Generalized parking function polytopes

A classical parking function of length $n$ is a list of positive integers $(a_1, a_2, \ldots, a_n)$ whose nondecreasing rearrangement $b_1 \leq b_2 \leq \cdots \leq b_n$ satisfies $b_i \leq i$. The convex hull of all parking functions of length $n$ is an $n$-dimensional polytope in $\mathbb{R}^n$, which we refer to as the classical parking function polytope. Its geometric properties have been explored in (Amanbayeva and Wang 2022) in response to a question posed in (Stanley 2020). We generalize this family of polytopes by studying the geometric properties of the convex hull of $\mathbf{x}$-parking functions for $\mathbf{x}=(a,b,\dots,b)$, which we refer to as $\mathbf{x}$-parking function polytopes. We explore connections between these $\mathbf{x}$-parking function polytopes, the Pitman-Stanley polytope, and the partial permutahedra of (Heuer and Striker 2022). In particular, we establish a closed-form expression for the volume of $\mathbf{x}$-parking function polytopes. This allows us to answer a conjecture of (Behrend et al. 2022) and also obtain a new closed-form expression for the volume of the convex hull of classical parking functions as a corollary.


Introduction
A classical parking function of length n is a list (a 1 , a 2 , . . . , a n ) of positive integers whose nondecreasing rearrangement b 1 ≤ b 2 ≤ · · · ≤ b n satisfies b i ≤ i. It is well-known that the number of classical parking functions of length n is (n + 1) n−1 ; this number surfaces in a variety of places, for example, it counts the number of planted forests on n vertices and the number of regions of a Shi arrangement (see [24] for further discussion). Let PF n denote the convex hull of all parking functions of length n in R n . In 2020, Stanley [18] asked for the number of vertices, the number of faces, the number of lattice points PF n ∩ Z n , and the volume of PF n . These questions were first answered by Amanbayeva and Wang [1] and Stong [20].
In Section 2, we revisit the classical parking function polytope PF n . We provide new results on the appearance of both lower-dimensional parking function polytopes and permutahedra as facets of PF n . We connect the classical parking function polytope with the recent work of Heuer and Striker [9] and Behrend et al. [2] on partial permutahedra, and show when they are integrally equivalent. We collect different characterizations of the normalized volume of PF n in Theorem 2.9 and give a new simple, closed-form answer to Stanley's original question on the volume of PF n .
A natural direction is to extend these results for generalizations of parking functions. We consider x-parking functions, where x = (x 1 , . . . , x n ) is a vector of positive integers, which have been explored from an enumerative perspective previously by Yan [22][23][24] and Pitman and Stanley [19]. In Section 3, we focus on the case when x = (a, b, b, . . . , b) and generalize results of Amanbayeva and Wang [1]. We establish in Theorem 1.1 a closed-form normalized volume formula for all positive integers a, b. Theorem 1.1. For any positive integers a, b, n, the normalized volume of the x-parking function polytope NVol(X n (a, b)) is given by Hanada is partially supported by the Funai Overseas Scholarship. Lentfer is supported by the National Science Foundation Graduate Research Fellowship DGE-2146752. Vindas-Meléndez is partially supported by the National Science Foundation under Award DMS-2102921. 1 The proof of Theorem 1.1 uses tools from analytic combinatorics and analytic number theory, e.g., Ramanujan's Master Theorem. By determining when partial permutohedra are integrally equivalent to x-parking functions, we also prove a conjecture of Behrend et al. [2], as Corollary 3.29.
In Section 4, we introduce weakly increasing x-parking functions, which are x-parking functions that are in weakly increasing order a 1 ≤ a 2 ≤ · · · ≤ a n without any rearrangement. We explore a subpolytope of the x-parking function polytope which is constructed as the convex hull of weakly increasing x-parking functions. We show that the convex hull of weakly increasing x-parking functions is integrally equivalent to certain Pitman-Stanley polytopes. In the literature, the Pitman-Stanley polytope has been called the "parking function polytope" as its volume is related to parking functions. However, in this paper, we call the convex hull of any generalization of parking functions a parking function polytope.
We conclude with Section 5 where we present some routes for further study on unimodular triangulations, Ehrhart theory, other generalizations where x = (a, b, . . . , b), and rational parking functions.

The classical parking function polytope
The classical parking function polytope was first studied in [1,20]. As mentioned in the introduction, a classical parking function of length n is a list (a 1 , a 2 , . . . , a n ) of positive integers whose nondecreasing rearrangement b 1 ≤ b 2 ≤ · · · ≤ b n satisfies b i ≤ i. Let PF n denote the convex hull of all parking functions of length n in R n , which we call the classical parking function polytope.  A convex polytope can be described by its vertex and hyperplane descriptions, which we briefly state here. Both were known to Stanley (see [1]). The vertex description of PF n is the convex hull of all vertices given by permutations of (1, . . . , 1 k , k + 1, k + 2, . . . , n − 2, n − 1 n−k ), for 1 ≤ k ≤ n. For a proof, this follows from the more general Proposition 3.4 (taking a = b = 1). 2 The hyperplane description of PF n : 1 ≤ x i ≤ n, for 1 ≤ i ≤ n, x i + x j ≤ n + (n − 1), for i < j, x i + x j + x k ≤ n + (n − 1) + (n − 2), for i < j < k, . . .
