: A155485 A155181 A082761 * A120978 A035028 A104550
Adjacent sequences: A076032 A076033 A076034 * A076036 A076037 A076038
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KEYWORD
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nonn
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AUTHOR
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N. J. A. Sloane, Oct 29 2002
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STATUS
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approved
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