# Stack-sorting simplices: geometry and lattice-point enumeration

Eon Lee  School of Electrical Engineering and Computer Science, Gwangju
Institute of Science and Technology<EMAIL_ADDRESS>,  Carson Mitchell
Department of Mathematics, UC Berkeley<EMAIL_ADDRESS>and  Andrés R.
Vindas-Meléndez  Department of Mathematics, UC Berkeley & Harvey Mudd College,
https://math.berkeley.edu/~vindas<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

We study the polytopes that arise from the convex hulls of stack-sorting on
particular permutations. We show that they are simplices and proceed to study
their geometry and lattice-point enumeration. First, we prove some enumerative
results on $Ln1$ permutations, i.e., permutations of length $n$ whose
penultimate and last entries are $n$ and $1$, respectively. Additionally, we
then focus on a specific permutation, which we call $L^{\prime}n1$, and show
that the convex hull of all its iterations through the stack-sorting algorithm
share the same lattice-point enumerator as that of the $(n-1)$-dimensional
unit cube and lecture-hall simplex. Lastly, we detail some results on the real
lattice-point enumerator for variations of the simplices arising from stack-
sorting $L^{\prime}n1$ permutations. This then allows us to show that
$L^{\prime}n1$ simplices are Gorenstein of index $2$.

##  1\. Introduction

The study of sorting a permutation using stacks was first introduced by Knuth
in the 1960s [7]. Classically, the aim of the stack-sorting problem is to sort
a permutation using a last-in/first-out algorithm. In its simplest form, a
stack is used to rearrange a permutation
$\boldsymbol{\pi}={\pi}_{1}{\pi}_{2}\cdots{\pi}_{n-1}{\pi}_{n}$ as follows.
The elements of $\boldsymbol{\pi}$ are pushed onto an originally empty stack
and an output permutation is formed by popping elements from the stack. In
this paper, we consider the convex hull of all the permutations that arise in
each step of the stack-sorting algorithm, given a particular kind of input
permutation. This paper is organized as follows. All undefined terms or
notation will be introduced in the relevant sections.

In Section 2, we present necessary preliminaries and background. Subsections
2.1 and 2.2 present the stack-sorting algorithm and a brief overview of
polyhedral geometry and Ehrhart theory, respectively. We continue with
background on integral and unimodular equivalence and lecture-hall simplices,
in Subsections 2.3 and 2.4, repsectively. We conclude this section with a
notation table in Subsection 2.5.

Section 3 presents some enumerative and structural results on $Ln1$
permutations. For $S_{n}$, the group of all permutations on $[n]$, we label a
permutation $\boldsymbol{\pi}\in S_{n}$ as a $Ln1$ permutation if it has the
form $Ln1$, where $L$ is a permutation of $\\{2,3,\ldots,n-1\\}$. We denote
the set of all permutations in $S_{n}$ of this form as $\mathcal{L}n1$. The
main theorem of this section is the following:

######  Theorem 3.12.

A permutation $\boldsymbol{\pi}\in\mathcal{L}n1$ if and only if
$\boldsymbol{\pi}$ is exactly $(n-1)$-stack sortable.

In Section 4, we initiate the study of the geometry of $Ln1$ families. We take
a _$Ln1$ family_, $\mathcal{F}_{\boldsymbol{\pi}}$, to be the set of $n$
permutations obtained from a fixed $\boldsymbol{\pi}\in\mathcal{L}n1$ and all
its iterations under the stack-sorting algorithm. We denote the special
$\mathcal{L}n1$ permutation where $L=23\ldots(n-1)$ as $L^{\prime}n1$. The
main results of this section are the following proposition and theorem:

######  Proposition 4.2.

The convex hull of any $Ln1$ family forms a $(n-1)$-simplex in
$\mathbb{R}^{n}$.

######  Theorem 4.3.

The $L^{\prime}n1$ simplex is hollow. Specifically, all non-vertex integer
points in $\triangle_{n}$ lie on the facet formed from the convex hull of
$\mathcal{F}_{L^{\prime}n1}-\\{12\ldots n\\}$.

Lastly, Section 5 deals with the Ehrhart theory of $L^{\prime}n1$ simplices.
The main result of this section is the following:

######  Theorem 5.2.

The $L^{\prime}n1$ simplex $\triangle_{n}$ and $(n-1)$-dimensional lecture
hall simplex $P_{n-1}$ are integrally equivalent. In particular,

$L_{\mathbb{Z}}(\triangle_{n};t)=L_{\mathbb{Z}}(P_{n-1};t)=(t+1)^{n-1}.$

We conclude by exploring the following additional results, including
developing a recursive relationship and proving the Gorenstein property for
translates of the $L^{\prime}n1$ simplex. In what follows, allow
$\boldsymbol{\pi}_{n}$ to denote $L^{\prime}n1$.

######  Theorem 5.9.

For all $\lambda\in\mathbb{R}_{\geq 0}$,

$L_{\mathbb{R}}(\triangle_{n+1}-\boldsymbol{\pi}_{n+1};\lambda)=\sum_{k=0}^{\lfloor
n\lambda\rfloor}\
L_{\mathbb{R}}\left(\triangle_{n}-\boldsymbol{\pi}_{n};\frac{k}{n}\right).$

######  Theorem 5.12.

For all $\lambda\in\mathbb{R}_{\geq 0}$,

$L_{\mathbb{R}}(\triangle_{n}-\boldsymbol{\pi}_{n};\lambda)=L_{\mathbb{R}}((\triangle_{n}-\boldsymbol{\pi}_{n})^{\circ};\lambda+2).$

Hence, any integer translate of $\triangle_{n}-\boldsymbol{\pi}_{n}$, in
particular $\triangle_{n}$, is Gorenstein of index $2$.

##  2\. Background & Preliminaries

###  2.1. The stack-sorting algorithm

The following algorithm for sorting an input sequence was first outlined in
Donald Knuth’s influential work, _The Art of Computer Programming_ [7].

######  Definition 2.1.

Given a permutation $\boldsymbol{\pi}$ on an ordered set, the _stack-sorting
algorithm_ is defined as follows:

  1. (1)

Initialize an empty _stack_ (i.e., a last-in, first-out collection of
elements).

  2. (2)

For each input value $x\in\boldsymbol{\pi}$, starting with the first element,
$\pi_{1}$:

     * •

If the stack is non-empty and $x$ is greater than the _top_ (most-recently
added) element $y$ of the stack, then _pop $y$ from the stack_ (i.e., move $y$
to the output) and repeat.

     * •

Otherwise, _push $x$ onto the stack_ (move $x$ to the top of the stack).

  3. (3)

Once every element has been pushed and there is no input left to consider, pop
all remaining in the stack from last-in to first-in.

For a permutation $\boldsymbol{\pi}$, we denote the output of the stack-
sorting algorithm as $s(\boldsymbol{\pi})$.

$(i)$$2$$1$$3$

$(ii)$$2$$1$$3$

$(iii)$$2$$1$$3$

$(iv)$$2$$1$$3$

Figure 1. A visualization of step $2$ of the stack-sorting algorithm as
presented in Example 2.2.

######  Example 2.2.

The algorithm would sort the permutation $\boldsymbol{\pi}=213$ as follows:

  1. (1)

We begin with an empty stack $S=\\{\\}$.

  2. (2)

The first element in the permutation is $2$. The stack is empty; thus, we push
$2$ onto the stack and obtain $S=\\{2\\}$.

