Functional Puzzle
Paul D. Hanna
pauldhanna at juno.com
Sat Mar 10 05:33:33 CET 2007
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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