Sequences A120588 - A120607 : Formulas?

[Paul D. Hanna]

Seqfans,
      Final note regarding solutions for A(x) that satisfies:  
(1) r*A(x) = c + b*x + A(x)^n  
  
Max Alekseyev wrote:
> 
> > (2) A(x) = Sum_{i>=0} C(n*i,i)/(n*i-i+1)*(c+bx)^(n*i-i+1)/r^(n*i+1)
> [...]
> > Yet I am still interested in an explicit formula for a(n) ...
> 
> Why not simply express a(k) as the coefficient of x^k in (2)? That is
> 
> a(k) = Sum_{i>=0} C(n*i,i) * C(n*i-i+1,k) * b^k * c^(n*i-i+1-k) /
(n*i-i+1) / r^(n*i+1)
> 
> Max
 
I did not prefer that infinite series expression for a(n) because 
it converges quite slowly, and also I was looking for a finite sum.  
 
By inspecting the triangle of coefficients of t generated by: 
   A(x) = 1 + Series_Reversion[ (1 + (t+n)*x - (1+x)^n )/b ] 
it seems that a finite expression for a(n) is not far away. 
In the above, setting r=t+n simplifies the series rather substantially. 
 
Note that in the cases that I am studying, c=r-1 since A(0)=1.
 
Well, enough said on this subject. 
Thanks, 
     Paul 
 
P.S.:  
To derive the series reversion formula:
   A(x) = 1 + Series_Reversion[ (1 + r*x - (1+x)^n )/b ] 
from 
(1) r*A(x) = c + b*x + A(x)^n  
one may proceed as follows. 
 
Let A=A(x) and since A(0)=1 then set c=r-1 in (1): 
   r*A = r-1 + bx + A^n 
or 
   1 + r(A - 1) = bx + A^n .
 
Now let y = A-1 so that 
   x = [ (1 + ry - (1+y)^n )/b ] .
Thus
   y = Series_Reversion[ (1 + rx - (1+x)^n )/b ]
and so 
   A = 1 + Series_Reversion[ (1 + rx - (1+x)^n )/b ]. 
END. 
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