     * •

The next element is $1$. The stack is non-empty but $1$ is not greater the the
top of the stack, $2$. Therefore, we push $1$ onto the stack and obtain
$S=\\{2,1\\}$.

     * •

The last element is $3$. The stack is non-empty and $3$ is greater than the
top, $1$. Thus, we pop $1$ from the stack and obtain the first element of
$s(\boldsymbol{\pi})$ to be $1$. We repeat and the exact same thing occurs
with $2$. We deduce that $s(\boldsymbol{\pi})$ begins with $12$.

  3. (3)

We push $3$ onto an empty stack and then pop it to obtain that
$s(\boldsymbol{\pi})=123$. $\diamondsuit$

######  Definition 2.3.

A permutation $\boldsymbol{\pi}$ on $[n]:=\\{1,2,\dots,n\\}$ is _$t$ -stack-
sortable_ if $t$ iterations of the stack-sorting algorithm yields the identity
permutation, i.e., $s^{t}(\boldsymbol{\pi})=12\cdots n$. Furthermore,
$\boldsymbol{\pi}$ is _exactly_ $t$-stack-sortable if
$s^{t}(\boldsymbol{\pi})=12\cdots n$ and $m<t$ implies
$s^{m}(\boldsymbol{\pi})\neq 12\cdots n$.

The identity permutation itself is defined to be $0$-stack-sortable, and
$s(12\cdots n)=12\cdots n$ for all $n$. All permutations on $[n]$ will reach
the identity after at most $n-1$ iterations of the algorithm. If
$\boldsymbol{\pi}$ is exactly $(n-1)$-stack-sortable, then $\boldsymbol{\pi}$
is said to be _maximal_ with respect to the algorithm. Thus, revisiting
Example 2.2, $\boldsymbol{\pi}=213$ is exactly $1$-stack-sortable, is also
$t$-stack-sortable for any $t\geq 1$, and thus is not maximal.

###  2.2. Polyhedral geometry & Ehrhart theory

In this paper, we investigate the properties of a family of _convex polytopes_
, in particular, a family of lattice simplices. We present some preliminaries
on polyhedral geometry and Ehrhart theory, i.e., the study of lattice-points
in dilations of polytopes.

A _convex polytope_ $P$ is the _convex hull_ of a set of points
$C=\\{\mathbf{x}_{1},\ldots,\mathbf{x}_{k}\\}$, that is,

(1)
$P=\operatorname{conv}(C):=\left\\{\sum_{i=1}^{k}\lambda_{i}\mathbf{x}_{i}:\lambda_{i}\in\mathbb{R}_{\geq
0}\text{ and }\sum_{i=1}^{k}\lambda_{i}=1\right\\}.$

Any expression of the form $\sum_{i=1}^{k}\lambda_{i}\mathbf{x}_{i}$, with the
same conditions as imposed on the $\lambda_{i}$ in (1), is called a _convex
combination_ of the points $\mathbf{x}_{i}$. The convex hull of $C$ is the set
of all convex combinations of points in $C$.

An _affine subspace_ or affine space is a linear subspace with a possible
translation by a fixed vector to its elements. In other words, it is not
required that the origin be part of a set for it to be an affine subspace.
Every linear subspace is an affine subspace.

######  Example 2.4.

Any line in $\mathbb{R}^{2}$ is an affine subspace of $\mathbb{R}^{2}$. The
same applies for any plane in $\mathbb{R}^{3}$. Notice these objects partition
their respective spaces into two parts, those on one side of the line (plane)
and those on the other. A line in $\mathbb{R}^{2}$ is defined by a linear
equation $ax+by+c=0$ and a plane in $\mathbb{R}^{3}$ is defined by
$ax+by+cz+d=0$, where $a,b,c,d\in\mathbb{R}$ and $x,y,z$ are variables. At
least one coefficient attached to a variable must be nonzero. $\diamondsuit$

The concept from Example 2.4 can be generalized to define a _hyperplane_ in
$\mathbb{R}^{n}$ as an $(n-1)$-dimensional affine subspace determined by the
solution space of a linear equation

$a_{0}+a_{1}x_{1}+\cdots+a_{n}x_{n}=0,$

where each $a_{i}\in\mathbb{R}$ and at least one of $a_{1}$ through $a_{n}$ is
nonzero. Any hyperplane of $\mathbb{R}^{n}$ partitions $\mathbb{R}^{n}$ into
two parts called _halfspaces_ ; namely, without loss of generality,

$a_{0}+a_{1}x_{1}+\cdots+a_{n}x_{n}\geq 0\quad\text{ and }\quad
a_{0}+a_{1}x_{1}+\cdots+a_{n}x_{n}<0.$

A polytope $P$ can also be defined as the bounded intersection of finitely
many _halfspaces_ ; equivalently, this is the bounded solution set of a system
of linear inequalities.

A polytope $P$ is contained in the _affine hull_ of $C$, denoted by
$\operatorname{aff}(C)$, which is similar to (1), but with the looser
condition that $\lambda_{i}\in\mathbb{R}$. We can characterize
$\operatorname{aff}(C)$ as the smallest affine subspace containing
$\operatorname{conv}(C)$. Moreover, the set
$C=\\{\mathbf{x}_{1},\ldots,\mathbf{x}_{k}\\}$ is _affinely independent_ if
$\operatorname{aff}(C)$ is a $(k-1)$-dimensional space. Equivalently, $C$ is
affinely independent if and only if vectors,

$\mathbf{x}_{1}-\mathbf{x}_{i},\dots,\mathbf{x}_{i-1}-\mathbf{x}_{i},\mathbf{x}_{i+1}-\mathbf{x}_{i},\dots,\mathbf{x}_{k}-\mathbf{x}_{i}\
\in\mathbb{R}^{n},$

are linearly independent in $\mathbb{R}^{n}$ for any $i\in[k]$.

The _dimension_ of a polytope $P$, denoted $\dim(P)$, is the dimension of its
affine hull. We say a polytope $P\subset\mathbb{R}^{n}$ is _full-dimensional_
if $\dim(P)=n$. The convex hull of $n+1$ affinely independent points forms a
special polytope called a _simplex_ of dimension $n$. A simplex generalizes
the concept of a triangle in $\mathbb{R}^{2}$ or the tetrahedron in
$\mathbb{R}^{3}$.

Changing every inequality to strict inequalities in the minimal halfspace
description of $P$ yields the _interior_ of $P$, denoted $P^{\circ}$. Changing
every inequality to an equality in a minimal halfspace description yields the
_boundary_ of $P$, denoted $\partial P$. Note that $P=P^{\circ}\uplus\partial
P$, where $\uplus$ denotes a disjoint union. A polytope is _hollow_ if all its
integer points lie on its boundary or, equivalently, it has no integer points
in its interior.

A polytope is said to be a _lattice polytope_ if all of its vertices have
integral coordinates. The _Ehrhart function_ (or _lattice-point enumerator_ or
_discrete volume_) of any polytope $P$ is defined as

$L_{\mathbb{Z}}(P;t):=|tP\cap\mathbb{Z}^{n}|,$

where $tP=\\{t\mathbf{x}:\ \mathbf{x}\in P\\}$ and $t$ is an integer. When $P$
is a lattice polytope, the Ehrhart function is a polynomial in $t$ of degree
equal to the dimension of $P$, which we refer to as the _Ehrhart polynomial_.

