Functional Puzzle

[Paul D. Hanna]

Seqfans, 
     I wish to state the conjecture more generally. 
 
Suppose functions {A(x,n), n=0..infinity} satisfy: 
   A(x,n) = 1 + x*[Product_{k=0..n} A(x,k)]^m*A(x,n+1) 
then 
   A(x, n+1) = A(x*A(x,0)^m, n)  for n>=0 
where G(x) = A(x,0) satisfies: 
   G(x) = 1 + x*G(x)^m*G(x*G(x)^m). 
   
Below I give an example for m=2.
    Paul

EXAMPLE: m=2. 
Suppose A=A(x), B=B(x), C=C(x),... satisfy: 
  A = 1 + xA^2*B 
  B = 1 + x(AB)^2*C 
  C = 1 + x(ABC)^2*D 
  D = 1 + x(ABCD)^2*E 
  ...
then 
  B(x) = A(x*A(x)^2) 
  C(x) = B(x*A(x)^2) 
  D(x) = C(x*A(x)^2)
  ...
where A(x) satisfies:
  A(x) = 1 + x*A(x)^2*A(x*A(x)^2)
 
and is the g.f. of A088717: 
[1, 1, 3, 14, 84, 596, 4785, 42349, 406287, 4176971, ...].
END.
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