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

[Max Alekseyev]

I've got the following o.g.f. for a(n) but I was not able to simplify it much:

SUM a(n) * t^n =

(1+sqrt(1-4*t)) * (-b+sqrt(b^2-4*c)) / sqrt(1-4*t) / ( 8*t*c - 2*b^2*t
+ b^2 + b^2*sqrt(1-4*t) + 2*b*t*sqrt(b^2-4*c) - b*sqrt(b^2-4*c) -
b*sqrt((b^2-4*c)*(1-4*t)) - b - b*sqrt(1-4*t) - 2*c - 2*sqrt(1-4*t)*c
+ sqrt(b^2-4*c) + sqrt((b^2-4*c)*(1-4*t)) )

+

(1+sqrt(1-4*t*c)) * (-b+sqrt(b^2-4*c)) / sqrt(1-4*t*c) / ( 2*b^2*t -
2*b*t*sqrt(b^2-4*c) - 8*t*c - b - b*sqrt(1-4*t*c) + sqrt(b^2-4*c) +
sqrt((b^2-4*c)*(1-4*t*c)) + 2 + 2*sqrt(1-4*t*c) )

Regards,
Max

On 8/18/07, Paul D. Hanna <pauldhanna at juno.com> wrote:
> Seqfans,
>      Can anyone provide a general formula (by inspection) for
> the g.f. of:
>
> (1) a(n) = C(2n,n) * Sum_{k=0..2n} T(n,k) / C(2n,k)
> where T(n,k) = [x^k] (1 + b*x + c*x^2)^n.
>
> See A132310 (copied below) for the special case b=c=1.
>
> I suspect that for integer b, c, that the g.f. will be of the form:
>
> (2) G.f.: A(x) = 1/sqrt(1 + d*x + e*x^2 + f*x^3).
>
> Objective: from integers b, c, in (1) find d, e, and f in (2).
>
> I do not know the formula, but (1) generates various sequences
> already in the OEIS.
> Thanks,
>      Paul
> ------------------------------------------------------
> A132310
>
> a(n) = C(2n,n) * Sum_{k=0..2n} trinomial(n,k) / C(2n,k)
> where trinomial(n,k) = [x^k] (1 + x + x^2)^n.
>
> 1,5,21,83,319,1209,4551,17085,64125,240995,907741,3428655,12990121,
> 49370963,188229489,719805987,2760498351,10615101273,40920439119,
> 158106581157,612166272291,2374756691313,9228369037659,35918537840577,
>
> FORMULA.
> G.f.: A(x) = 1/sqrt(1 - 10*x + 33*x^2 - 36*x^3).
>
> EXAMPLE.
> a(1) = C(2,1)*(1/1 + 1/2 + 1/1) = 2*(5/2) = 5 ;
> a(2) = C(4,2)*(1/1 + 2/4 + 3/6 + 2/4 + 1/1) = 6*(7/2) = 21 ;
> a(3) = C(6,3)*(1/1 + 3/6 + 6/15 + 7/20 + 6/15 + 3/6 + 1/1) = 20*(83/20) =
> 83.
>
> END.
>