\bigl( \frac{1-x}{1-xy^r}\bigr)^m
-y\Bigr)^{-1}.\end{aligned}$$ From Equation  and the previous equation, and after the substitution $(x,y)  \leftarrow  (xy^r(1-z) , y^{-1} ) $, we reach:

\[ogfC\] $$\sum_{n\geq 0 } C_n^{(r)}(y,z) x^n = 
  \frac{(1-z)(1-y)}{1-xy^r(1-z)}
 \sum_{m\ge 0}z^m\,
 \Bigl(1-
 y\bigl( \frac {1-xy^r(1-z) }{ 1-x(1-z)} \bigr)^m
 \Bigr)^{-1}.$$

Besides, from Equation , we have: $$\begin{aligned}
    \sum_{n\geq0} x^n \frac{A_n^{(r)}(y,z)}{ (1-y^r)^n(1-y)(1-z)^{n+1}}
      &=\sum_{i,j \geq0 } \Bigl(\frac{1}{1-x}\Bigr)^{ij} y^i z^j
       =  \sum_{j\geq0} z^j \Bigl(1-y\bigl( \frac{1}{1-x}\bigr)^j   \Bigr)^{-1}.\end{aligned}$$ After the substition $x\leftarrow x(1-y^r)(1-z)$, we obtain:

\[ogfA\] $$\sum_{n\geq0} {A_n^{(r)}(y,z)} x^n =
 (1-y)(1-z)\sum_{m\geq 0} z^m \Bigl(1-y\bigl( \frac{1}{1-x(1-y^r)(1-z)}\bigr)^m   \Bigr)^{-1}.$$

Since Theorem \[relAC\] is a relation on exponential generating functions, it is convenient to use the Laplace transform. It sends a function $f(x)$ to $$\mathcal{L}(f(x),x,s) = \int_0^\infty f(x) e^{-xs} {\rm d}x,$$ in particular, $$\mathcal{L} \Big( \frac{x^k}{k!} , x , s \Big) = \frac{1}{s^{k+1}}.$$ We have: $$\begin{aligned}
  \mathcal{L} \big( e^{(1-z)y^rx} A^{(r)}(x,y,z) , x, s \big)  & = \int_0^{\infty} e^{(1-z)y^rx} A^{(r)}(x,y,z) e^{-xs} {\rm d}x \\
                                                               & = \int_0^{\infty}  A^{(r)}(x,y,z) e^{(1-z)y^rx-xs} {\rm d}x \\
                                                               & = \mathcal{L}\big( A^{(r)}(x,y,z) , x , s-(1-z)y^r \big).\end{aligned}$$ By Theorem \[ogfA\], with $s'=s-(1-z)y^r$, the latter expression is equal to $$\frac{(1-y)(1-z)}{s'}
  \sum_{m\geq0} z^m \Bigl(1-y\bigl(\frac{1}{1-\frac{1}{s'}(1-y^r)(1-z)}\bigr)^m   \Bigr)^{-1}.$$ Since $$\frac{1}{1-\frac{1}{s'}(1-y^r)(1-z)}
=
\frac{1}{1-\frac{1}{s-(1-z)y^r}(1-y^r)(1-z)}
=
\frac{s-(1-z)y^r}{s-(1-z)},$$ we get: $$\mathcal{L} \big( e^{(1-z)y^rx} A^{(r)}(x,y,z) , x, s \big) = 
  \frac{(1-y)(1-z)}{s-(1-z)y^r}\sum_{m\geq 0} z^m \Bigl(1-y\bigl(\frac{s-(1-z)y^r}{s-(1-z)}\bigr)^m   \Bigr)^{-1}.$$ Besides, 