G.f. for a(n) = C(2n,n)*Sum_{k=0..2n} T(n,k)/C(2n,k) ?

[Max Alekseyev]

On 8/18/07, Max Alekseyev <maxale at gmail.com> wrote:
> On 8/18/07, Paul D. Hanna <pauldhanna at juno.com> wrote:
>
> > By inspection, the formula for the g.f. of (1), when c=1 is:
> >
> > (2) G.f.: A(x) = 1/sqrt(1 + d*x + e*x^2 + f*x^3)  where
> > d = -2*b - 8 ;
> > e =  b^2 + 12*b + 20 ;
> > f = -4*b^2 - 16*b - 16 ;
> > (proof, anyone?)
>
> That follows from the general formula for o.g.f.
> I've verified the identity in maple.

Actually, the square root in the formula above can be simplified:

sqrt(1 + d*x + e*x^2 + f*x^3) = (1-b*x-2*x) * sqrt(1-4*x)

Correspondingly, the o.g.f. (2) is simply

A(x) = 1 / ( (1-b*x-2*x) * sqrt(1-4*x) )

Max