We can encode the information of an Ehrhart polynomial in a generating series
to obtain the _Ehrhart series_ of a polytope:

$\mathrm{Ehr}_{\mathbb{Z}}(P;z):=1+\sum_{n\in\mathbb{Z}_{+}}L_{\mathbb{Z}}(P;n)z^{n}=\frac{h_{\mathbb{Z}}^{*}(P;z)}{(1-z)^{n+1}},$

where $n$ is the dimension of $P$ and
$h^{*}_{\mathbb{Z}}(P;z)=1+h^{*}_{1}z+\cdots+h^{*}_{n}z^{n}$ is a polynomial
in $z$ of degree at most $n$ called the _$h^{*}$ -polynomial_ of $P$. The
coefficients $h^{*}_{i}$ of this polynomial are nonnegative integers and have
relevant combinatorial interpretations. The interested reader can consult [2]
for more information.

A fundamental result in Ehrhart theory is one of _reciprocity_ , known as
_Ehrhart-Macdonald reciprocity_. For any convex lattice polytope $P$ of
dimension $d$:

$L_{\mathbb{Z}}(P;-t)=(-1)^{n}L_{\mathbb{Z}}(P^{\circ};t).$

This not only provides a relationship between a polytope and its interior, but
also provides an interpretation for negative values of $t$. Evaluating an
Ehrhart polynomial at $-t$ gives the lattice count for the $t^{\text{th}}$
dilate of its interior up to a sign.

A lattice polytope $P\subset\mathbb{R}^{d}$ is said to be _Gorenstein of index
$k$_ if there exists a positive integer $k$ such that

(2) $L_{\mathbb{Z}}(P^{\circ};k-1)=0,\quad
L_{\mathbb{Z}}(P^{\circ};k)=1,\quad\text{ and
}L_{\mathbb{Z}}(P^{\circ};t)=L_{\mathbb{Z}}(P;t-k)$

for all integers $t>k$ [2].

More generally, for any convex polytope $P\subseteq\mathbb{R}^{d}$ (i.e., $P$
has any vertices in $\mathbb{R}^{n}$), the real Ehrhart counting function for
all dilates $\lambda\in\mathbb{R}_{\geq 0}$ is defined as:

$L_{\mathbb{R}}(P;\lambda):=|\lambda P\cap\mathbb{Z}^{n}|.$

For more on Ehrhart theory, one can consult [2] for an in-depth treatment of
the material. A larger survey on the combinatorics of polytopes can be found
in [8]. Additionally, one can learn more about rational Ehrhart theory from
[1].

######  Example 2.5.

$\square^{2}$$2\square^{2}$$3\square^{2}$ Figure 2. The first three integral
dilates of the two-dimensional unit cube $\square^{2}$.

We define the _$n$ -dimensional unit cube_ as the convex hull over all binary
strings of length $n$, i.e., the binary strings form the vertex set. The two-
dimensional unit cube $\square^{2}$ contains $4$ points in its first dilate,
$9$ in its second, $16$ in its third, etc. Without much difficulty, one can
show directly or inductively that:

$L_{\mathbb{Z}}(\square^{2};t)=\left|t\square^{2}\cap\mathbb{Z}^{2}\right|=(t+1)^{2}.$

Analogous argumentation can be used to show
$L_{\mathbb{Z}}(\square^{3};t)=(t+1)^{3}$; and, in fact, that

$L_{\mathbb{Z}}(\square^{n};t)=(t+1)^{n}.$

We can encode the lattice-point enumerator in a generating function to obtain
the Ehrhart series of $\square^{n}$:

$\sum_{n\in\mathbb{N}}L_{\mathbb{Z}}(\square^{n};t)z^{n}=\frac{\sum_{k=0}^{n-1}A(n,k)z^{k}}{(1-z)^{n-1}},$

where $A(n,k)$ denotes the Eulerian number, i.e., the number of permutations
of $1$ through $n$ having exactly $k$ descents. Further, by reciprocity, we
have:

$L_{\mathbb{Z}}(\square^{n};-t)=(1-t)^{n}=(-1)^{n}(t-1)^{n}=(-1)^{n}L_{\mathbb{Z}}({\square^{n}}^{\circ};t),$

and can conclude the interior of the $t$-dilate has exactly as many lattice
points as the entirety of the $(t-2)$-dilate. For example, the interior of
$3\square^{2}$ has $4$ lattice points, the same number as all of
$\square^{2}$. We also notice that the interior of the first dilate of
$\square^{n}$ is empty and the interior of the second dilate has exactly one
integer point. These are the conditions from (2) for $k=2$, which implies
$\square^{n}$ is Gorenstein of index 2, an important property that will appear
again later in the paper. $\diamondsuit$

###  2.3. Integral and unimodular equivalence

We recall the definition of a unimodular transformation.

######  Definition 2.6.

A _unimodular transformation_ in $\mathbb{R}^{n}$ is a linear transformation
$U$, i.e., a $n$ by $n$ matrix, with coefficients in $\mathbb{Z}$ such that
$\det(U)=\pm 1$.

The following result from [6] provides a heuristic for determining whether two
polytopes have the same lattice-point count.

######  Proposition 2.7.

If a linear transformation on a lattice polytope $P$ is unimodular, then it
preserves the lattice.

In other words, the resulting polytope will have the same integer point count
as $P$. However, not all polytopes that have equivalent lattice counts have a
unimodular transformation between them. For example, if the two polytopes have
square matrix representations (vertices as columns of a matrix), they must
have the same determinant up to a sign for a unimodular transformation to
exist between them.

######  Definition 2.8.

Two lattice polytopes $P\subset\mathbb{R}^{m}$ and $Q\subset\mathbb{R}^{n}$
are _integrally equivalent_ if there exists an affine transformation
$\varphi:\mathbb{R}^{m}\rightarrow\mathbb{R}^{n}$ whose restriction to $P$
preserves the lattice. In other words, $\varphi$ is a bijection

$P\cap\mathbb{Z}^{m}\longleftrightarrow Q\cap\mathbb{Z}^{n}.$

In particular, $L_{\mathbb{Z}}(P;t)=L_{\mathbb{Z}}(Q;t)$ for all
$t\in\mathbb{N}$.

We now have an effective method to show two $n$-polytopes are integrally
equivalent: we can search for a unimodular transformation from the lattice
points of one polytope to the other.

###  2.4. Lecture-hall simplices

We begin this subsection by recalling the definition of lecture-hall
partitions, which were studied in [5]. They were further studied in the
context of lecture-hall simplices in [4].

######  Definition 2.9.

A _lecture-hall partition_ of length $n$ is a sequence
$\\{\alpha_{1},\ldots,\alpha_{n}\\}\in\mathbb{Z}^{n}$ such that

$0\leq\alpha_{1}\leq\frac{\alpha_{2}}{2}\leq\cdots\leq\frac{\alpha_{n}}{n}.$

We can construct the _lecture-hall simplex_ of dimension $n$ as

$P_{n}:=\operatorname{conv}\left\\{\boldsymbol{\alpha}\in\mathbb{Z}^{n}:0\leq\alpha_{1}\leq\frac{\alpha_{2}}{2}\leq\cdots\leq\frac{\alpha_{n}}{n}\leq
1\right\\}.$

Dilations of $P_{n}$ yields the result
$tP_{n}=\operatorname{conv}\left\\{\boldsymbol{\alpha}\in\mathbb{Z}^{n}:0\leq\alpha_{1}\leq\frac{\alpha_{2}}{2}\leq\cdots\leq\frac{\alpha_{n}}{n}\leq
t\right\\}$. We also have that $P_{n}\subset 2P_{n}\subset\cdots\subset
tP_{n}$ because $(0,0,\ldots,0)\in P_{n}$. This fixes a vertex and thus avoids
any translations of the polytope away from previous dilates, an observation
that will become useful later on.

