number of its facets, which was shown to be $N\binom{N-1}{(N-1)/2}$. So, we only need to consider when $N$ is even and would like to obtain an analogous formula.

For even $N$, let $\Lambda_n \subseteq {\{-1,1\}}^n$ be the set of sequences $\boldsymbol{\lambda} = (\lambda_1, \dots, \lambda_n)$ such that $\sum_{i=1}^n \lambda_i = 1$. The facets of $P_N$ are then $$\mathcal{F}(P_N) =
        \left\{
            \pm \operatorname{conv}\left\{
                \lambda_1 \mathbf{e}_1, \dots, \lambda_n \mathbf{e}_n,
                \sum_{i=1}^n \mathbf{e}_i
            \right\}
            \mid (\lambda_1,\dots,\lambda_n) \in \Lambda_n
        \right\}$$

First, observe that the vertices of a facet must consist of a subset of $$\label{eq: sign pattern}
        \{
            \lambda_1 \mathbf{e}_1,
            \dots,
            \lambda_n \mathbf{e}_n,
            \lambda_{n+1} ( \mathbf{e}_1 + \cdots + \mathbf{e}_n )
        \}$$ for some choice of $\lambda_1,\dots,\lambda_{n+1} \in \{-1,1\}$. Otherwise, two vertices $\pm \mathbf{v}$ of $P_N$ would be part of a facet, which is impossible since the line segment $\operatorname{conv}\{ -\mathbf{v},\mathbf{v}\}$ passes through the interior of $P_N$.

Next, note that if $F$ is a facet, then so is $-F$ since $P_N = -P_N$. One specific choice of facet is $$F_0 =
        \operatorname{conv}\left \{
            \mathbf{e}_1, \dots, \mathbf{e}_{(n-1)/2},-\mathbf{e}_{(n+1)/2},
            \dots,
            -\mathbf{e}_n, -\sum_{i=1}^n \mathbf{e}_i
        \right \}.$$ To see why this is true, observe that each of the vertices in $F_0$ lies on the hyperplane $$\{(x_1\dots,x_n) \in {\mathbb{R}}^n \mid \ell(x_1,\dots,x_n) 