For a proof, this follows from the more general Proposition 3.6 (taking a = b = 1).
2.1. Face structure. We are able to say more about its face structure by using the regular permutahedron, which we now define.
Definition 2.1 (Example 0.10, [26]; Definition 2.1, [14]). The regular permutahedron Π n ⊆ R n is the (n − 1)-dimensional polytope obtained as the convex hull of all vectors obtained by permuting the coordinates of the vector (1, 2, . . . , n). Its vertices can be identified with the permutations in S n by associating with (x 1 , x 2 , . . . , x n ) the permutation that maps x i → i such that two permutations are adjacent if and only if the corresponding permutations differ by an adjacent transposition. More generally, for r := (r 1 , . . . , r n ) ∈ R n , a permutahedron Π n (r) is the convex hull of all vectors that are obtained by permuting the coordinates of the vector r.
We have the following result on when the permutahedron appears as a facet of the classical parking function polytope.
Proposition 2.4. The regular permutahedron appears as a facet of the classical parking function polytope exactly once.
Proof. By the definition of the parking function polytope PF n , it is the convex hull of all vertices given by permutations of (1, . . . , 1 k , k + 1, k + 2, . . . , n − 2, n − 1 for 1 ≤ k ≤ n (here, k = 0 would be superfluous). Each permutation of (1, 2, . . . , n) appears as a vertex. Thus, the convex hull of these n! vertices, which is exactly Π n , is contained within PF n . 3 We now use the hyperplane description of PF n . Consider the hyperplane H defined by x 1 + x 2 + · · · + x n = n(n+1)

2
. We claim H ∩ PF n = Π n . Since all permutations of (1, 2, . . . , n) satisfy 1 + 2 + · · · + n = n(n+1) 2 , the vertices that give the vertex description of Π n are in H ∩ PF n . By taking their convex hull, it follows that Π n ⊆ H ∩ PF n . Now, suppose x := (x 1 , x 2 , . . . , x n ) is a point in H ∩ PF n . As we are dealing with a subset of a polytope intersecting a hyperplane, H ∩ PF n is a polytope of dimension (at most) n − 1. Suppose towards a contradiction that x is not in Π n . This means that by the defining inequalities for Π n , (1) x 1 + · · · + x n = n(n+1) 2 , a contradiction, or (2) there exists some nonempty subset {i 1 , . . . , i k } ⊆ {1, . . . , n} such that But (2) is not allowed by the defining inequalities for a parking function polytope. Hence, x is in Π n , so Π n = H ∩ PF n is a facet of PF n .
To show uniqueness, it suffices to show that the only hyperplane in the inequality description that corresponds to a facet with n! vertices is H. Assume a facet with n! vertices corresponds to a hyperplane of the form By the vertex description of PF n , the number of vertices that satisfy this equation is given by where m is the number of coordinates in the vertex that have value 1. Note that since 1 ≤ k ≤ n−2, it follows that n = n 1 ≤ n k . Rearranging the inequality, we get that n · k!(n − k)! ≤ n!. We can also see that Hence, Therefore, H is the only possible hyperplane in the inequality description of PF n that corresponds to a facet with n! many variables.
The following lemma allows us to establish a lower bound for the number of permutahedra of any dimension in the parking function polytope.
Proof. Consider the defining inequalities of Π n given in Definition 2.1. We claim that the hyperplanes that correspond to Π n−1 as facets are those of the form x i ≤ n and x i 1 + · · · + x i n−1 ≤ n + (n − 1) + · · · + 2.
We can see that for any i, the vertices of Π n that satisfy x i = n are those where the i-th coordinate is n and all the other coordinates can be written as a permutation of (1, 2, . . . , n − 1). This facet is exactly Π n−1 . Similarly, we can see the vertices that satisfy x i 1 + · · · + x i n−1 = n + (n − 1) + · · · + 2, where {i 1 , . . . , i n−1 } = {1, . . . , n} \ {k}, for some k, are vertices where the k-th coordinate is 1 and all the other coordinates can be written as a permutation of (2, . . . , n). Thus, facets that correspond to hyperplanes of these forms are Π n−1 .

4
To show that these are the only hyperplanes that can correspond to Π n−1 , we will show that the only hyperplanes that can correspond to a facet with (n − 1)! vertices are the ones mentioned above. The proof is similar to the uniqueness proof of Proposition 2.4.
For a hyperplane x i 1 + · · · + x i k = n + · · · + (n − k + 1) consisting of k many variables, we can see that there are k!(n − k)! vertices of Π n that satisfy it. If we have that k is neither equal to 1 nor n − 1, it follows that k!(n − k)! < (n − 1)!. Hence, if a hyperplane with k many variables corresponds to Π n−1 , which has (n − 1)! vertices, it follows that k equals 1 or n − 1.
Proposition 2.6. The (n − 1)-dimensional parking function polytope PF n−1 appears as a facet of the n-dimensional parking function polytope PF n exactly n times.