######  Example 2.10.

For $n=4$, we must have $0\leq\alpha_{1}\leq\cdots\leq\frac{\alpha_{4}}{4}\leq
1$. The first points we can identify in this simplex $P_{n}$ are $(0,0,0,0)$
as well as $(1,2,3,4)$. We can also take the four points $(0,0,0,n)$ for
$n\in[4]$. We have $(0,0,1,2)$, $(0,0,1,3)$, and $(0,0,1,4)$; $(0,0,2,3)$ and
$(0,0,2,4)$; as well as $(0,0,3,4)$. Finally, we can take $(0,1,2,3)$,
$(0,1,2,4)$, $(0,1,3,4)$, and $(0,2,3,4)$.

This consists of a total of $16$ lattice points and these are the only points
in the simplex in the first dilate. Note, not all of these points are
vertices, for example, $(0,0,0,2)$ will be on the line between the origin and
$(0,0,0,4)$. This gives $L_{\mathbb{Z}}(P_{4};1)=16=2^{4}=(1+1)^{4}$. In fact,
we will later find the distribution of points in the lecture-hall simplex and
our $L^{\prime}n1$ simplex to be remarkably similar. $\diamondsuit$

In general, the vertex set for $P_{n}$ consists of exactly the $n+1$ affinely
independent points:

$(0,0,\ldots,0),(0,0,\ldots,n),\ldots(0,2,\ldots,n),\text{ and
}(1,2,\ldots,n).$

More information about $P_{n}$, including the proof of the following
proposition can be found in [4].

######  Proposition 2.11 (Theorems 1 and 2.2,[4]).

Let $P_{n}$ be the lecture-hall simplex of dimension $n$.

$L_{\mathbb{Z}}(P_{n};t)=(t+1)^{n}.$

Observe that the lattice-point enumerator is equivalent for both the lecture-
hall simplex $P_{n}$ and the $n$-dimensional unit cube $\square^{n}$ from
Example 2.5. We will show that our $L^{\prime}n1$ simplices (defined in
Section 4) are also in bijection with these polytopes, and hence have the same
lattice-point-enumerator.

###  2.5. Notation

We conclude our preliminaries with a notation table for reference. Note that
it includes notation for terms that will be defined in later sections.

Table 1. Notation Table Symbol | Meaning
---|---
$[n]$ | the set $\\{1,2,\ldots,n\\}$
$S_{n}$ | the set of all permutations on $[n]$
$s(\boldsymbol{\pi})$ | the output from stack-sorting a permutation $\boldsymbol{\pi}$
$\operatorname{conv}$ | convex hull
$\operatorname{aff}$ | affine hull
$P^{\circ}$ | the interior of the polytope $P$
$\partial P$ | the boundary of the polytope $P$
$L_{\mathbb{Z}}(P;t)$ | the number of lattice points in $tP$ where $t\in\mathbb{Z}$
$\mathrm{Ehr}_{\mathbb{Z}}(P;z)$ | the ordinary generating function for $L_{\mathbb{Z}}(P;t)$
$h^{*}_{\mathbb{Z}}(P;z)$ | the polynomial numerator in the rational form of $\mathrm{Ehr}_{\mathbb{Z}}(P;z)$
$\square^{n}$ | the $n$-dimensional unit cube
$P_{n}$ | the $n$-dimensional lecture-hall simplex
$\mathcal{L}n1$ | the set of permutations on $[n]$ ending in $n1$
$L^{\prime}n1$ or $\boldsymbol{\pi}_{n}$ | the permutation $(2,3,\ldots,n,1)$
$\mathcal{F}_{L^{\prime}n1}$ | the $L^{\prime}n1$ family
$\triangle_{n}$ | the $(n-1)$-dimensional $L^{\prime}n1$ simplex
$\mathbf{e}$ | the permutation $(1,2,\ldots,n)$
$P^{\prime}$ | the projection of $P\subset\mathbb{R}^{n}$ onto the hyperplane defined by $x_{n}=0$
$\overline{P}$ | the projection of $P\subset\mathbb{R}^{n}$ into $\mathbb{R}^{n-1}$ by removing each point’s last coordinate
$\rule{0.0pt}{12.91663pt}\hat{P}$ | the lift of $P\subset\mathbb{R}^{n}$ into $\mathbb{R}^{n+1}$ by appending a last coordinate of $0$ to each point

##  3\. Stack-Sorting on $Ln1$ permutations

In this section we explore the behavior of the stack-sorting algorithm on a
special family of permutations, one we will find to act in an interesting way
with respect to the algorithm.

######  Definition 3.1.

A permutation $\boldsymbol{\pi}\in S_{n}$ is a _$Ln1$ permutation_ if
$\boldsymbol{\pi}$ has the form $Ln1$, such that $L$ is a permutation of
$\\{2,3,\ldots,n-1\\}$, i.e., a permutation that exactly ends with $n$ and
then $1$. We denote the set of all permutations of such form as
$\mathcal{L}n1$ and write $\boldsymbol{\pi}\in\mathcal{L}n1$ if it takes the
appropriate form.

Notice that there are $(n-2)!$ distinct permutations in $\mathcal{L}n1$ as
they can be counted by the number of different possibilities for $L$, the
number of permutations on a set of $n-2$ elements. We denote the special $Ln1$
permutation where $L=23\ldots(n-1)$, i.e., where $L$ is already sorted in
ascending order, as $\mathrm{}{L^{\prime}n1}$.

######  Example 3.2.

For $n=4$, $2341$ and $3241$ are the two permutations that comprise the set
$\mathcal{L}41$. $L^{\prime}n1$ will always refer to a unique permutation; as
an example, $L^{\prime}51=23451$. We have that $L^{\prime}51$ is one of
$(5-2)!=6$ permutations in the set $\mathcal{L}51$.

######  Lemma 3.3.

Let $\boldsymbol{\pi}=\pi_{1}\pi_{2}\dots\pi_{n}$ be a permutation of $n$
distinct ordered elements and let $x=\max\\{\pi_{1},\pi_{2},\dots,\pi_{n}\\}$.
If $L$ and $R$ are the sub-permutations of $\boldsymbol{\pi}$ such that
$\boldsymbol{\pi}=LxR$, then the stack-sorting algorithm yields
$s(\boldsymbol{\pi})=s(L)s(R)x$.

This result is a generalization of [3, Lemma 1.2]. In other words, the
algorithm sorts everything to the left of $x$ (the maximum element) first,
regardless of the contents to its right. Then, everything to the right of $x$
is sorted and $x$ takes the $n^{\text{th}}$ and final place in the output
permutation.

###### Proof.

Consider an arbitrary $\pi_{i}\in L$. Since $x\not\in L$ and thus $\pi_{i}\neq
x$, we must have $\pi_{i}<x$. This implies that when all of $L$ has been
pushed on the stack and $x$ is taken as input by the algorithm, all of $L$
will be popped _before_ $x$ is pushed on the stack. After all of $L$ has been
popped, $x$ will be pushed onto an empty stack.

Similarly, it can be determined that if $\pi_{j}\in R$, then $\pi_{j}<x$.
Because there is no element greater than $x$ that can pop it from the stack,
all of $R$ will be pushed and then subsequently popped in some order before
$x$ (at the bottom) is finally popped. Hence, the algorithm will sort $R$
entirely into $s(R)$ before popping $x$; by combining these two results, we
obtain that $s(\boldsymbol{\pi})=s(L)s(R)x$. ∎

As a consequence of this lemma, we obtain the following two corollaries.

