Formula for A = (1 + x*(1 + x*A)^q)^p ?

[pauldhanna at juno.com]

Seqfans, 
    Can someone find a formula for a(n,p,q), the coefficient of x^n 
of the g.f. A(x,p,q) that satisfies: 
 
A(x,p,q) = (1 + x*(1 + x*A(x,p,q) )^q )^p 
 
Note that when q=p, then we have the simple formula: 
 
a(n,p,p) = C(p*(n+1),n)/(n+1) 
 
but what is a(n,p,q) in general?  
I sense there must be a nice formula in terms of n, p, and q. 
 
 
Example 1. p=2, q=3: 
A(x) = (1 + x*(1 + x*A(x))^3)^2 is the g.f. of:
1, 2, 7, 24, 95, 386, 1641, 7150, 31844, 144216, 662228, ...
 
Example 2. p=3, q=2: 
A(x) = (1 + x*(1 + x*A(x))^2)^3 is the g.f. of:
1, 3, 9, 34, 132, 546, 2327, 10191, 45534, 206788, 951723,...
 

Example 3. p=2, q=4: 
A(x) = (1 + x*(1 + x*A(x))^4)^2 is the g.f. of:
1, 2, 9, 36, 172, 842, 4310, 22676, 121896, 666884,...
 
Example 4. p=4, q=2: 
A(x) = (1 + x*(1 + x*A(x))^2)^4 is the g.f. of:
1, 4, 14, 64, 301, 1500, 7738, 40948, 221278, 1215284,...


Below are the initial terms: n!*a(n,p,q), n=0..6. 
 
Thanks, 
   Paul 
 
A=1+x;for(n=0,20,A = (1 + x*(1+x*A)^q +O(x^21))^p);B=Vec(A);
for(n=1,7,print(B[n]*(n-1)!);print(""))

1

p

2*p*q + (p^2 - p)

3*p*q^2 + (12*p^2 - 9*p)*q + (p^3 - 3*p^2 + 2*p)

4*p*q^3 + (72*p^2 - 36*p)*q^2 + (48*p^3 - 108*p^2 + 44*p)*q + (p^4 -
6*p^3 + 11*p^2 - 6*p)

5*p*q^4 + (320*p^2 - 110*p)*q^3 + (810*p^3 - 1290*p^2 + 355*p)*q^2 +
(160*p^4 - 750*p^3 + 870*p^2 - 250*p)*q + (p^5 - 10*p^4 + 35*p^3 - 50*p^2
+ 24*p)

6*p*q^5 + (1200*p^2 - 300*p)*q^4 + (8640*p^3 - 10080*p^2 + 2010*p)*q^3 +
(6480*p^4 - 21600*p^3 + 17700*p^2 - 3480*p)*q^2 + (480*p^5 - 4020*p^4 +
9360*p^3 - 7320*p^2 + 1644*p)*q + (p^6 - 15*p^5 + 85*p^4 - 225*p^3 +
274*p^2 - 120*p)

...
END.
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