The $n$-dimensional permutohedron $P\_n$ is the polytope given by the convex hull of all the possible permutations of the vector $(1,2,\dots,n+1)\in\mathbb{R}^{n+1}$. So it has $(n+1)!$ vertexes.


I would like to ask if there is a formula for the the number of integer points of $P\_n$ and whether it is known that $P\_n$ is a very ample polytope for any $n\geq 1$.


For instance, $P\_2$ is an hexagon in $\mathbb{R}^3$ contained in the plane $x\_1+x\_2+x\_3 = 6$. It has $7$ integer points and it is very ample.



Two questions on the permutohedron
The number of integer points in $P\_n$ is the number of forests on $[n]$; see Section 3 of Stanley's [Decompositions of rational convex polytopes](http://www-math.mit.edu/~rstan/pubs/pubfiles/40.pdf). In fact you can see there a simple description of its entire Ehrhart polynomial in terms of forests. (See also Section 9.3 of Beck and Robins's book [Computing the continuous discretely](http://math.sfsu.edu/beck/papers/ccd.pdf).)


The toric variety corresponding to the permutohedron is also very well-studied, and is often called the "permutohedral variety," but I don't know off the top of my head an answer to your question on ampleness. I would take a look at the book by Gelfand-Kapranov-Zelevinsky, which is a seminal source for a lot of this.


**EDIT**: Okay, I looked up the definition of very ample polytope. A lattice polytope $\mathcal{P}$ is *very ample* if for every $N \gg 0$, and $\mathbf{a} \in \mathbb{Z}^{n}\cap N\mathcal{P}$, there are $\mathbf{a}\_1,\ldots,\mathbf{a}\_N\in \mathbb{Z}\cap \mathcal{P}$ such that $\mathbf{a}=\mathbf{a}\_1+\cdots+\mathbf{a}\_N$. In fact, the (standard) permutohedron has a stronger property than this. It has the *Integer Decomposition Property (IDP)* which says that for every $N\geq 1$, and $\mathbf{a} \in \mathbb{Z}^{n}\cap N\mathcal{P}$, there are $\mathbf{a}\_1,\ldots,\mathbf{a}\_N\in \mathbb{Z}\cap \mathcal{P}$ such that $\mathbf{a}=\mathbf{a}\_1+\cdots+\mathbf{a}\_N$ (this property is also sometimes called *normal*, in the context of algebraic geometry, I think). The reason the permutohedron is IDP is because any lattice zonotope is IDP: every zonotope has a tiling by half-open parallelepipeds, and these parallelepipeds are easily seen to be IDP; and the permutohedron is a lattice zonotope- see Chapter 9 of the Beck-Robins book mentioned above.


As John Machacek linked to below, a recent paper of Higashitani and Ohsugi shows that in fact the class of Minkowski sums of unit simplices (which includes the regular permutohedra and many of the generalized permutohedra of Postnikov) have the IDP property.


