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E.g.f. satisfies A(x) = 1 - x*A(x)^2 * log(1 - x).
3

%I #15 Mar 12 2024 08:53:26

%S 1,0,2,3,56,270,5064,47040,984416,14116032,336538080,6589416240,

%T 179336461248,4446985514400,137520942168960,4112410749501600,

%U 143445512622458880,5004065722611594240,195260931334478223360,7762385328551718796800,336051947630616458065920

%N E.g.f. satisfies A(x) = 1 - x*A(x)^2 * log(1 - x).

%F E.g.f.: 2/(1 + sqrt(1+4*x*log(1-x))).

%F a(n) = n! * Sum_{k=0..floor(n/2)} (2*k)!/(k+1)! * |Stirling1(n-k,k)|/(n-k)!.

%F a(n) ~ sqrt(2 + 8*r^2/(1-r)) * n^(n-1) / (exp(n) * r^n), where r = 0.436224579489690436773045325306926562580857950193340891933383996... is the root of the equation 4*r*log(1-r) = -1. - _Vaclav Kotesovec_, Mar 12 2024

%t nmax = 20; CoefficientList[Series[(-1 + Sqrt[1 + 4*x*Log[1 - x]])/(2*x*Log[1 - x]), {x, 0, nmax}], x] * Range[0, nmax]! (* _Vaclav Kotesovec_, Mar 12 2024 *)

%o (PARI) my(N=30, x='x+O('x^N)); Vec(serlaplace(2/(1+sqrt(1+4*x*log(1-x)))))

%o (PARI) a(n) = n!*sum(k=0, n\2, (2*k)!/(k+1)!*abs(stirling(n-k, k, 1))/(n-k)!);

%Y Cf. A052830, A371141.

%Y Cf. A052803.

%K nonn

%O 0,3

%A _Seiichi Manyama_, Mar 12 2024