Proof. Consider the inequality description of PF n : . . .
for i 1 < i 2 < · · · < i n−2 < n, This is exactly the inequality description for PF n−1 . Hence, PF n−1 is a facet of PF n . Now, as the choice of i was arbitrary, there are n choices for i, so there are n copies of PF n−1 appearing as facets of PF n . Now we will show that these are the only occurrences. We know that PF n−1 has (n − 1)! 1 m! many vertices that satisfy it. If k is not equal to 1 or n, we have n ≤ n k , thus k!(n − k)! ≤ (n − 1)!. Then since (n−1)! m! many vertices that satisfy the equation, hence it also cannot correspond to PF n−1 . Thus, the only hyperplanes that support a facet of the form PF n−1 are the n mentioned above.

2.2.
Volume. Next, we consider a collection of related polytopes, called partial permutahedra, which were first introduced in [9] in terms of partial permutation matrices; a recursive volume formula was presented in [2]. For our purposes, it suffices to take their vertex description as the definition. Two integral polytopes P ⊆ R m and Q ⊆ R n are integrally equivalent if there exists an affine transformation Φ : R m → R n whose restriction to P preserves the lattice. We establish the following result relating partial permutahedra and classical parking functions.
Proposition 2.8. The classical parking function polytope PF n is integrally equivalent to the partial permutahedron P(n, n−1). In particular, they are related by a translation by the vector (1, 1, . . . , 1).
We can now establish the equivalence of several different formulas throughout the literature, as they count the normalized volume of the same polytope. Previously, [1] found a generating function and recursive formula for the volume of PF n , and [20] found a closed-form volume formula which uses an alternating sum (inclusion-exclusion). Our new contribution to the following theorem, in part (iii), gives a more simple closed-form volume formula, since it does not contain an alternating sum.
We finish by showing that (iii) ⇐⇒ (v). This is equivalent to showing that their difference is 0. From (v), subtract (iii), which gives We now use Wilf-Zeilberger theory. We need only show that f (n) : Note that as F (n, i) contains a binomial coefficient, the sum f (n) = i∈Z F (n, i). We use Zeilberger's creative telescoping algorithm ct in the package EKHAD as described in Chapter 6.5 of [13] and available from [25]. Calling The sequence for the normalized volume of PF n begins 0, 1, 24, 954, 59040, 5295150, 651354480, 105393619800, 21717404916480, . . . and is OEIS sequence A174586 [11].

The convex hull of x-parking functions
Next, we discuss a generalization of the classical parking functions. Let x = (x 1 , . . . , x n ) ∈ Z n >0 . Define an x-parking function to be a sequence (a 1 , . . . , a n ) of positive integers whose nondecreasing Throughout Section 3, we will specialize to vectors of the form x = (a, b, b, . . . , b) following the work of Yan [22,23]. Additionally the volume calculations in Section 3.2 become more difficult for arbitrary x = (x 1 , . . . , x n ), which we discuss more in Section 5.2. On the other hand, the vertex and hyperplane descriptions along with some of the basic enumerative results can be easily generalized to the x = (x 1 , . . . , x n ) setting, and we leave those details to the interested reader.
As mentioned in [23], from work of Pitman and Stanley [19], the number of x-parking functions for x = (a, b, b, . . . , b) is the following: >0 , the number of x-parking functions is given by a(a + nb) n−1 .
The following definition introduces an n-dimensional polytope associated to x-parking functions of length n for the specific sequence x = (a, b, b, . . . , b) ∈ Z n >0 , which is one of the main objects of study in this paper.

Definition 3.2.
Define the x-parking function polytope X n (a, b) as the convex hull of all x-parking functions of length n in R n for x = (a, b, b, . . . , b) ∈ Z n >0 . See Figure 2 for examples.

Remark 3.3.
Note that if n = 1, x = (a), so no b is used. As a result, we may denote the x-parking function polytope X 1 (a, b) alternatively by X 1 (a). Figure 2. The x-parking function polytopes, from left to right: . Note that when a > 1, there are new facets that do not appear when a = 1.
3.1. Face structure. In this subsection we describe the face structure of the x-parking function polytope X n (a, b). From Theorem 3.1, we obtain an upper bound for the number of x-parking functions which arise as vertices of X n (a, b) since it is well-known that if a polytope can be written as the convex hull of a finite set of points, then the set contains all the vertices of the polytope (Proposition 2.2, [26]). We now give a vertex description of X n (a, b).
Proof. Consider an x-parking function x = (x 1 , . . . , x n ) for which there is a term x i > 1 such that (x 1 , . . . , x i−1 , x i + 1, x i+1 , . . . , x n ) is also an x-parking function. Then x is a convex combination of two other x-parking functions. Second, if is a convex combination of y, z ∈ X n (a, b), then x = y = z, as the first k coordinates of x are minimal at 1 and the last (n − k) coordinates are maximized due to the condition on the nondecreasing rearrangement of x-parking functions. Thus, x is a vertex of X n (a, b).
Next, we enumerate the vertices of X n (a, b). In the case that a = 1, we have the same number of vertices as in the case of the classical parking function polytope PF n , as b is a parameter that increases the lengths of the edges of the polytope. However, when a > 1, new vertices and edges come into play.
Proposition 3.5. The number of vertices of X n (a, b) is Proof. The vertices are the permutations of the following: Observe that there is one permutation of v 0 , n permutations of v 1 , n(n − 1) permutations of v 2 , and in general, n(n − 1) · · · (n − k + 1) permutations of v k . If a > 1, then the vertices v n−1 and v n are distinct, so we count the contribution of both. However, if a = 1, then v n−1 = v n , and we only count one. For a > 1, this gives 1 + n + n(n − 1) + · · · + (n(n − 1) · · · 2) + (n(n − 1) · · · 2 · 1) and for a = 1, We continue by presenting an inequality description of X n (a, b).
