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

[Paul D. Hanna]

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.