\Delta \} $. The sets $\mathcal{H}_\Delta$ form a partition of $\mathcal{B}^{(r)}_{n,k}$, so that $$B^{(r)}_{n,k}(z) = \sum_{\Delta \subset \{1,\dots,n\}}  E^*( \mathcal{H}_\Delta , z ).$$ By removing the coordinates equal to $r$, we see that $\mathcal{H}_\Delta$ is in bijection with $\mathcal{A}^{(r)}_{n-j,k-rj}$ where $j=\#\Delta$. The bijection preserves integral points, and this holds with the convention that both polytopes are empty if $k-rj<0$. Hence: $$E^*(  \mathcal{H}_\Delta , z ) = (1-z)^j A^{(r)}_{n-j,k-rj}(z).$$

The previous proposition is conveniently rewritten in terms of the generating functions. Let $$A^{(r)}(x,y,z) = \sum_{n\geq 0} \left( \sum_{k=0}^{rn} A^{(r)}_{n,k}(z) y^k \right) \frac{x^n}{n!},
  \quad
  B^{(r)}(x,y,z) = \sum_{n\geq 0} \left( \sum_{k=0}^{rn} B^{(r)}_{n,k}(z) y^k \right) \frac{x^n}{n!},$$ then these two series are related as stated below.

\[relAB\] The following identity holds: $$B^{(r)}(x,y,z) = e^{(1-z)y^rx} A^{(r)}(x,y,z).$$

From Proposition \[relab\], we get: $$\sum_{k \geq 0} y^k B^{(r)}_{n,k}(z) = \sum_{j=0}^n \binom nj ((1-z)y^r)^j \sum_{k \geq 0}  y^{k-rj} A^{(r)}_{n-j,k-rj}(z)$$ and the result follows.

Together with Proposition \[ehrsA\], the relation in the previous theorem shows that a generalization of Stanley’s conjecture can be obtained via an identity on generating functions. This identity will be presented in the next sections. We first need some definitions to state the result.

\[[@bagno2; @foatahan]\] The [*flag excedence number*]{} of a colored permutation is: $${\rm 