Proposition 3.6. The x-parking function polytope X n (a, b) is given by the following minimal inequality description: for all 1 ≤ i < j ≤ n, for all 1 ≤ i < j < k ≤ n, if a > 1, for all 1 ≤ i 1 < i 2 < · · · < i n−1 ≤ n, and (regardless of a), Proof. We use the vertex description given in Proposition 3.4, which states that the vertices of X n (a, b) are all permutations of First consider that since a, b ≥ 1, all coordinates in a vertex of X n (a, b) are at least 1, i.e., 1 ≤ x i for all 1 ≤ i ≤ n. Now, we turn our attention to inequalities that are solely upper bounds, so we can assume that k = 0 and only consider the largest possible coordinates. The largest a single coordinate could be is a Next, summing the largest two coordinates is at most ((n − 2)b + a) + ((n − 1)b + a), so Repeating the same process with summing the largest possible 3, . . . , n variables, completes the inequality description.
Note that in the case that a = 1, the inequality that bounds the sum of any n − 1 variables is not needed in the minimal inequality description. We justify this as follows. Given that a = 1, the inequality is that Also, we always need the final inequality , the only value that is not used is a, which in this case equals 1. Hence, the final inequality is sufficient, because removing one coordinate from the sum on its left hand side subtracts a = 1 from the right hand side, exactly yielding the inequality Note that this only happens in the case that a = 1 because if a > 1, we are able to get a tighter inequality when we subtract a > 1 from the right hand side as before. We now show that the inequality description is minimal. Observe that by construction for each of these defining inequalities, there exists an extremal point of the polytope which gives equality. It is clear that the inequalities 1 ≤ x i in each coordinate are necessary, as they are the only lower bounds on the coordinates. We only need to consider the minimality of the set of inequalities which are upper bounds on partial sums of the coordinates. A general inequality on 1 ≤ k ≤ n variables says Observe that for each of the inequalities on k variables (and of course excluding the case of a = 1 and k = n − 1), it is strictly stronger on at least one point than all inequalities on 1, . . . , k − 1 variables. Explicitly, a general inequality on 1 ≤ k − 1 ≤ n variables says Note that adding one coordinate x k to the left hand side of (6) and a + (n − 1)b to the right hand side gives which is a worse bound for all k > 1 due to the b(k − 1) term. As we constructed this inequality description in increasing order on the number of variables, each one refining the previous, the inequality description is an irredundant system of inequalities and is therefore minimal.
Since we have a minimal inequality description of X n (a, b), by enumerating the number of defining inequalities we obtain a count for the number of facets of X n (a, b). Proof. It suffices to count the number of minimal defining inequalities given in Proposition 3.6. Note that there are n k different inequalities with 1 < k < n − 1 or k = n distinct variables. We have 2 n 1 many inequalities with one variable since we have 1 ≤ x i and x i ≤ (n − 1)b + a. If a = 1, there are no inequalities in n − 1 variables. Then the number of inequalities is n 1 + n 1 + n 2 + · · · + n n − 2 + n n = 2 n − 1.
Next we study the edges of X n (a, b).
Definition 3.8. Let x be a vertex of X n (a, b). Then it is a permutation of (1, . . . , 1, a + kb, a + (k + 1)b, . . . , a + (n − 2)b, a + (n − 1)b), for some unique 1 ≤ k ≤ n if a = 1 or 0 ≤ k ≤ n if a > 1. We say that x is on layer n − k. For x = (1, . . . , 1), we say that it is on layer 0. Lemma 3.9. If v, u are two vertices of X n (a, b) such that vu is an edge, then the layers of v and u differ by at most 1.
Proof. The proof follows almost identically to that of Proposition 2.1 in [1] in the case that a = 1.
In the case that a > 1, there is just one more layer to address. Let c · x = c 1 x 1 + · · · + c n x n be the dot product of vectors c, x ∈ R n . If vu is an edge, then there exists c ∈ R n such that c · v = c · u > c · w for any vertex w of X n (a, b) such that w = v, u.
Note that X n (a, b) is invariant under permutation of the coordinates, so without loss of generality, assume that c 1 ≤ c 2 ≤ · · · ≤ c n .
In the case that a > 1, there is one more layer to address, as (1, a + b, a + 2b, . . . , a + (n − 2)b, a + (n − 1)b) = (a, a + b, a + 2b, . . . , a + (n − 2)b, a + (n − 1)b) are distinct now. By modifying the proof from the case of a = 1, one only needs to check that layers n − 2, n − 1 (the new layer, meaning the layer that causes new facets to appear), and n have vertices with the correct number of edges.
A polytope is smooth if it is simple and at each vertex v, the primitive edge directions from v form a lattice basis of Z n . More background on smooth polytopes, as well as their connections to toric ideals and projective geometry, can be found in [6]. Proof. By symmetry, it suffices to check this at some vertex v on each layer n − k for 0 ≤ k ≤ n (or 1 ≤ k ≤ n in the case where a = 1). Without loss of generality, let v = (1, . . . 1, a + kb, . . . , a + (n − 1)b). From the proof of Proposition 3.11, for any given vertex v of X n (a, b), we have determined the n vertices u 1 , . . . , u n such that vu i is an edge.
