$\sigma = \operatorname{std}( v_1 \mod 1,\dots, v_n \mod 1 )$.

\[lem1\] $ \lfloor\operatorname{fdes}(v)\rfloor = \operatorname{fdes}( \operatorname{cstd}(v) )$.

This follows straighforwardly from the definitions.

\[lem1b\] $\mathcal{A}^{(r)}_{n,k} = \phi(\mathcal{F}^{(r)}_{n,k})$.

From and keeping the notation we get $$\sum_{i=1}^n b_i = r.\operatorname{des}(a_1,\dots,a_n) + a_n = \operatorname{fdes}(a_1,\dots,a_n)$$ and the result follows.

\[lem2\] $E^*( \phi( \mathcal{T}_{(\sigma,c)} ) , z ) = z^{\operatorname{des}(\sigma^{-1})+1}$.

Let $v\in\mathcal{T}_{(\sigma,c)}$, then the condition $v_i\leq v_{i+1}$ or $v_i > v_{i+1}$ only depends on $(\sigma,c)$. So from its definition in , we see that the restriction of $\phi$ to $\mathcal{T}_{(\sigma,c)}$ is equal to an affine map that sends $\mathbb{Z}^n$ to itself. It follows that $\phi(\mathcal{T}_{(\sigma,c)})$ has the same Ehrhart series as $\mathcal{T}_{(\sigma,c)}$. Besides, since $\mathcal{T}_{(\sigma,c)}$ is a translation of $\mathcal{T}_{\sigma}$ by an integer vector, they have the same Ehrhart series, which is therefore $z^{\operatorname{des}(\sigma^{-1})+1}$ by Lemma \[ehrtsigma\].

\[ehrsA\] $$E^*(\mathcal{A}^{(r)}_{n,k} , z ) 
   =  \sum_{\substack{(\sigma,c) \in \mathfrak{S}_n^{(r)} \\ \operatorname{fdes}(\sigma,c) = k-1  }}  z^{\operatorname{des}(\sigma^{-1})+1}.$$

Using Lemma \[lem1\], we get: $$\mathcal{F}^{(r)}_{n,k} = \biguplus_{\substack{ (\sigma,c) \in \mathfrak{S}^{(r)}_n \\ \operatorname{fdes}(\sigma,c)=k-1  }} \mathcal{T}_{(\sigma,c)}.$$ From Lemma \[lem1b\] and the fact that $\phi$ is a bijection, we have: $$\mathcal{A}^{(r)}_{n,k} = \phi(\mathcal{F}^{(r)}_{n,k}) 
  = \biguplus_{\substack{ (\sigma,c) \in \mathfrak{S}^{(r)}_n \\ \operatorname{fdes}(\sigma,c)=k-1  }}  \phi( \mathcal{T}_{(\sigma,c)} ).$$ From Lemma \[lem2\] and the fact that the Ehrhart series 