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%I #16 Jul 24 2017 01:57:33
%S 1,-1,0,2,-2,-5,14,5,-72,68,278,-726,-520,4691,-3514,-21758,50374,
%T 56185,-374566,194596,1962618,-3956504,-6258320,33057877,-8974630,
%U -190822072,330170022,710487590,-3088268200,18008739,19398384974,-28292606291,-81631282280,298546543220,84094857302,-2028216574806,2428288153424,9450205225145
%N Series reversion of A028310.
%H G. C. Greubel, <a href="/A185966/b185966.txt">Table of n, a(n) for n = 0..1000</a>
%F a(n) = A185962(2*n,n)/(n+1) = A185965(n)/(n+1).
%F Given g.f. A(x) then B(x) = x * A(x) satisfies B(x) = (1 - B(x)) * (x + B(x) * (B(x) - x)). - _Michael Somos_, Apr 05 2012
%F Conjecture: 6*n*(n+1)*a(n) -n*(n-14)*a(n-1) +2*n*(14*n-19)*a(n-2) -4*(n-2)*(17*n-48)*a(n-3) +6*(2*n-5)*(n-4)*a(n-4)=0. - _R. J. Mathar_, Nov 15 2012
%F Recurrence (of order 3): 3*n*(n+1)*(19*n-27)*a(n) = -2*n*(38*n^2 - 73*n + 9)*a(n-1) - 20*(19*n^3 - 65*n^2 + 66*n - 18)*a(n-2) + 2*(n-3)*(2*n-3)*(19*n-8)*a(n-3). - _Vaclav Kotesovec_, Jan 22 2014
%F Lim sup n->infinity |a(n)|^(1/n) = sqrt(20/9 + 1/27*(272376 - 12312 * sqrt(57))^(1/3) + 2/9*(1261 + 57 * sqrt(57))^(1/3)) = 2.637962913244886521522... - _Vaclav Kotesovec_, Jan 22 2014
%e 1 - x + 2*x^3 - 2*x^4 - 5*x^5 + 14*x^6 + 5*x^7 - 72*x^8 + 68*x^9 + ...
%t CoefficientList[1/x*InverseSeries[Series[x*(1-x+x^2) /(1-x)^2, {x, 0, 20}], x],x] (* _Vaclav Kotesovec_, Jan 22 2014 *)
%o (PARI) {a(n) = if( n<0, 0, polcoeff( serreverse( x * ((1 - x + x^2) / (1 - x)^2 + x * O(x^n))) / x, n))} /* _Michael Somos_, Apr 05 2012 */
%o (PARI) {a(n) = local(B); if( n<0, 0, B = O(x); for( k=0, n, B = (1 - B) * (x + B * (B - x))); polcoeff( B / x, n))} /* _Michael Somos_, Apr 05 2012 */
%Y Cf. A028130, A185962, A185965.
%K sign,easy
%O 0,4
%A _Paul Barry_, Feb 07 2011