######  Corollary 3.4.

If $\boldsymbol{\pi}\in\mathcal{L}n1$, then $s(\boldsymbol{\pi})=s(L)1n$.

######  Corollary 3.5.

If $\boldsymbol{\pi}\in\mathcal{L}n1$, then $s(\boldsymbol{\pi})$ exactly ends
with $n-1$, $1$, then $n$.

Note that Corollary 3.5 is simply an extra application of Lemma 3.3 to
Corollary 3.4, specifically to $s(L)$. Both corollaries can be readily
extended to a biconditional.

######  Theorem 3.6.

Let $\boldsymbol{\pi}\in\mathcal{L}n1$. For all $i\in[n-2]$,
$s^{i}(\boldsymbol{\pi})$ exactly ends with

$(n-i)1(n-i+1)\ldots(n-2)(n-1)n.$

###### Proof.

The base case when $i=1$ is the statement of Corollary 3.5. Assume the theorem
holds for $s^{i-1}(\boldsymbol{\pi})$, i.e., it begins with some permutation
$\boldsymbol{\rho}$ of $\\{2,3,\ldots,n-i\\}$ and ends exactly with
$(n-i+1)1(n-i+2)\ldots(n-1)n$.

Consider another application of the stack-sorting algorithm to
$s^{i-1}(\boldsymbol{\pi})$. Notice that $(n-i+1)$ is greater than all
elements in $\boldsymbol{\rho}$. Thus, when $(n-i+1)$ is considered by the
algorithm, all elements in $\boldsymbol{\rho}$ will be popped before $(n-i+1)$
is pushed. By Lemma 3.3, we know the final element in $s(\boldsymbol{\rho})$
will be $n-i$. Now, the algorithm sorts $(n-i+1)1(n-i+2)\ldots(n-1)n$ into
$1(n-i+1)(n-i+2)\ldots(n-1)n$.

Therefore, we obtain that $s^{i}(\boldsymbol{\pi})$ exactly ends with
$(n-i)1(n-i+1)\ldots(n-1)n$ as desired. Further, note that when $i=n-1$ we
finally obtain the identity permutation after the maximal number of
iterations. ∎

######  Example 3.7.

For $n=6$, Theorem 3.6 tells us that $\boldsymbol{\pi}\in\mathcal{L}n1$ ends
with $61$, as expected; $s(\boldsymbol{\pi})$ ends with $516$, as stated in
Corollary 3.5. Further, $s^{2}(\boldsymbol{\pi})$ ends with $4156$,
$s^{3}(\boldsymbol{\pi})$ ends with $31456$, and thus
$s^{3}(\boldsymbol{\pi})=231456$. We conclude $s^{4}(\boldsymbol{\pi})=213456$
and $s^{5}(\boldsymbol{\pi})=123456$.

######  Remark 3.8.

For all $n\geq 4$, if $\boldsymbol{\pi}\in\mathcal{L}n1$, then
$s^{n-3}(\boldsymbol{\pi})=2314\cdots n$, $s^{n-2}(\boldsymbol{\pi})=213\cdots
n$, and of course $s^{n-1}(\boldsymbol{\pi})=123\cdots n$.

This tells us that regardless of what $\boldsymbol{\pi}\in\mathcal{L}n1$ is,
the final three iterations of the algorithm are all the same. The permutations
“meet” at this point. However, before those last few iterations, we cannot
explicitly describe what $s^{i}(\boldsymbol{\pi})$ will start with without
specifying $L$. Specifying $L=L^{\prime}$ yields the following useful result.

######  Corollary 3.9.

Let $\boldsymbol{\pi}=L^{\prime}n1$. For $i\in[n-1]$,

$s^{i}(\boldsymbol{\pi})=23\cdots(n-i)1(n-i+1)\cdots(n-1)n.$

Inputting any sequence in ascending order will have the stack-sorting
algorithm output the original sequence. Thus, the initial segment
$\boldsymbol{\rho}$ of the permutation will always be $23\ldots(n-i-1)$. We
effectively learn, not only about the maximality of $Ln1$, but also enumerate
the general form of each algorithm iteration in the special case of
$L^{\prime}n1$.

######  Definition 3.10.

We define the _$L^{\prime}n1$ family_ to be the set of $n$ permutations
obtained from $L^{\prime}n1$ and its iterations under the stack-sorting
algorithm. We denote this family by $\mathcal{F}_{L^{\prime}n1}$. Generally,
any $Ln1$ permutation with its iterations up to and including the identity
form a _$Ln1$ family_.

######  Example 3.11.

Consider $L^{\prime}51=23451$. By Corollary 3.9 we obtain that
$s(L^{\prime}51)=23415$, $s^{2}(L^{\prime}51)=23145$,
$s^{3}(L^{\prime}51)=21345$, and $s^{4}(L^{\prime}51)=12345$. Notice that
$L^{\prime}51$ is maximal and

$\mathcal{F}_{L^{\prime}51}=\\{23451,23415,23145,21345,12345\\}.$

We are essentially just moving the ‘1’ one place to the left each time.

The following theorem tells us that maximal permutations and $Ln1$
permutations are equivalent.

######  Theorem 3.12.

A permutation $\boldsymbol{\pi}\in\mathcal{L}n1$ if and only if
$\boldsymbol{\pi}$ is exactly $(n-1)$-stack sortable.

###### Proof.

If $\boldsymbol{\pi}\in\mathcal{L}n1$, the desired consequence is a result of
Theorem 3.6, which dictates that $s^{n-2}(\boldsymbol{\pi})=213\cdots n$. It
follows $s^{n-1}(\boldsymbol{\pi})=123\cdots n$, therefore $\boldsymbol{\pi}$
is exactly $(n-1)$-stack-sortable.

If $\boldsymbol{\pi}\not\in\mathcal{L}n1$, then $1$ is not in the last place
of $\boldsymbol{\pi}$ _or_ some $k<n$ is directly left of $1$ in
$\boldsymbol{\pi}$. We consider these two cases.

######  Case 1.

If $1$ is not the last digit in $\boldsymbol{\pi}$, then $1$ will reach the
first position of the output in less than $n-1$ iterations of the algorithm.
This is because the $1$ will always move at least one place to the left if it
is not already leftmost; the top of the stack will always be greater than $1$,
so we will always push it onto the stack where it will be immediately popped
before the previous top element. The remaining $n-1$ elements will also have
been sorted in less than $n-1$ iterations because the maximal possible number
of iterations it takes to sort the remaining sequence of $n-1$ elements is
$n-2$. Thus, $\boldsymbol{\pi}$ can be stack-sorted in less than $n-1$
iterations and is not exactly $(n-1)$-stack-sortable.

######  Case 2.

If $\boldsymbol{\pi}=Lk1$ such that $k<n$ and $L$ is a permutation of
$\\{1,2,\ldots,k-1,k+1,\ldots,n\\}$, then $1$ will move more than one space
left after the first iteration of the algorithm. This is because $n$ will
remain on the bottom of the stack (possibly more elements) while $k$ and $1$
are both pushed and then cleared first, therefore $s(\boldsymbol{\pi})$ will
contain $1$ to the left of both $k$ and $n$. Similar to the reasoning in the
prior case, $1$ will reach the first position of the output in less than $n-2$
more iterations, while the remaining elements will also be sorted in ascending
order in less than $n-2$ more iterations by maximality. As before,
$\boldsymbol{\pi}$ can be stack-sorted in less than $n-1$ iterations and is
not exactly $(n-1)$-stack-sortable.

