some help is needed Re: SEQ+# A127157 FROM Emeric Deutsch (fwd)

[Max Alekseyev]

On 2/28/07, Paul D. Hanna <pauldhanna at juno.com> wrote:

> Restating:
> (7) t*z/(1-z)^2 = A*( 1 - z/(1-z)*A )^2
>
>
> Now, from your nice formula for [t^k z^n] A(z,t):
> (2) [t^k z^n] = 2*binomial(3k-1,2k)*binomial(n-1+k,3k-2)/(3k-1)
> we can easily derive A to be:
>
> (8) A(z,t) = t*z/(1-z)^2 * F( t*z^2/(1-z)^3 )^2
>
> where F = 1 + x*F^3 is the g.f. of A001764 = [1,1,3,12,55,273,...].
>
>
> So the problem now is to obtain (8) from (7).

It is easier to go an opposite way around, i.e., to derive (7) from (8).

 From (8) we have
A(z,t) * z/(1-z) * F( t*z^2/(1-z)^3 ) = t*z^2/(1-z)^3 * F( t*z^2/(1-z)^3 )^3
Let x=t*z^2/(1-z)^3, then
A(z,t) * z/(1-z) * F(x) = x * F(x)^3 = F(x) - 1
by the definition of the function F.
 From the last identity we can express
F(x) = 1 / (1 - A(z,t) * z/(1-z))
and substituting this into (8), we have:
A(z,t) = t*z/(1-z)^2 / (1 - A(z,t) * z/(1-z))^2
which is equivalent to (7).

Max