%I #10 Dec 23 2018 12:46:11
%S 1,1,2,14,210,5572,245224,16484608,1592692724,211735948032,
%T 37486076895064,8611994418091904,2512364155208956104,
%U 913526595412940173952,407407936880027138109376
%N Numerators of coefficients of a formal power series solution of f''(x) = f(f(x)).
%C Define f(x)=sum for n>=0 of a(n)/(2n+1)!*x^(2n+1). Formally this satisfies f''(x) = f(f(x)), but the series diverges.
%H sci.math thread, 2/18/02 by William Elliot: <a href="https://groups.google.com/forum/?hl=en#!searchin/sci.math/x5lmdpx5jn.fsf@tupik.goethe.zz%7Csort:date">f''(x)=f(f(x))</a>
%e f(x) = x + 1/6*x^3 + 2/120*x^5 + 14/5040*x^7 + ...
%t b[1]=1; b[n_] := Module[{f, bn}, If[EvenQ[n], Return[b[n]=0]]; f=Series[Sum[b[k]*x^k, {k, 1, n-2, 2}]+bn*x^n, {x, 0, n}]; b[n]=Solve[Coefficient[D[f, {x, 2}]-(f/.x->f), x, n-2]==0, bn][[1, 1, 2]]]; a[n_] := (2n+1)!b[2n+1]
%K nonn
%O 0,3
%A Joe Keane (jgk(AT)jgk.org), Mar 01 2002
%E Edited by _Dean Hickerson_, Aug 06 2002