∎

##  4\. Geometry of $Ln1$ families

In this section we interpret our permutations as points in $\mathbb{R}^{n}$,
which leads us to ask: What is the geometry arising by taking the convex hull
of stack-sorting iterations?

######  Proposition 4.1.

Every $Ln1$ family forms an affinely independent set in $\mathbb{R}^{n}$.

###### Proof.

Consider $\boldsymbol{\pi}\in\mathcal{L}n1$. Let $\mathbf{e}=123\cdots
n=s^{n-1}(\boldsymbol{\pi})$. Subtracting $\mathbf{e}$ component-wise from
each of the other family members and utilizing Theorem 3.6, we deduce that
$s^{i}(\boldsymbol{\pi})-\mathbf{e}$ must end with $1(1-n+i)0\ldots 000$,
where the $1$ is in the $(n-i-1)^{\text{th}}$ position.

Note that

$s^{n-2}(\boldsymbol{\pi})-\mathbf{e}=1(-1)00\cdots 00\text{ and
}s^{n-3}(\boldsymbol{\pi})-\mathbf{e}=11(-2)0\cdots 00$

by Remark 3.8. As vectors in $\mathbb{R}^{n}$, these two are linearly
independent as no scalar multiple of one vector can produce the other – there
is no way to obtain a nonzero number in the third coordinate of
$s^{n-2}(\boldsymbol{\pi})-\mathbf{e}$.

Assume
$S=\\{s^{n-2}(\boldsymbol{\pi})-\mathbf{e},s^{n-3}(\boldsymbol{\pi})-\mathbf{e},\ldots,s^{n-k+1}(\boldsymbol{\pi})-\mathbf{e}\\}$
is a linearly independent set in $\mathbb{R}^{n}$ and consider
$S\cup\\{s^{n-k}(\boldsymbol{\pi})-\mathbf{e}\\}$. We know that
$s^{n-k}(\boldsymbol{\pi})-\mathbf{e}$ is the only vector in this set with a
nonzero number in the $k^{\text{th}}$ coordinate. Therefore, no linear
combination of the vectors in $S$ yields
$s^{n-k}(\boldsymbol{\pi})-\mathbf{e}$ and
$S\cup\\{s^{n-k}(\boldsymbol{\pi})-\mathbf{e}\\}$ is linearly independent.

We can conclude the original $Ln1$ family is an affinely independent set in
$\mathbb{R}^{n}$. ∎

As a corollary we obtain the following proposition.

######  Proposition 4.2.

The convex hull of any $Ln1$ family forms a $(n-1)$-simplex in
$\mathbb{R}^{n}$.

When the family is $\mathcal{F}_{L^{\prime}n1}$, we will denote its convex
hull as $\triangle_{n}$ and refer to it as the _$L^{\prime}n1$ simplex_.
Notice that the simplex formed is one dimension less than the dimension of its
ambient space. This is not a trivial detail, it will necessitate some caution
as we proceed to develop the Ehrhart theory of $\triangle_{n}$, especially
when comparing it to the unit cube and lecture-hall simplex which are both
full-dimensional.

######  Theorem 4.3.

The $L^{\prime}n1$ simplex is hollow. Specifically, all non-vertex integer
points in $\triangle_{n}$ lie on the facet formed from the convex hull of
$\mathcal{F}_{L^{\prime}n1}\setminus\\{\mathbf{e}\\}$.

###### Proof.

Note that all elements of $\mathcal{F}_{L^{\prime}n1}$, treated as points in
$\mathbb{R}^{n}$, have $2$ as their fist coordinate entry except for
$\mathbf{e}=(1,2,\ldots,n)$. This implies that if we take a convex combination
of points in $\mathcal{F}_{L^{\prime}n1}$, the first coordinate entry is of
the form

$\lambda_{1}+2(\lambda_{2}+\dots+\lambda_{n})=k,$

where $k$ is a real number and $0<k\leq 2$ because of the constraints on
$\lambda$ from Equation (1). Further note that
$\lambda_{2}+\dots+\lambda_{n}=1-\lambda_{1}$. Hence, $2-\lambda_{1}=k$, which
we can rewrite as $\lambda_{1}=2-k$. For our first-coordinate entry to be an
integer, we can only consider $k=1$ and $2$. If $k=1$, then $\lambda_{1}=1$
and we obtain the vertex point $\mathbf{e}$. If $k=2$, then $\lambda_{1}=0$
and the integer point exists on the facet containing all vertices except
$\mathbf{e}$. ∎

Hollowness is not a property special to $\triangle_{n}$. The unit cube and
lecture-hall simplex also satisfy this property. Because we know their Ehrhart
polynomials, we can argue this much more efficiently using reciprocity:

$(-1)^{d}L_{\mathbb{Z}}(P_{n}^{\circ};1)=L_{\mathbb{Z}}(P_{n};-1)=(-1+1)^{n}=0.$

######  Remark 4.4.

The lecture-hall simplex is even more so like $\triangle_{n}$ because it is
hollow in the exact same way, all non-vertex points are on the facet formed
from the convex hull of its vertices excluding $(1,2,\ldots,n)$. An almost
identical argument to Theorem 4.3 can show this to be the case.

######  Remark 4.5.

A lattice polytope $P$ is hollow if and only if $-1$ is a root of its Ehrhart
polynomial. More generally, the dilate $tP$ is hollow if and only if $-t$ is a
root of the Ehrhart polynomial of $P$.

##  5\. Ehrhart theory of $L^{\prime}n1$ simplices

We now develop the Ehrhart theory of our family of $L^{\prime}n1$ simplices.
There is a slight caveat that prevents us from constructing unimodular
transformations directly between $\triangle_{n}$ and $\square^{n-1}$ or
$P_{n-1}$ because $\triangle_{n}$, unlike the other objects being full-
dimensional in $\mathbb{R}^{n-1}$, is $(n-1)$-dimensional in $\mathbb{R}^{n}$.
This means a matrix representation of $\triangle_{n}$ will be $n$ by $n$,
while that of say $P_{n-1}$ will be $n-1$ by $n$ or $\square^{n-1}$ will be
$n-1$ by $2^{n-1}$. Any transformation between them is a non-square matrix and
thus cannot be unimodular.

We transform from $\triangle_{n}$ to $P_{n-1}$ instead of $\square^{n-1}$
because the former is also a simplex of the same dimension, which implies they
both have $n$ vertices. We then need to make a correction for the difference
in their ambient spaces.

One option, the one we will proceed with, is to lift $P_{n-1}$ to
$\mathbb{R}^{n}$ by appending a new last coordinate that does not affect the
lattice point count. We will establish a unimodular transformation from this
new construction to $\triangle_{n}$. Another option is to project
$\triangle_{n}$ onto a hyperplane, ensuring the lattice count remains
preserved, and then transform this construction in $\mathbb{R}^{n-1}$ to
$P_{n-1}$. Regardless, to construct a unimodular transformation we need to
compare two polytopes living in the same ambient space.

Before we proceed, a quick observation about the recursive structure of the
lecture-hall simplex will be insightful. Let $\mathbf{0_{n}}$ be the zero
vector in $\mathbb{R}^{n}$.

