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E.g.f. satisfies: A(x) = exp(x*A(((x+1)^4-1)/4)).
3

%I #9 Dec 20 2014 13:47:20

%S 1,1,3,25,329,6471,175747,6222259,277683681,15206462497,1000136567591,

%T 77666331244239,7021789807671817,730394622232111747,

%U 86529393614846902371,11573498785704862459891,1734360074041552070631713

%N E.g.f. satisfies: A(x) = exp(x*A(((x+1)^4-1)/4)).

%H Alois P. Heinz, <a href="/A143635/b143635.txt">Table of n, a(n) for n = 0..100</a>

%p A:= proc(n,k::nonnegint) option remember; if n<=0 or k=0 then 1 else A(n-1,k)(((x+1)^k-1)/k) fi; unapply(convert(series(exp(x*%), x,n+1), polynom), x) end: a:= n-> coeff(A(n,4)(x), x,n)*n!: seq(a(n), n=0..21);

%t A[n_, k_] := A[n, k] = Module[{f}, f = Function[If[n <= 0 || k == 0, 1, A[n-1, k][((#+1)^k-1)/k]]]; Function[Normal[Series[Exp[x*f[x]], {x, 0, n+1}]] /. x -> #]]; a[n_] := SeriesCoefficient[A[n, 4][x], {x, 0, n}]*n!; Table[a[n], {n, 0, 21}] (* _Jean-François Alcover_, Feb 13 2014, after Maple *)

%Y Cf. 4th column of A143632.

%K nonn

%O 0,3

%A _Alois P. Heinz_, Aug 27 2008