If k = n, we have v = (1, . . . , 1) and the u i are all the permutations of (1, . . . , 1, n). Thus, all the primitive edge directions, determined by v and the first lattice point along each edge containing it, clearly form an integral basis of Z n . Now, assume a > 1 and consider 2 ≤ k ≤ n − 1. Let u 1 , . . . , u k be on the layer above v, where u i is obtained by changing the i-th coordinate of v from 1 to a + (k − 1)b. Let u k+1 , . . . , u n−1 be on the same layer as v, where u j is obtained by switching the j-th and (j + 1)-st coordinates of v. The last vertex u n is on the layer below v and is obtained by changing the k-th coordinate of v from a + kb to 1. We can see that the primitive edge directions from v to u 1 , . . . , u n form a basis of Z n , where the first k vectors are basis for the first k-coordinates of Z n and the last (n − k) are a basis of the last n − k coordinates.
If k = 1, we have v = (1, a + b, . . . , a + (n − 1)b). Let u 1 = (a, a + b, . . . , a + (n − 1)b), which is on the layer above v. Let u 2 , . . . , u n−1 be on the same layer as v where u j is obtained by switching the j-th and (j + 1)-th coordinates of v. Finally, let u n = (1, 1, a + 2b, . . . , a + (n − 1)b), which is on the layer below v. It is clear that the primitive edge directions from v to these u i form an integral basis of Z n .
If k = 0, we have v = (a, a + b, . . . , a + (n − 1)b). Let u 1 , . . . , u n−1 be on the same layer as v where u j is obtained by switching the j-th and (j + 1)-st coordinates of v. Finally, let u n = (1, a + b, a + 2b, . . . , a + (n − 1)b), which is on the layer below v. It is clear that the primitive edge directions from v to these u i form an integral basis of Z n .
A similar proof holds for a = 1: note that when a = 1 we have one less layer.
Proposition 3.13. The number of edges of X n (a, b) is n 2 Ver(X n (a, b)), where Ver(X n (a, b)) denotes the number of vertices X n (a, b).
Proof. By Proposition 3.11, the graph of X n (a, b) is an n-regular graph. This gives the desired formula, where the number of vertices is given by Proposition 3.5.
Up to this point we have the number of 0-dimensional faces (vertices), the 1-dimensional faces (edges), and (n − 1)-dimensional facets (facets). In what comes next, we study the faces of higher dimension.
Proposition 3.14. Let f k be the number of k-dimensional faces of X n (a, b) for k ∈ {0, . . . , n}. Then if a = 1, and if a > 1, where S(n, k) are the Stirling numbers of the second kind.
In the case that a > 1, there is one more layer to address, as (1, a + b, a + 2b, . . . , a + (n − 2)b, a + (n − 1)b) = (a, a + b, a + 2b, . . . , a + (n − 2)b, a + (n − 1)b) are distinct now. By modifying the proof from the case of a = 1, the restriction given in Lemma 3.1 in [1] which causes m = 1 to be excluded from the sum in the proof of Theorem 3.1 in [1] is removed due to the additional layer.
The faces of a convex polytope P form a lattice called its face lattice, denoted by F(P ), where the partial ordering is given by set inclusion of the faces. Two polytopes whose face lattices are isomorphic (as unlabeled partially ordered sets) are combinatorially equivalent. We continue with a characterization of which X n (a, b) are combinatorially equivalent.
Proposition 3.15. For fixed n and for a = 1, the X n (1, b) are combinatorially equivalent for all b ≥ 1. Additionally, for fixed n, the X n (a, b) are combinatorially equivalent for all a > 1 and b ≥ 1.
Proof. By Proposition 3.14, for each b, the face lattice of X n (1, b) has the same number of elements, and the same number of elements at each level corresponding to the dimension of the face. For a face lattice, the join of two elements in the same layer will be in the layer above it, and the meet of two elements in the same layer will be in the layer below it, as face inclusions pass through faces of one dimension more/less (face inclusions of faces of the same dimension result in equality). Thus, it only remains to construct the isomorphism. Define ϕ : F(X n (1, b 1 )) → F(X n (1, b 2 )) by replacing b 1 by b 2 in the vertex description of each face in the face lattice. It is clear that this is an isomorphism, as the inverse is given by replacing b 2 by b 1 . Therefore, the X n (1, b) are combinatorially equivalent for a = 1 and for any b ≥ 1.
Similarly as before, for all a > 1, we define ϕ : F(X n (a 1 , b 1 )) → F(X n (a 2 , b 2 )) by replacing a 1 by a 2 and b 1 by b 2 in the vertex description of each face in the face lattice. Hence, the X n (a, b) are combinatorially equivalent for any a > 1 and for all b ≥ 1.

3.2.
Volume. We turn our attention to calculating the volume of x-parking function polytopes X n (a, b). We extend previous work of [1] to find a recursive volume formula in Theorem 3.23. Then, by using exponential generating functions, we find a closed-form expression for the volume of X n (a, b) in Theorem 1.1. We also consider the relationship between the x-parking function polytopes and partial permutahedra, allowing us to expand known results on the volume of partial permutahedra. The relationship between partial permutahedra and x-parking function polytopes is given by the following proposition, which answers the question of when a partial permutahedra is an x-parking function polytope and vice versa. (ii) if p < n − 1, then P(n, p) is not integrally equivalent to any X n (a, b).
