[seqfan] Triangle T(n, k) = C(n, k)^m (m-th Powers of Binomial Coefficients)

[Paul D Hanna]

SeqFans,
      In researching series related to Franel numbers (A000172), I was surprised to find that the triangle of cubed binomial coefficients, T(n,k) = C(n,k)^3, was not in the OEIS.  
I have submitted the triangle as A181543, and arrived at this formula: 
 
(1) G.f.: A(x,y) = Sum_{n>=0} (3n)!/n!^3 * x^(2n)*y^n/(1-x-x*y)^(3n+1). 
 
We can see the cofficients C(n,k)^3 in the expansion of (1):  
A(x,y) = 1 + 
(y + 1)*x + 
(y^2 + 8*y + 1)*x^2 + 
(y^3 + 27*y^2 + 27*y + 1)*x^3 + 
(y^4 + 64*y^3 + 216*y^2 + 64*y + 1)*x^4 + 
(y^5 + 125*y^4 + 1000*y^3 + 1000*y^2 + 125*y + 1)*x^5 + 
(y^6 + 216*y^5 + 3375*y^4 + 8000*y^3 + 3375*y^2 + 216*y + 1)*x^6 +...
 
However, I was unable to find such a g.f. for the triangle T(n,k) = C(n,k)^m where m>3. 
 
Can anyone find a similar g.f. for triangle T(n,k) = C(n,k)^4 ? 
 
 
>From (1), we now have this g.f. for Franel numbers (A000172): 
  
(2) Sum_{n>=0} A000172(n)*x^n = Sum_{n>=0} (3n)!/n!^3 * x^(2n)/(1-2x)^(3n+1) 
 
where A000172(n) = Sum_{k=0..n} C(n,k)^3 forms the row sums of A181543. 
  
 
Also, (2) leads me to a rather complex functional equation for the g.f. of the Franel numbers (A000172): 
 
(3) G.f.: A(x) = G( x^2/(1-2*x)^3 )/(1-2*x) where G(x) satisfies:
    G(x^3) = G( x*(1+3*x+9*x^2)/(1+6*x)^3 )/(1+6*x) 
    and G(x) is the g.f. of A006480.
 
Can anyone improve/simplify the functional equation of (3)? 
 
Thanks, 
    Paul