%I #36 Sep 25 2024 09:34:40
%S 1,1,4,40,664,15424,460576,16808320,724904896,36072438016,
%T 2034328297984,128223244372480,8932539799788544,681536817951791104,
%U 56521548341146402816,5062454448656689500160,487013865350356256137216,50082306316236214342844416,5482502331779770770018893824
%N E.g.f. satisfies: A(x) = 1 + A(x)^4 * Integral 1/A(x)^4 dx.
%F E.g.f.: 1 + Series_Reversion( 4*log(1+x) - 3*x ).
%F E.g.f.: -4/3*LambertW(-3/4*exp((x-3)/4)).
%F E.g.f.: 1 / ( d/dx Series_Reversion( Integral G(x)^4 dx ) )^(1/4), where G(x) = 1 + x*G(x)^4 is the g.f. of A002293.
%F E.g.f.: 1 / sqrt( d/dx Series_Reversion( Integral (1+2*x*C(2*x))^2/(1+x)^2 dx ) ), where C(x) = 1 + x*C(x)^2 = (1 - sqrt(1-4*x))/(2*x), is the Catalan function of A000108.
%F O.g.f.: 1 + x/(1-x - 3*x/(1-2*x - 3*2*x/(1-3*x - 3*3*x/(1-4*x - 3*4*x/(1-...))))), a continued fraction.
%F a(n) ~ 2 * n^(n-1) / (3 * exp(n) * (8*log(2)-4*log(3)-1)^(n-1/2)). - _Vaclav Kotesovec_, Dec 26 2013
%e E.g.f.: A(x) = 1 + x + 4*x^2/2! + 40*x^3/3! + 664*x^4/4! + 15424*x^5/5! +...
%e where A(4*log(1+x) - 3*x) = 1+x.
%e Related series:
%e A(x)^4 = 1 + 4*x + 28*x^2/2! + 328*x^3/3! + 5752*x^4/4! + 137056*x^5/5! +...
%e 1/A(x)^4 = 1 - 4*x + 4*x^2/2! - 40*x^3/3! - 536*x^4/4! - 13216*x^5/5! +...
%e (d/dx Series_Reversion(Integral 1/A(x)^4 dx))^(1/4) begins:
%e G(x) = 1 + x + 4*x^2 + 22*x^3 + 140*x^4 + 969*x^5 +...+ A002293(n)*x^n +...
%e where G(x) = 1 + x*G(x)^4.
%t CoefficientList[1 + InverseSeries[Series[4*Log[1+x]-3*x, {x, 0, 20}], x],x]* Range[0, 20]! (* _Vaclav Kotesovec_, Dec 26 2013 *)
%o (PARI) {a(n)=local(A=1+x); for(i=1, n, A=1+A^4*intformal(1/(A^4+x*O(x^n)))); n!*polcoeff(A, n)}
%o for(n=0, 25, print1(a(n), ", "))
%o (PARI) {a(n)=local(A=1, X=x+x^2*O(x^n)); A=1+serreverse(4*log(1+X) - 3*X); n!*polcoeff(A, n)}
%o for(n=0, 25, print1(a(n), ", "))
%o (PARI) /* O.g.f. continued fraction: */
%o {a(n)=local(CF=1+x*O(x)); for(k=0, n, CF=1-(n-k+1)*x-3*(n-k+1)*x/CF); polcoeff(1+x/CF, n, x)}
%o for(n=0, 25, print1(a(n), ", "))
%Y Cf. A006351, A058562, A234295.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Dec 24 2013