The <a href="https://en.wikipedia.org/wiki/H-vector">$h$-vector</a> of the (simplicial complex dual to the) <a href="https://en.wikipedia.org/wiki/Permutohedron">permutohedron</a> is given by the sequence of <a href="https://en.wikipedia.org/wiki/Eulerian_number">Eulerian numbers</a>.

**Example**: The (dual of the) 2-dimensional permutohedron is a hexagon, with 6 1-cells, 6 0-cells, and 1 -1-cell. So we compute the $h$-vector by
$$\begin{array}{c c}
(1,6,6) &= 1 \times (1,2,1)\\
& +4 \times (0,1,1) \\
&+1 \times (0,0,1) \end{array}$$
so that its $h$-vector is $(1,4,1)$.

Meanwhile the <a href="https://en.wikipedia.org/wiki/Ehrhart_polynomial#Ehrhart_series">$h^*$-vector</a> of the hypercube is also given by the Eulerian numbers. This means that the $h$-vector of a unimodular triangulation of the hypercube is the Eulerian numbers, padded with two trailing 0's.

**Example**: Looking at

[![enter image description here][1]][1]

we see that a unimodular triangulation of the cube has 6 tetrahedra, 18 triangles, 19 edges, and 8 vertices. So we compute the $h$-vector of this triangulation by
$$\begin{array}{c c}
(1,8,19,18,6) &= 1 \times (1,4,6,4,1)\\
& +4 \times (0,1,3,3,1) \\
&+1 \times (0,0,1,2,1) \\
&+0 \times (0,0,0,1,1) \\
&+0 \times (0,0,0,0,1)
\end{array}$$
so that the $h$-vector is $(1,4,1,0,0)$.

**Question**: Is there some deeper *geometric* connection between a (unimodular) triangulation of the hypercube and the (dual to) the permuotohedron suggested by this numerology?


  [1]: https://i.sstatic.net/zc7iJ.png