######  Proposition 5.1.

Let $V_{n}$ denote the vertex set of $P_{n}$ and
$Q_{n}:=\operatorname{conv}(V_{n}\setminus\\{\mathbf{0_{n}}\\})$. Then

$L_{\mathbb{Z}}(P_{n};t)=L_{\mathbb{Z}}(Q_{n+1};t).$

In other terms, the polytopes $\operatorname{conv}(V_{n})$ and
$\operatorname{conv}(V_{n+1}\setminus\\{\mathbf{0_{n+1}}\\})$ are integrally
equivalent.

###### Proof.

Consider $V_{n}$, which consists of the points
$\mathbf{0_{n}},(0,0,\ldots,n),\ldots,(0,2,\ldots,n),(1,2,\ldots,n)$. Append a
last coordinate of $n+1$ to each vertex to obtain a polytope in
$\mathbb{R}^{n+1}$ that is integrally equivalent to the original. Notice this
new set is $V_{n+1}$ without $\mathbf{0_{n+1}}$. ∎

######  Theorem 5.2.

The $L^{\prime}n1$ simplex $\triangle_{n}$ and $(n-1)$-dimensional lecture
hall simplex $P_{n-1}$ are integrally equivalent. In particular,

$L_{\mathbb{Z}}(\triangle_{n};t)=L_{\mathbb{Z}}(P_{n-1};t).$

We employ the notation $S+\mathbf{v}$ when appropriate to denote adding the
point $\mathbf{v}$ to each point in the set $S$. If $M$ is a matrix,
$M+\mathbf{v}$ adds $\mathbf{v}$ to each column of $M$.

###### Proof.

Let $M$ be the $n$ by $n$ matrix representation of $\triangle_{n}$ with column
$i$ as $s^{n-i}(L^{\prime}n1)$. Further, let $L$ be the $n$ by $n$ lower-
triangular matrix of $-1$’s, let $v_{k}=\sum_{i=2}^{k}i$, and let
$\mathbf{v}_{k}=(v_{2},v_{3},\ldots,v_{k+1})$.

We have that $L\cdot M+\mathbf{v}_{n}$ has columns
$(1,2,\ldots,n),(0,2,\ldots,n),\ldots(0,0,\ldots,n)$. This is the vertex set
$V_{n}$ of $P_{n}$ without $\mathbf{0_{n}}$. Recall from Proposition 5.1 that
$Q_{n}=\operatorname{conv}(V_{n}\setminus\\{\mathbf{0_{n}}\\})$ yields a
polytope integrally equivalent to $P_{n-1}$. We now have a unimodular
transformation from $\triangle_{n}$ to an integrally equivalent polytope to
$P_{n-1}$, from which we obtain:

$L_{\mathbb{Z}}(\triangle_{n};t)=L_{\mathbb{Z}}(Q_{n}-\mathbf{v};t)=L_{\mathbb{Z}}(Q_{n};t)=L_{\mathbb{Z}}(P_{n-1};t).$

∎

######  Example 5.3.

Observe in Figure 3 that the three polytopes, the $2341$ simplex
$\triangle_{4}\subseteq\mathbb{R}^{4}$, the 3-unit cube $\square^{3}$, and the
3-dimensional lecture-hall simplex $P_{3}$, all share the same lattice-point
count in their first dilate. Also note that $\triangle_{4}$ is a 3-dimensional
polytope in $\mathbb{R}^{4}$, whereas the other two polytopes are full-
dimensional in their ambient space. Notice the similarity between how points
are distributed in $P_{3}$ and $\triangle_{3}$.

Figure 3. From Left to Right: the $2341$-simplex
$\triangle_{4}\subseteq\mathbb{R}^{4}$, the 3-unit cube $\square^{3}$, and the
3-dimensional lecture-hall simplex $P_{3}$.

Since the Ehrhart polynomial for the unit cube and the lecture-hall simplex is
known, we quickly obtain the following result as a corollary.

######  Corollary 5.4.

The lattice point enumerator of the $L^{\prime}n1$ simplex is

$L_{\mathbb{Z}}(\triangle_{n};t)=(t+1)^{n-1}.$

######  Corollary 5.5.

The Ehrhart series of the $L^{\prime}n1$ simplex is

$\mathrm{Ehr}_{\mathbb{Z}}(\triangle_{n};z)=\frac{\sum_{k=0}^{n-1}A(n,k)z^{k}}{(1-z)^{n-1}}$

where $A(n,k)$ is the number of permutations of $[n]$ with $k$ descents.

######  Remark 5.6.

By Ehrhart-Macdonald reciprocity, the following holds:

$\displaystyle L_{\mathbb{Z}}(\triangle_{n}^{\circ};t+2)$
$\displaystyle=(-1)^{n-1}L_{\mathbb{Z}}(\triangle_{n};-t-2)$
$\displaystyle=(-1)^{n-1}(-t-1)^{n-1}=(t+1)^{n-1}=L_{\mathbb{Z}}(\triangle_{n};t).$

Also note that $L_{\mathbb{Z}}(\triangle_{n}^{\circ};1)=0$ and
$L_{\mathbb{Z}}(\triangle_{n}^{\circ};2)=1$. Therefore, $\triangle_{n}$ is
Gorenstein of index $2$.

###  5.1. Using real lattice-point enumerators for recursive structures

Initial efforts to obtain the lattice-point enumerator for $L^{\prime}n1$
simplices resulted in differing techniques than those used previously in the
section. While a less direct method, the approach used in the following
results yielded some connections to _real_ lattice-point enumerators of
various translations of our simplices which we present now.

Note that an integral equivalence of two polytopes, as defined in Definition
2.8, does not imply that the two polytopes have the same _real_ lattice-point
count. For example, $\triangle_{n}$ and $\triangle_{n}-\boldsymbol{\pi}_{n}$
are integrally equivalent, but have different real-lattice point enumerators,
as we will later see. This is in notable contrast to the fact integer
translations are lattice invariant for lattice polytopes.

Taking the last coordinate in each lattice point of a polytope
$P\subset\mathbb{R}^{n}$ and setting it equal to $0$ is the _projection_ of
$P$ to the hyperplane defined by $x_{n}=0$. We denote this operation as
$\operatorname{proj}_{x_{n}=0}(P)$ and use an apostrophe $P^{\prime}$ as
shorthand.

######  Lemma 5.7.

Let $\mathbf{p}=(p_{1},\ldots,p_{n})\in\operatorname{aff}(\triangle_{n})$, and
$\mathbf{p}^{\prime}=(p_{1},\ldots,p_{n-1},0)$. Consider the $L^{\prime}n1$
simplex $\triangle_{n}$ and let
$\triangle_{n}^{\prime}:=\operatorname{proj}_{(x_{n}=0)}(\triangle_{n})$.
Then, for any $\lambda\in\mathbb{R}_{\geq 0}$:

$L_{\mathbb{R}}(\triangle_{n}-\mathbf{p};\lambda)=L_{\mathbb{R}}(\triangle_{n}^{\prime}-\mathbf{p}^{\prime};\lambda).$

We shift $\triangle_{n}$ by $\mathbf{p}$ (and thus $\triangle_{n}^{\prime}$ by
$\mathbf{p}^{\prime}$) as a slight extra step in order to show the much
stronger result that these two shifted polytopes are in bijection for _all_
real dilates $\lambda$ instead of just integer dilates. Our desired result for
the unshifted polytopes will come as a brief corollary.

