this section is to prove the following relation between the two generating functions for colored permutations:

\[relAC\] $ C^{(r)}(x,y,z) = e^{(1-z)y^rx} A^{(r)}(x,y,z)$.

Let us define $$W_n(y,z) := \sum_{(\sigma,c) \in \mathfrak{S}_n^{(r)}}  y^{ \operatorname{fexc}(\sigma,c)} z^{\operatorname{fdes}^*(\sigma,c)}.$$ The particular case $q=1$ of [@foatahan Theorem 5.11], after an easy simplification, gives the following formula: $$\label{formulawn}
  \sum_{n\geq 0  }  W_n(y,z) \frac{x^n}{(1-z^r)^n}  = (1-z)  \sum_{ k \geq 0 }  z^k F_k(x,y) ,$$ where $$F_k(x,y) = \frac{ (1-xy^r)^{\lfloor k/r\rfloor} } {(1-x)^{\lfloor k/r\rfloor+1}} (1-y^r)
 \times
\Bigl(  \frac{1-y^r}{1-y} - \sum_{i=1}^{r} y^i 
{  \frac{ (1-xy^r)^{\lfloor (k-i)/r\rfloor+1} } { (1-x)^{\lfloor (k-i)/r\rfloor+1}} } \Bigr)^{-1}.$$ Next, we define a linear operator $\beta$ on power series in $z$ by $\beta(z^k)= z^{\lceil k/r \rceil }$. So: $$\beta( z^k - z^{k+1} )  = \begin{cases}
                                0 & \text{ if } k \not\equiv 0 \mod r, \\
                                z^m-z^{m+1} & \text{ if } k=rm.
                             \end{cases}$$ From Equation , we get $$\label{formulabetawn}
  \sum_{n\geq 0  }  \beta( W_n(y,z) ) \frac{x^n}{(1-z)^n}  = \sum_{ m \geq 0 } ( z^m - z^{m+1} ) F_{rm}(x,y),$$ and from the definition of $F_k$ we get $$\begin{aligned}
  F_{rm}(x,y) &= 
     \frac{(1-xy^r)^m}{ (1-x)^{m+1}} (1-y^r)
\Bigl(
 \frac{1-y^r}{1-y}
- \sum_{i=1}^{r}
y^i \frac {(1-xy^r)^{m} }
  { (1-x)^{m}} \Bigr)^{-1} 
\\
 & = \Bigl(  \frac{(1-x)^{m+1}}{(1-xy^r)^m(1-y)}  - \sum_{i=1}^r y^i \frac{1-x}{1-y^r}   \Bigl)^{-1}
\\
 & = \frac{1-y}{1-x}
 \Bigl(
 