%I #21 Jan 26 2014 08:50:49

%S 1,2,12,128,1968,39488,977088,28742912,979744512,37968868352,

%T 1648597834752,79272057049088,4181485522464768,240067201819885568,

%U 14902137637759008768,994529776192394166272,71009035425186633940992,5401058272888913168433152,435991257271370763778916352

%N E.g.f. A(x) satisfies: A'(x) = A(x)^2 + A(x)^4.

%H Vaclav Kotesovec, <a href="/A228608/b228608.txt">Table of n, a(n) for n = 0..350</a>

%F E.g.f. A(x) satisfies:

%F (1) A(x) = exp( Integral A(x) + A(x)^3 dx ) with A(0)=1.

%F (2) A(x) = (1 + B(x))/(1 - B(x)) where B(x) = tan(1-x - 1/A(x)).

%F (3) log(A(x)) = Series_Reversion( 1-exp(-x) - atan(tanh(x/2)) ).

%F (4) A( 1-exp(-x) - atan(tanh(x/2)) ) = exp(x).

%F a(n) ~ n! / (GAMMA(1/3) * 3^(1/3) * n^(2/3) * (1-Pi/4)^(n+1/3)). - _Vaclav Kotesovec_, Jan 26 2014

%e E.g.f.: A(x) = 1 + 2*x + 12*x^2/2! + 128*x^3/3! + 1968*x^4/4! + 39488*x^5/5! +...

%e Related expansions.

%e A(x)^2 = 1 + 4*x + 32*x^2/2! + 400*x^3/3! + 6848*x^4/4! + 149056*x^5/5! +...

%e A(x)^4 = 1 + 8*x + 96*x^2/2! + 1568*x^3/3! + 32640*x^4/4! + 828032*x^5/5! +...

%e The logarithm of e.g.f. A(x) begins:

%e log(A(x)) = 2*x + 8*x^2/2! + 72*x^3/3! + 992*x^4/4! + 18336*x^5/5! +...

%e and equals Integral A(x) + A(x)^3 dx, where

%e A(x)^3 = 1 + 6*x + 60*x^2/2! + 864*x^3/3! + 16368*x^4/4! + 385344*x^5/5! +...

%t CoefficientList[Exp[InverseSeries[Series[1-Exp[-x]-ArcTan[Tanh[x/2]], {x, 0, 20}], x]],x]*Range[0, 20]! (* _Vaclav Kotesovec_, Dec 20 2013 *)

%o (PARI) /* Explicit formula: */

%o {a(n)=local(A,X=x+x^2*O(x^n));A=exp(serreverse(1-exp(-X) - atan(tanh(X/2))));n!*polcoeff(A,n)}

%o for(n=0,20,print1(a(n),", "))

%o (PARI) /* By definition: A'(x) = A(x)^2 + A(x)^4: */

%o {a(n)=local(A=1+x); for(i=1, n, A=1+intformal(A^2+A^4+x*O(x^n))); n!*polcoeff(A, n)}

%o for(n=0,20,print1(a(n),", "))

%o (PARI) /* From: A(x) = exp( Integral A(x) + A(x)^3 dx ): */

%o {a(n)=local(A=1+x); for(i=1, n, A=exp(intformal(A+A^3)+x*O(x^n))); n!*polcoeff(A, n)}

%o for(n=0,20,print1(a(n),", "))

%Y Cf. A112487, A124214.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Dec 18 2013