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%I #29 Apr 07 2019 19:05:03
%S 1,4,112,8608,1295104,322018816,119597651968,62037189087232,
%T 42842215767801856,38001850792907505664,42106195262186260529152,
%U 56992583802129636291248128,92535374062287540141093289984,177509548409832461509746497683456,397176345992538727622693418988208128
%N E.g.f. A(x) = Sum_{n>=0} a(n)*x^(2*n+1)/(2*n+1)! is inverse to f(x) = 2*arctan(x) - x.
%H Vladimir Kruchinin, <a href="http://arxiv.org/abs/1211.3244">The method for obtaining expressions for coefficients of reverse generating functions</a>, arXiv:1211.3244 [math.CO], 2012.
%F a(n) = sum(k=1..2*n, (2*n+k)!*sum(j=1..k, sum(l=0..j-1, ((-1)^(n+l+j)*sum(i=j-l..2*n-l+j, (2^i*Stirling1(i,j-l)*binomial(2*n-l+j-1,i-1))/i!))/l!)/(k-j)!)), n>0, a(0)=1.
%F a(n) ~ 2^(2*n+1) * n^(2*n) / (exp(2*n) * (Pi/2-1)^(2*n+1/2)). - _Vaclav Kotesovec_, Jan 02 2014
%t Table[n!*SeriesCoefficient[InverseSeries[Series[2*ArcTan[x]-x,{x,0,41}],x],{x,0,n}],{n,1,41,2}] (* _Vaclav Kotesovec_, Jan 02 2014 *)
%o (Maxima) a(n):=if n=0 then 1 else sum((2*n+k)!*sum(sum(((-1)^(n+l+j)*sum((2^i*stirling1(i,j-l)*binomial(2*n-l+j-1,i-1))/i!,i,j-l,2*n-l+j))/l!,l,0,j-1)/(k-j)!,j,1,k),k,1,2*n);
%K nonn
%O 0,2
%A _Vladimir Kruchinin_, Feb 05 2012
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