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A097188
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G.f. A(x) satisfies A057083(x*A(x)) = A(x) and so equals the ratio of the g.f.s of any two adjacent diagonals of triangle A097186.
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14
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1, 3, 15, 90, 594, 4158, 30294, 227205, 1741905, 13586859, 107459703, 859677624, 6943550040, 56540336040, 463630755528, 3824953733106, 31724616256938, 264371802141150, 2212374554760150, 18583946259985260, 156636118477018620
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OFFSET
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0,2
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LINKS
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FORMULA
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G.f.: A(x) = (1 - (1-9*x)^(1/3))/(3*x).
G.f.: A(x) = (1/x)*(series reversion of x/A057083(x)).
x*A(x) is the compositional inverse of x-3*x^2+3*x^3. - Ira M. Gessel, Feb 18 2012
a(n) = 1/(n+1) * Sum_{k=1..n} binomial(k,n-k) * 3^(k)*(-1)^(n-k) * binomial(n+k,n), if n>0; a(0)=1. - Vladimir Kruchinin, Feb 07 2011
Conjecture: (n+1)*a(n) +3*(-3*n+1)*a(n-1)=0. - R. J. Mathar, Nov 16 2012
a(n) = 9^n * Gamma(n+2/3) / ((n+1) * Gamma(2/3) * Gamma(n+1)). - Vaclav Kotesovec, Feb 09 2014
Sum_{n>=0} 1/a(n) = 21/16 + 3*sqrt(3)*Pi/64 - 9*log(3)/64. - Amiram Eldar, Dec 02 2022
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MAPLE
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seq(coeff(series((1-(1-9*x)^(1/3))/(3*x), x, n+2), x, n), n = 0..25); # G. C. Greubel, Sep 17 2019
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MATHEMATICA
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Table[FullSimplify[9^n * Gamma[n+2/3] / ((n+1) * Gamma[2/3] * Gamma[n+1])], {n, 0, 20}] (* Vaclav Kotesovec, Feb 09 2014 *)
CoefficientList[Series[(1-(1 - 9 x)^(1/3))/(3 x), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 10 2014 *)
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PROG
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(PARI) a(n)=polcoeff((1-(1-9*x+x^2*O(x^n))^(1/3))/(3*x), n, x)
(Magma) R<x>:=PowerSeriesRing(Rationals(), 25); Coefficients(R!( (1 - (1-9*x)^(1/3))/(3*x) )); // G. C. Greubel, Sep 17 2019
(Sage)
P.<x> = PowerSeriesRing(QQ, prec)
return P((1 - (1-9*x)^(1/3))/(3*x)).list()
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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