%I #16 Oct 12 2017 19:18:12
%S 1,1,4,18,112,880,8256,90384,1131264,15927552,249164800,4287669760,
%T 80490393600,1636924403712,35850727342080,841260590499840,
%U 21056773882052608,559992309313503232,15768699458743959552
%N E.g.f. satisfies: A(x) = 1 + x*exp(2*Integral A(x) dx).
%C Compare the definition of e.g.f. A(x) to this trivial statement:
%C if F(x) = 1/(1-2x) then F(x) = 1 + 2*x*exp(2*Integral F(x) dx).
%F E.g.f. satisfies: A'(x) = [1 + 2*x*A(x)]*(A(x) - 1)/x where A'(x) = d/dx A(x).
%F E.g.f.: (1+exp(2*x)) / (1+exp(2*x)*(1-2*x)). - _Vaclav Kotesovec_, Jan 05 2014
%F a(n) ~ n! * 2^n / (1 + LambertW(exp(-1)))^(n+1). - _Vaclav Kotesovec_, Jan 05 2014
%F E.g.f.: -1/E(0), where E(k)= 4*k-1 + x/(1 - x/(4*k+1 + x/(1 - x/E(k+1) ))); (continued fraction). - _Sergei N. Gladkovskii_, Jan 22 2015
%e E.g.f. A(x) = 1 + x + 4*x^2/2! + 18*x^3/3! + 112*x^4/4! +...
%e CONVERGENCE AND ASYMPTOTICS.
%e Let r be the radius of convergence of the power series A(x), then:
%e a(n)/n! ~ (1/2)/r^(n+1) where
%e r=0.63923227138053689755467936951149007771973874430987288272658905276...
%e so that the power series A(x) diverges at x=r.
%e Note: A(-r) is evaluated as 1/(2r) since Integral A(x) dx is a
%e convergent alternating series at x=-r having the sum:
%e Sum_{n>=0} a(n)*(-r)^(n+1)/(n+1)! = log(r - 1/2)/2 - log(r);
%e however, as N approaches infinity, the N-th partial sum of A(x) at x=-r,
%e Sum_{n>=0..N} a(n)*(-r)^n/n!, oscillates between 1/(4r) and 3/(4r).
%e Thus the power series A(x) converges only for |x| < r.
%e In closed form, r = 1/2 + LambertW(exp(-1))/2. - _Vaclav Kotesovec_, Jan 05 2014
%t CoefficientList[Series[(1+E^(2*x))/(1+E^(2*x)*(1-2*x)), {x, 0, 20}], x] * Range[0, 20]! (* _Vaclav Kotesovec_, Jan 05 2014 *)
%t max = 20; Clear[g]; g[max + 2] = 1; g[k_] := g[k] = 4*k-1 + x/(1 - x/(4*k+1 + x/(1 - x/g[k+1]))); gf = -1/g[0]; CoefficientList[Series[gf, {x, 0, max}], x] * Range[0, max]! (* _Vaclav Kotesovec_, Jan 22 2015, after _Sergei N. Gladkovskii_ *)
%o (PARI) {a(n)=local(A=1+x+x*O(x^n)); for(i=0,n,A=1+x*exp(2*intformal(A)+x*O(x^n)));n!*polcoeff(A,n)}
%K nonn
%O 0,3
%A _Paul D. Hanna_, Sep 05 2008
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