Proof. First, recall from Remark 3.3 that when n = 1 for an x-parking function, b is unused. The result where n = 1 is clear, as these polytopes are just 1-dimensional line segments of length a − 1 = p. Thus, we reduce to the case where n ≥ 2.
For two polytopes to be integrally equivalent, they must have the same dimension (relative to its affine span). All partial permutahedra (recall that p must be a positive integer) are full dimensional (Remark 5.5 [9]). As all X n (a, b) are also full-dimensional, we are justified in using n to denote the dimension for both.
We will show that the conditions b = 1, a = p − n + 2 are necessary for P(n, p) and X n (a, b) to be integrally equivalent by comparing "maximal" vertices. Apply the coordinate transformation to increase the each coordinate by one in P(n, p) (as in the proof of Proposition 2.8) to match the minimal vertices of the two polytopes. Consider the "maximal" vertices v = (a, a + b, . . . , a + (n − 2)b, a + (n − 1)b) of X n (a, b) and u = (max(p − n + 2, 1), . . . , p, p + 1) of the shifted P(n, p), where "maximal" means the maximal vertex where all the coordinates are weakly increasing. If the two polytopes are integrally equivalent, we must have v = u. If b = 1, the difference between the last two coordinates of v is b ≥ 2 while the difference between the last two coordinates of u is 1, thus u = v. Hence, b = 1 is a necessary condition.
Furthermore, consider a = p − n + 2. If n ≤ p + 1, the first coordinate of u is max(p − n + 2, 1) = p − n + 2, which gives us u = v. If n > p + 1, we can see that max(p − n + 2, 1) = 1. Additionally, the second coordinate of u is at most 1, since (p − n + 2) + 1 ≤ 1. Thus we have the first two coordinates of u must be 1. However, the first two coordinates of u are a, a + b, where b is nonzero, thus u = v.
Hence, the conditions b = 1, and a = p − n + 2 are necessary for P(n, p) = X n (a, b). 15 We next show that they are sufficient. Consider X n (a, b) where b = 1, and a = p − n + 2. Note that this second condition implies n ≤ p + 1, as a ≥ 1. Using the vertex description of Proposition 3.4, we can consider X n (a, b) to be the convex hull of all permutations of (1, . . . , 1 k , p − n + k + 2, p − n + k + 3, . . . , p, p + 1 for all 0 ≤ k ≤ n. Since n ≤ p + 1 implies min(n, p) = n, this is exactly the vertex description of shifted P(n, p).
Finally, we saw that the necessary condition for integral equivalence a = p − n + 2 implies that n ≤ p + 1. Hence if p < n − 1, then P(n, p) is not integrally equivalent to any X n (a, b).
We now recall the recursive volume formula for the classical parking function polytope.
We are able to obtain some immediate corollaries of these results by considering dilations of PF n .
Definition 3.18. For any positive integer d, the d-dilate of an x-parking function polytope X n (a, b) is given by the following map on points ϕ d : In general, a d-dilate of any polytope is the image of a map which (up to translation) multiplies all coordinates by d. (1) The x-parking function polytope X n (1, b) is a b-dilate of X n (1, 1).
Proof. This follows from applying the b-dilate map ϕ b onto all the vertices of X n (1, 1) and X n (a, 1) given by Proposition 3.4, which results in all the vertices of X n (1, b) and X n (a+(b−1)(a−1), b).
Knowing that one polytope is a dilate of another allows for more direct approach to finding volumes for some of these parking function polytopes.
Proof. Note that the dilation of an n-dimensional polytope by a factor of b increases the volume by a factor of b n . The result then follows from Lemma 3.19.
Using the following result on partial permutahedra, we can then calculate the volume of more x-parking function polytopes.
These results, however, do not give Vol (X n (a, b)) for all a, b ≥ 1. To do so, we need a more general construction, which is given by the following recursive volume formula.  (X n (a, b)) for all positive integers n. Then V a,b 1 = a − 1 and for n ≥ 2, V a,b n is given recursively by In the proof of this theorem, we will use the following decomposition lemma.
where K ′ 1 , . . . , K ′ n−m denote the orthogonal projections of K 1 , . . . , K n−m onto U ⊥ , respectively. We will also use the following fact.  1, [2]). The Euclidean volume of the regular permutahedron Π n ⊆ R n is n n−2 √ n.
