A_n(z) = (1-z)^{n+1} \sum_{t \geq 0 } (t+1)^n z^t.$$

Let us turn to the case of the half-open hypercube $[0,1)^n$. Note that we are now dealing with a half-open polytope, i.e., a polytope where some of the $(n-1)$-dimensional faces are removed. In this case, it is not [*a priori*]{} clear that the Ehrhart series is a polynomial with nonnegative integral coefficients. We can decompose $[0,1)^n$ as the disjoint union of the polytopes: $$\mathcal{T}_{\sigma} = \{ v \in [0,1)^n \, : \, \operatorname{std}(v)=\sigma \}.$$ These are also simplices where some facets are removed, and we get:

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

It is similar to the one of Lemma \[ehrssigma\].

Thus, the half-open hypercube has the Ehrhart series $zA_n(z)$. Besides, its Ehrhart polynomial is clearly $E([0,1)^n,t)=t^n$. Once again we get Identity , with an additional factor $z$.

Let us present another example of a Ehrhart series that will be used in the sequel. It is presented in [@stanley2 Section 7.19] in the context of quasi-symmetric functions. Let $\lambda$ be a Young diagram (we use the French notation). Let $\mathbb{Z}^{\lambda}$ (respectively, $\mathbb{R}^\lambda$) denote the set of fillings of $\lambda$ with integers (respectively, real numbers). And let $\mathcal{Y}_{\lambda}$ denote the set of 