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%I #71 Aug 06 2021 07:30:56
%S -1,2,6,32,210,1536,12012,98304,831402,7208960,63740820,572522496,
%T 5209363380,47915728896,444799488600,4161823309824,39209074920090,
%U 371626340253696,3541117629057540
%N Expand the real root of y^3 - y + x in powers of x, then multiply coefficient of x^n by -4^n to get integers.
%C Also coefficients of the series S(u) for which (-sqrt(3u))*S converges to the larger of the two real roots of x^3 - 3ux + 4u for u >= 4. Specifically, S(u)=Sum_{n>=0} a(n)/(27*u)^(n/2). - _Dixon J. Jones_, Jun 24 2021
%D G. E. Andrews, Number Theory, 1971, Dover Publications New York, pp. 41 - 43.
%F G.f.: -(12*x)/(2*sin(arcsin(216*x^2-1)/3)+1). - _Vladimir Kruchinin_, Oct 30 2014
%F G.f.: -x/Revert((x*sqrt(1-4*x))). - _Thomas Baruchel_, Jul 02 2018
%F G.f.: - (1/x) * Revert( x*sqrt(c(4*x)) ), where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. of the Catalan numbers A000108 and sqrt(c(4*x)) is the g.f. of A048990. - _Peter Bala_, Mar 05 2020
%F From _Dixon J. Jones_, Jun 24 2021: (Start)
%F a(n) = 2*A085614(n) for n>=1.
%F a(n) = 2^(2*n - 1) Gamma((3*n - 1)/2)/(Gamma((n + 1)/2)*n!).
%F a(n) = (2^(2*n - 1)*((n + 1)/2)_(n-1))/n!, where (x)_k is the Pochhammer symbol. (End)
%t p[x_] = y /. Solve[y^3 - y + x == 0, y][[1]]
%t b = Table[-4^n*FullSimplify[ExpandAll[SeriesCoefficient[ Series[p[x], {x, 0, 30}], n]]], {n, 0, 30}]
%t (* From _Dixon J. Jones_, Jun 24 2021: (Start) *)
%t Clear[a]; a=Table[2^(2n - 1) Gamma[(3n - 1)/2]/(Gamma[(n + 1)/2]n!), {n, 0, 20}]
%t Clear[a]; a=Table[2^(2n - 1) Pochhammer[(n + 1)/2, (n-1)]/n!, {n, 0, 20}] (* End *)
%o (PARI) -x/serreverse((x*sqrt(1-4*x))) \\ _Thomas Baruchel_, Jul 02 2018
%Y Cf. A000108, A048990, A224884 (signed version).
%Y Cf. A085614.
%K sign,easy
%O 0,2
%A _Roger L. Bagula_, Feb 05 2012
%E Edited by _N. J. A. Sloane_, Feb 09 2012
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