Proof of Theorem 3.23. The case where a = 1 follows from Theorem 3.17 and Corollary 3.20(1). They imply that For the case where a > 1, we generalize the proof of Theorem 4.1 in [1]. Divide X n (a, b) into ndimensional (full dimensional) pyramids with facets of X n (a, b) which do not contain I = (1, . . . , 1) as the base, and point I as a vertex. For an example, see Figure 3. Recall from Theorem 3.7 that there are 2 n − 1 + n facets since a > 1. Of those, exactly n have I as a vertex, so the number of pyramids is 2 n − 1. Each pyramid has a base which is a facet F with points of X n (a, b) satisfying the equation for some k ∈ {1, 2, . . . , n − 1, n} and distinct i 1 < · · · < i k , due to the defining inequalities of X n (a, b). Now let {j 1 , j 2 , . . . , j n−k } = {1, 2, . . . , n} \ {i 1 , i 2 , . . . , i k }. Let X ′ n−k (a, b) be the polytope containing all points x ′ such that x ′ p = 0 for all p ∈ {i 1 , i 2 , . . . , i k } and for some x ∈ F , x ′ p = x p for all p ∈ {j 1 , j 2 , . . . , j n−k }. Then X ′ n−k (a, b) is an (n − k)-dimensional polytope, with the following defining inequalities: , By comparing with the inequality description given in Proposition 3.6, we see that X ′ n−k (a, b) ∼ = X n−k (a, b), as the defining inequalities are the same. Hence the (n − k)-dimensional volume Vol n−k (X ′ n−k (a, b)) = Vol n−k (X n−k (a, b)) = V a,b n−k . Let Q a,b k be the polytope containing all points x ′ such that for all p ∈ {j 1 , j 2 , . . . , j n−k }, we have x ′ p = 0, and for some x ∈ F , we have x ′ p = x p for all p ∈ {i 1 , i 2 , . . . , i k }. Then the coordinate values of (x ′ i 1 , x ′ i 2 , . . . , x ′ i k ) of the vertices of Q a,b k are the permutations of ((n − k)b + a, . . . , (n − 2)b + a, (n − 1)b + a). Note that Q a,b k is a translate of the polytope Q 1,b k (both are (k − 1)-dimensional polytopes), so Q a,b k and Q 1,b k have the same (k − 1)-dimensional volume. Furthermore, Q 1,b k is a b-dilate of Q 1,1 k , which is integrally equivalent to Π(1, . . . , k) = Π k−1 , the regular permutahedron. As Π k−1 has volume k k−2 √ k by Lemma 3.25, then Q a,b k has volume b k−1 k k−2 √ k. Thus, F is a Minkowski sum of two polytopes X ′ n−k (a, b) and Q a,b k which lie in two orthogonal subspaces of R n . Therefore, by Lemma 3.24, the volume of F is where K 1 = X ′ n−k,a,b and K 2 = Q a,b k . Then the volume of Pyr(I, F ), the pyramid with I as the vertex over base F is Vol (Pyr(I, F ) where h k denotes the minimum distance from point I to the face F . We calculate that Thus, By the definition of the sequence, V a,b 0 = 1. Note that V a,b 1 = a − 1, as the 1-dimensional volume (in this case length) of the convex hull of colinear points 1, . . . , a in R is of length a − 1. Thus for n ≥ 2, Figure 3. The x-parking function polytope X 3 (2, 1) where, as in the proof of Theorem 3.23, it has been split up into pyramids with I = (1, 1, 1) as the vertex over each face F which does not contain I. By summing the volume of all pyramids, we obtain the volume of the whole polytope.
Theorem 1.1. For any positive integers a, b, n, the normalized volume NVol(X n (a, b)) is given by To prove this, we use the following lemma.
We also need the following exponential generating function for the (non-normalized) volume of X n,a,b . This is a generalization of Proposition 4.2 in [1].  n! x n be its exponential generating function. Then is the exponential generating function given in Lemma 3.26.
Proof. By Theorem 3.23, we have that which by summing over all x n gives Note that by Equation (9), .
Thus by Equation (13), , where the last line follows by Equation (14). Then by dividing both sides by Equation (15), we get This differential equation has solution Our proof will also use the following theorem (see for example Chapter 4, Entry 11, [4]).
Theorem 3.28 (Ramanujan's Master Theorem). Let Γ(s) denote the gamma function. If then the Mellin transform of f (x) is given by Now we are ready to prove our theorem.
Proof of Theorem 1.1. By Ramanujan's Master Theorem, for ϕ(n) : where Γ(s) denotes the gamma function. By taking the limit s → −n, we get By Equation (9), we have that By Equations (15) and (9), we have that By Proposition 3.27, we have Thus by Equations (18) and (19), we have that and by replacing g b by −t and dg b by −dt, which implies that Note that as a formal power series, where (− 1 2 ) ℓ is the Pochhammer symbol, defined by (λ) ℓ := λ(λ + 1) · · · (λ + ℓ − 1), for any positive integer ℓ, and by (λ) 0 := 1. So, which is a multiple of a Poisson-Charlier polynomial. See for example [12] for the following facts about the Poisson-Charlier polynomial. Thus, and to get the normalized volume, we simply multiply by n!, which gives the desired formula.
As a corollary, we prove the following conjecture of Behrend et al. [2]. 1 where the sequence c k satisfies c k = 2(k − 1)(c k−1 − c k−2 ) and has exponential generating function Proof. The two statements are shown to be equivalent in [2], so we prove the latter. By Proposition 3.16, for p ≥ n − 1, NVol(P(n, p) = NVol(X n (p − n + 2, 1)). Then by Theorem 1.1, as desired.
4. The convex hull of weakly increasing x-parking functions Definition 4.1. A weakly increasing x-parking function associated to a positive integer vector x is a weakly increasing sequence (a 1 , a 2 , . . . , a n ) of positive integers which satisfy a i ≤ x 1 + · · · + x i .
