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A072350 E.g.f. A(x) satisfies A(A(x)) = tan(x), where A(x) = Sum_{n>=1} a(n)*x^(2n-1)/(2n-1)!.

%I A072350 #28 Jan 15 2024 11:06:42
%S A072350 1,1,3,17,225,3613,-42997,8725357,2116966081,-549193907111,
%T A072350 -114757574954509,117893333517545097,14433599120070484321,
%U A072350 -65568697910890921624715,2968238619232726100394235,86999609037195113208781248165
%N A072350 E.g.f. A(x) satisfies A(A(x)) = tan(x), where A(x) = Sum_{n>=1} a(n)*x^(2n-1)/(2n-1)!.
%C A072350 The inverse of this g.f. A(x) is the g.f. of A095885. - _Paul D. Hanna_, Dec 09 2004
%H A072350 Paul D. Hanna, <a href="/A072350/b072350.txt">Table of n, a(n) for n = 1..40</a>
%H A072350 Dmitry Kruchinin, Vladimir Kruchinin, <a href="http://arxiv.org/abs/1302.1986">Method for solving an iterative functional equation A^{2^n}(x)=F(x)</a>, arXiv:1302.1986 [math.CO], 2013.
%F A072350 a(n)=(2*n-1)!*T(2*n-1,1), T(n,k)=if n=k then 1 else 1/2*(T059419(n,k)*k!/n!-sum(i=k+1..n-1, T(n,i)*T(i,k))). [_Vladimir Kruchinin_, Nov 11 2011]
%e A072350 a(x) = x/1!+x^3/3!+3*x^5/5!+17*x^7/7!+225*x^9/9!+3613*x^11/11!-42997*x^13/13!+...
%t A072350 a[n_] := Module[{A, B, F}, F = Tan[x + O[x]^(2n+1)]; A = F; For[i = 0, i <= 2n-1, i++, B = InverseSeries[A, x]; A = (A + (B /. x -> F))/2]; If[n<1, 0, (2n-1)!*SeriesCoefficient[A, {x, 0, 2n-1}]]]; Table[a[n], {n, 1, 16}] (* _Jean-François Alcover_, Oct 29 2015, adapted from PARI *)
%o A072350 (PARI) {a(n)=local(A,B,F);F=tan(x+O(x^(2*n+1)));A=F; for(i=0,2*n-1,B=serreverse(A);A=(A+subst(B,x,F))/2); if(n<1,0,(2*n-1)!*polcoeff(A,2*n-1,x))} \\ _Paul D. Hanna_, Dec 09 2004
%Y A072350 Cf. A048602, A048603, A052134, A052135.
%Y A072350 Cf. A095885 (inverse).
%K A072350 sign
%O A072350 1,3
%A A072350 _Vladeta Jovovic_, Jul 17 2002
%E A072350 More terms from _Paul D. Hanna_, Dec 09 2004