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%I #7 Jan 10 2014 17:03:31
%S 1,2,9,68,720,9804,163184,3210192,72870120,1874721360,53905894152,
%T 1713195438624,59633476003920,2256257009704320,92196226214092800,
%U 4046446853549201664,189845257963376620800,9481546020840245199360,502242773970728703225600,28124368575613839072714240
%N E.g.f. equals the series reversion of x - x*log(1+x).
%F a(n) = Sum_{k=0..n-1} k! * Stirling1(n-1,k) * binomial(n+k-1,n-1). [From a formula in A052819 due to _Vladimir Kruchinin_]
%F E.g.f. A(x) satisfies:
%F (1) A(x - x*log(1+x)) = x.
%F (2) A(x) = x/(1 - log(1+A(x))).
%F (3) A(x) = x + Sum_{n>=1} d^(n-1)/dx^(n-1) x^n * log(1+x)^n / n!.
%F (4) A(x) = x*exp( Sum_{n>=1} d^(n-1)/dx^(n-1) x^(n-1) * log(1+x)^n / n! ).
%F a(n) ~ n^(n-1) * (1-c) / (c*sqrt(1+c) * exp(n) * (c-2+1/c)^n), where c = LambertW(1) = 0.5671432904... (see A030178). - _Vaclav Kotesovec_, Jan 10 2014
%e E.g.f.: A(x) = x + 2*x^2/2! + 9*x^3/3! + 68*x^4/4! + 720*x^5/5! +...
%e where A(x) = x/(1 - log(1+A(x))).
%e The e.g.f. satisfies:
%e (3) A(x) = x + x*log(1+x) + d/dx x^2*log(1+x)^2/2! + d^2/dx^2 x^3*log(1+x)^3/3! + d^3/dx^3 x^4*log(1+x)^4/4! +...
%e (4) log(A(x)/x) = log(1+x) + d/dx x*log(1+x)^2/2! + d^2/dx^2 x^2*log(1+x)^3/3! + d^3/dx^3 x^3*log(1+x)^4/4! +...
%t Rest[CoefficientList[InverseSeries[Series[x - x*Log[1+x],{x,0,20}],x],x] * Range[0,20]!] (* _Vaclav Kotesovec_, Jan 10 2014 *)
%o (PARI) {a(n)=n!*polcoeff(serreverse(x-x*log(1+x +x*O(x^n))), n)}
%o for(n=1,25,print1(a(n),", "))
%o (PARI) {Dx(n, F)=local(D=F); for(i=1, n, D=deriv(D)); D}
%o {a(n)=local(A=x); A=x+sum(m=1, n, Dx(m-1, x^m*log(1+x+x*O(x^n))^m/m!)); n!*polcoeff(A, n)}
%o for(n=1,25,print1(a(n),", "))
%o (PARI) {Dx(n, F)=local(D=F); for(i=1, n, D=deriv(D)); D}
%o {a(n)=local(A=x+x^2+x*O(x^n)); A=x*exp(sum(m=1, n, Dx(m-1, x^(m-1)*log(1+x+x*O(x^n))^m/m!)+x*O(x^n))); n!*polcoeff(A, n)}
%o for(n=1,25,print1(a(n),", "))
%o (PARI) {Stirling1(n, k)=n!*polcoeff(binomial(x, n), k)}
%o {a(n)=sum(k=0,n-1,k!*Stirling1(n-1,k)*binomial(n+k-1,n-1))}
%o for(n=1,25,print1(a(n),", "))
%Y Cf. A052819, A030178.
%K nonn
%O 1,2
%A _Paul D. Hanna_, Jul 12 2013
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