\sum_{i,j \geq 0} \binom{ij+n-1}{n} y^i z^j.$$

Let $\text{Par}(n)$ denote the set of integer partitions of $n$. By the Robinson-Schensted correspondence, we have: $$\begin{aligned}
  A^{(1)}_n(y,z) &= \sum_{\sigma \in \mathfrak{S}_n } y^{\operatorname{des}(\sigma)+1} z^{\operatorname{des}(\sigma^{-1})+1} \\
                 &= \sum_{ \lambda \in {\rm Par}(n) } \sum_{P,Q \in SYT(\lambda)} y^{\operatorname{des}(P)+1} z^{\operatorname{des}(Q)+1}.\end{aligned}$$

So, using Equation , we get: $$\frac{A^{(1)}(y,z)}{ (1-y)^{n+1}(1-z)^{n+1} } 
  =  \sum_{ \lambda \in {\rm Par}(n) } \sum_{ s,t \geq 0 } s_{\lambda}(1^s) s_{\lambda}(1^t) y^s z^t.$$ By the Cauchy identity on Schur functions, we have $$\sum_{ \lambda \in {\rm Par}(n) } s_\lambda(1^s) s_\lambda(1^t) = [x^n] \Big( \frac{1}{1-x} \Big)^{st}
  = \binom{st+n-1}{n}.$$ This ends the proof.

From Equation  and the previous proposition, we deduce:

$$\label{formuleanr}
  \frac{A^{(r)}_n(y,z)}{ (1-y^r)^n (1-y) (1-z)^{n+1} } = \sum_{i,j \geq 0} \binom{ij+n-1}{n} y^i z^j.$$

Note that another consequence of Equation , together with Equation , is the following (which is not a new result, see for example [@bagno1]).

$$\label{flagpol}
  \frac{A^{(r)}_n(y,1)}{ (1-y^r)^n (1-y) } = \sum_{i \geq 1} i^n y^i.$$

Identities on bi-Eulerian generating functions {#sec:genfunc}
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We keep the definition of $A_n^{(r)}(y,z)$ and $A^{(r)}(x,y,z)$ as before, but in this section we only need the formula in Equation . We recall that $C_{n,k}^{(r)}(y,z)$, $C_n^{(r)}(y,z)$, and $C(x,y,z)$ were defined in the proof of Theorem \[theobnr\]. The goal of 