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E.g.f. A(x) satisfies: d/dx A(x) = 1 + A(x/(1 + x)).
1

%I #15 May 04 2019 21:51:41

%S 1,1,-1,-1,23,-197,1093,5377,-374863,9934889,-195976201,2134159519,

%T 67270069831,-6730482201869,365726633654957,-15494292929032063,

%U 458469986808144737,2828723973314497873,-2067360599320208561297,238687490584532161293631,-20425982116564721266720009

%N E.g.f. A(x) satisfies: d/dx A(x) = 1 + A(x/(1 + x)).

%F Recurrence: a(n+1) = Sum_{k=1..n} (-1)^(n-k)*binomial(n-1,k-1)*a(k)*n!/k!.

%t terms = 21; A[_] = 0; Do[A[x_] = Normal[Integrate[1 + A[x/(1 + x) + O[x]^(terms + 1)], x] + O[x]^(terms + 1)], terms]; Rest[CoefficientList[A[x], x] Range[0, terms]!]

%t a[n_] := a[n] = Sum[(-1)^(n - k - 1) Binomial[n - 2, k - 1] a[k] (n - 1)!/k!, {k, 1, n - 1}]; a[1] = 1; Table[a[n], {n, 1, 21}]

%Y Cf. A001063.

%K sign

%O 1,5

%A _Ilya Gutkovskiy_, May 04 2019