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%I #15 Nov 18 2023 05:17:09
%S 1,1,5,43,553,9501,204961,5330599,162432593,5677941817,224018814241,
%T 9848702243931,477481361216377,25309471236379669,1456206709854725921,
%U 90387017392004356591,6020486941130334199201,428348710658269120403313
%N E.g.f. satisfies A(x) = exp( x/(1 - x*A(x))^2 ).
%H Seiichi Manyama, <a href="/A161635/b161635.txt">Table of n, a(n) for n = 0..360</a>
%F a(n) = n!*Sum_{k=0..n} (n-k+1)^(k-1)/k! * C(n+k-1,n-k).
%F Let A(x)^m = Sum_{n>=0} a(n,m)*x^n/n!, then
%F a(n,m) = n!*Sum_{k=0..n} m*(n-k+m)^(k-1)/k! * C(n+k-1,n-k).
%F ...
%F E.g.f.: A(x) = (1/x)*Series_Reversion[ (1-x)^2*LambertW(x/(1-x)^2) ].
%F a(n) ~ sqrt(s*(1+r*s)/(3+2*r-6*r*s+3*r^2*s^2)) * n^(n-1) / (exp(n) * r^n), where r = 0.219092370374602436272454670371..., s = 1.952248277910295452167538973654... are the roots of the equations 2*s*r^2 = (1-r*s)^3 and r = log(s) * (1-r*s)^2. - _Vaclav Kotesovec_, Jan 10 2014
%e E.g.f: A(x) = 1 + x + 5*x^2/2! + 43*x^3/3! + 553*x^4/4! + 9501*x^5/5! +...
%e log(A(x))/x = 1 + 2*x*A(x) + 3*x^2*A(x)^2 + 4*x^3*A(x)^3 + 5*x^4*A(x)^4 +...
%t Flatten[{1,Table[n!*Sum[(n-k+1)^(k-1)/k! * Binomial[n+k-1,n-k],{k,0,n}],{n,1,20}]}] (* _Vaclav Kotesovec_, Jan 10 2014 *)
%t FindRoot[{2*s*r^2 == (1-r*s)^3, r == Log[s]*(1-r*s)^2},{r,1/2},{s,1}, WorkingPrecision->50] (* program for numerical values of constants r and s, _Vaclav Kotesovec_, Jan 10 2014 *)
%o (PARI) {a(n,m=1)=if(n==0,1,sum(k=0,n,n!/k!*m*(n-k+m)^(k-1)*binomial(n+k-1,n-k)))}
%o (PARI) {a(n,m=1)=my(A=1+x+x*O(x^n));for(i=1,n,A=exp(x/(1-x*A)^2));n!*polcoeff(A^m,n)}
%Y Cf. A161630, A364980.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Jun 19 2009
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