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A097184
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G.f. A(x) satisfies A097182(x*A(x)) = A(x) and so equals the ratio of the g.f.s of any two adjacent diagonals of triangle A097181.
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6
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1, 7, 70, 805, 9982, 129766, 1742572, 23960365, 335445110, 4763320562, 68418604436, 992069764322, 14499481170860, 213349508656940, 3157572728122712, 46968894330825341, 701770538825272742, 10526558082379091130, 158452400608443161220
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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-(1-16*x)^(1/8))/(2*x).
G.f.: A(x) = (1/x)*(series reversion of x/A097182(x)).
D-finite with recurrence: (n+1)*a(n) +2*(-8*n+1)*a(n-1)=0. - R. J. Mathar, Nov 16 2012
a(n) = 16^n * Gamma(n+7/8) / (Gamma(7/8) * Gamma(n+2)). - Vaclav Kotesovec, Feb 09 2014
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MAPLE
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seq(coeff(series((1-(1-16*x)^(1/8))/(2*x), x, n+2), x, n), n = 0..20); # G. C. Greubel, Sep 17 2019
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MATHEMATICA
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CoefficientList[Series[(1-(1-16*x)^(1/8))/(2*x), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 09 2014 *)
Table[FullSimplify[16^n*Gamma[n+7/8]/(Gamma[7/8]*Gamma[n+2])], {n, 0, 20}] (* Vaclav Kotesovec, Feb 09 2014 *)
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PROG
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(PARI) a(n)=polcoeff((1-(1-16*x+x^2*O(x^n))^(1/8))/(2*x), n, x)
(Magma) R<x>:=PowerSeriesRing(Rationals(), 20); Coefficients(R!( (1-(1-16*x)^(1/8))/(2*x) )); // G. C. Greubel, Sep 17 2019
(Sage)
P.<x> = PowerSeriesRing(QQ, prec)
return P((1-(1-16*x)^(1/8))/(2*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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EXTENSIONS
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STATUS
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approved
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