We will only focus on weakly increasing x-parking functions associated to vectors x = (a, b, b, . . . , b). Denote the weakly increasing x-parking function polytope associated to a positive integer vector of the form (a, b, b, . . . , b) by X w n (a, b), where n is the length of the vector. Note that the weakly increasing x-parking functions are just the subset of the x-parking functions which are weakly increasing. Note that X w n (1, b) is an (n − 1)-dimensional polytope in R n since the first coordinate of any weakly increasing parking function is 1, that is x 1 = 1, which brings down the dimension of the polytope by 1. See Figure 4 for examples.
Next, we reveal a connection between the weakly increasing x-parking function polytope and the Pitman-Stanley polytope. The Pitman-Stanley polytope is a well-studied polytope, which has connections to flow polytopes, parking functions, and many other combinatorial objects.
Proof. Let T : R n → R n be the linear transformation defined Note that if x ∈ X w n (a, b) ∩ Z n , it follows that x is also a weakly increasing x-parking function. We can see T (x) = y ∈ PS n (a − 1, b, . . . , b) ∩ Z n since 1 ≤ x i ≤ x i+1 implies y i ≥ 0 and 1 ≤ x i ≤ a + (i − 1)b implies k i=1 y i = x k − 1 ≤ a + (i − 1)b − 1 for all i. Next, define the linear transformation S : R n → R n by S(y) = (1 + y 1 , 1 + y 1 + y 2 , . . . , 1 + y 1 + · · · + y n ).
For y ∈ PS n ∩ Z n , we have that x = S(y) satisfies x i ≤ x i+1 and By construction, both T and S are injective. Corollary 4.4. Let t ∈ Z ≥0 . The number of lattice points in the t-dilate of X w n (a, b) is given by Proof. By Proposition 4.3, X w n (a, b) is integrally equivalent to PS n (a − 1, b, . . . , b). By substituting a − 1 for a in the equation of Theorem 13 in [19], we obtain the result.
Another consequence of Proposition 4.3 is the following: Corollary 4.5. For the special case when x = (a, b, . . . , b) = (1, 1, . . . , 1) ∈ R n , the weakly increasing classical parking function X w n (1, 1) has volume n n−2 and contains C n lattice points, where C n = 1 n + 1 2n n denotes the n-th Catalan number.
The following proposition known and can be deduced from the existing literature detailed above, but we provide a proof for completeness. Proposition 4.6. The weakly increasing classical parking function polytope X w n (1, 1) is an (n − 1)dimensional polytope given by the following equality and inequalities.
Proof. The inequality description follows similar to the proof of Proposition 3.6 restricting ourselves to the weakly-increasing x-parking functions.
(i) This follows from a straightforward enumeration of the inequalities in the description above. where v k = k and for each k ≤ i < n we have either v i+1 = v i ("min") or v i+1 = i + 1 ("max"). Hence, each vertex corresponds to a sequence of n − 1 binary choices. (iii) We claim that vertices v, u are connected by an edge if and only if their construction (given above by a sequence of binary choices) differs by exactly one choice. We show this inductively. Assume this is true for X w m (1, 1) where m < n and let u, v be two vertices in X w n (1, 1) that differ by the k-th choice. If k = n, we have that u ′ = v ′ where v ′ = (v 1 , . . . , v n−1 ), which is a vertex of X w n−1 (1, 1). Hence, there exists a c ′ ∈ R n−1 that maximizes it. Let c = (c ′ , 0) ∈ R n ; it is evident that c · u = c · v > c · w for any other vertex w ∈ X w n (1, 1). If k < n and we continue to choose min for the rest of the choices after k, it follows that Remark 4.7. It follows from the inequality description of X w n (1, 1) that all lattice points are weakly increasing parking functions.

Further Directions and Discussion
We conclude this paper by providing some directions for future research. 5.1. On the classical parking function polytope. Given a term order ≺, every non-zero polynomial f ∈ k[x] has a unique initial monomial, denoted by in ≺ (f ). If I is an ideal in k[x], then its initial ideal is the monomial ideal in ≺ (I) := in ≺ (f ) : f ∈ I . Let A = {a 1 , . . . , a k } ⊆ Z n and denote the toric ideal of A by I A .
Proposition 5.1 (Corollary 8.9, [21]). The initial ideal in ≺ (I A ) is square-free if and only if the corresponding regular triangulation ∆ ≺ of A is unimodular.
Computational evidence suggests the following conjecture, which could be approached using the theory of Gröbner bases.
Notice that the initial ideal is square-free; hence, there exists a unimodular triangulation of this parking function polytope using on the parking functions and an additional lattice point as vertices.
If this conjecture holds true, the following problem may be of interest.
Problem 5.4. Find a bijection between the simplices of a unimodular triangulation of PF n and (0, 1)-matrices with two 1's in each row with positive permanent, as discussed in Theorem 2.9.
The Ehrhart function of a polytope P ⊂ R n is ehr P (t) := |tP ∩ Z n |, where tP = {tx : x ∈ P }. When P is a lattice polytope (its vertices have integer coordinates), the Ehrhart function is a polynomial in t, with degree equal to the dimension of P , leading coefficient equal to its normalized volume, second-leading coefficient equal to half the surface area, and constant coefficient 1. The Ehrhart polynomial of a lattice polytope P of dimension n can always be written in the form ehr P (t) = n i=0 h * i t+n−i n ; the sequence (h * 0 , . . . , h * n ) is called the h * -vector. Equivalently,