###### Proof.

Let $X:=\lambda(\triangle_{n}-\mathbf{p})\cap\mathbb{Z}^{n}$ and
$X^{\prime}:=\lambda(\triangle_{n}^{\prime}-\mathbf{p}^{\prime})\cap\mathbb{Z}^{n}$.
Define

$\varphi:X\to X^{\prime}\text{ such that
}\varphi(x_{1},\ldots,x_{n-1},x_{n})=(x_{1},\ldots,x_{n-1},0).$

This map is well-defined because of our construction of $X^{\prime}$ from $X$;
it is immediate that if $(x_{1},\ldots,x_{n})\in X$ then
$(x_{1},\ldots,x_{n-1},0)\in X^{\prime}$ because $X^{\prime}$ is the
projection of each point in $X$ to the hyperplane $x_{n}=0$.

To prove injectivity, suppose $\mathbf{x},\mathbf{y}\in X$ such that
$\varphi(\mathbf{x})=\varphi(\mathbf{y})$. Thus,

$(x_{1},\ldots,x_{n-1},0)=(y_{1},\ldots,y_{n-1},0)\text{ and
}x_{i}=y_{i}\text{ for }i\in[n-1].$

Since the vertices of $\triangle_{n}$ are permutations of $[n]$, points
$\mathbf{z}=(z_{1},\dots,z_{n})$ in $\triangle_{n}$ exist on the hyperplane
defined by

$\sum_{k=1}^{n}z_{k}=\sum_{k=1}^{n}k=\frac{n(n+1)}{2}.$

Therefore, any point $\mathbf{z}$ in $\triangle_{n}-\mathbf{p}^{\prime}$ lives
in $\sum_{k=1}^{n}z_{k}=0$, as does any nonnegative real dilate
$\sum_{k=1}^{n}\lambda z_{k}=\lambda\sum_{k=1}^{n}z_{k}=0$. Hence,
$\sum_{k=1}^{n}x_{k}=\sum_{k=1}^{n}y_{k}$ and further $x_{n}=y_{n}$. We
conclude that $\mathbf{x}=\mathbf{y}$ and $\varphi$ is injective.

For surjectivity, consider $\mathbf{x}^{\prime}=(x_{1},\ldots,x_{n-1},0)\in
X^{\prime}$. We know there exists a unique convex combination of vertices of
$\lambda(\triangle_{n}^{\prime}-\mathbf{p}^{\prime})$ such that
$\sum_{k=1}^{n}c_{k}\mathbf{v}_{k}^{\prime}=\mathbf{x}^{\prime}$, where each
$c_{i}\in\mathbb{R}_{\geq 0}$ and $\sum_{k=1}^{n}c_{k}=1$. An integral point
in $\lambda(\triangle_{n}-\mathbf{p})$ is a convex combination of the vertices
of $\lambda(\triangle_{n}-\mathbf{p})$ and similarly for
$\lambda(\triangle^{\prime}_{n}-\mathbf{p}^{\prime})$. Note that the
corresponding vertices of both polytopes share the exact same first $n-1$
coordinates; thus, the convex combination will yield $x_{1},\ldots,x_{n-1}$ as
the first $n-1$ coordinates of the resulting point
$\mathbf{x}\in\lambda(\triangle_{n}-\mathbf{p})$. Further, this point lies in
the affine span of $\lambda(\triangle_{n}-\mathbf{p})$ and thus we know
$\sum_{i=1}^{n}x_{i}=0$. This implies $x_{n}=-\sum_{i=1}^{n-1}x_{i}$ and
therefore $x_{n}\in\mathbb{Z}$ because $x_{1},\ldots,x_{n-1}\in\mathbb{Z}$. We
conclude $\mathbf{x}\in X$ and $\varphi$ is surjective.

Hence, we have established a bijection between $X$ and $X^{\prime}$ for all
nonnegative real dilates $\lambda$ and can now deduce the polytopes
$\lambda(\triangle_{n}-\mathbf{p})$ and
$\lambda(\triangle_{n}^{\prime}-\mathbf{p}^{\prime})$ will always have the
same number of lattice points. ∎

######  Lemma 5.8.

Let $\triangle_{n}$ be the $L^{\prime}n1$ simplex,
$\boldsymbol{\pi}_{n}=L^{\prime}n1$, and
$\overline{\boldsymbol{\pi}}_{n}=L^{\prime}n=(\pi_{1},\ldots,\pi_{n-1})$. For
all $t\in\mathbb{N}$,

$L_{\mathbb{R}}\left(\triangle_{n}-\overline{\boldsymbol{\pi}_{n+1}};\frac{t}{n}\right)=L_{\mathbb{R}}\left(\triangle_{n}-\boldsymbol{\pi};\frac{t}{n}\right).$

###### Proof.

Recall integer translations preserve lattice count and note that
$\boldsymbol{\pi}_{n}=\overline{\boldsymbol{\pi}_{n+1}}-(0,\ldots,0,n)$. Thus,
translating $\triangle_{n}-\boldsymbol{\pi}_{n}$ by $(0,\ldots,0,n)$ yields
$\triangle_{n}-\overline{\boldsymbol{\pi}_{n+1}}$. We must conclude they have
equivalent lattice counts for all integer dilates $t$. Further, translation by
$(0,\ldots,0,\lambda n)$ will remain integral after dilation for all rational
dilates with form $\lambda=\frac{t}{n}$, making it lattice-preserving in these
special cases as well. ∎

The following result presents a recursive relationship that allows for the
enumeration of lattice points of _real_ dilates of $\triangle_{n+1}$ from
_rational_ dilates of $\triangle_{n}$.

######  Theorem 5.9.

For all $\lambda\in\mathbb{R}_{\geq 0}$,

$\displaystyle L_{\mathbb{R}}(\triangle_{n+1}-\boldsymbol{\pi}_{n+1};\lambda)$
$\displaystyle=\left|\lambda((\triangle_{n+1}-\boldsymbol{\pi}_{n+1})\cap\mathbb{Z}^{n+1})\right|$
$\displaystyle=\sum_{k=0}^{\lfloor n\lambda\rfloor}\
\left|\frac{k}{n}(\triangle_{n}-\boldsymbol{\pi}_{n})\cap\mathbb{Z}^{n}\right|=\sum_{k=0}^{\lfloor
n\lambda\rfloor}\
L_{\mathbb{R}}\left(\triangle_{n}-\boldsymbol{\pi}_{n};\frac{k}{n}\right).$

Before we present the proof of this result, we provide an example of the
theorem.

$\triangle_{3}-\boldsymbol{\pi}_{3}$“$\frac{1}{2}\left(\triangle_{2}-\boldsymbol{\pi}_{2}\right)$”“$\frac{2}{2}\left(\triangle_{2}-\boldsymbol{\pi}_{2}\right)$”“$\frac{3}{2}\left(\triangle_{2}-\boldsymbol{\pi}_{2}\right)$”“$\frac{4}{2}\left(\triangle_{2}-\boldsymbol{\pi}_{2}\right)$”“$\frac{5}{2}\left(\triangle_{2}-\boldsymbol{\pi}_{2}\right)$”(0,0,0)(0,-5,5)$\frac{5}{2}\left(\triangle_{3}-\boldsymbol{\pi}_{3}\right)$

Figure 4. A lattice translate of the $231$-simplex,
$\triangle_{3}-\boldsymbol{\pi}_{3}\subset\mathbb{R}^{3}$, is in blue, lattice
points are in magenta, $\frac{5}{2}(\triangle_{3}-\boldsymbol{\pi}_{3})$ is in
gray, and the (lifted) five rational dilates of
$\triangle_{2}-\boldsymbol{\pi}_{2}$ are in orange.

######  Example 5.10.

Take $n=2$, then $\boldsymbol{\pi}_{3}=231$ and $\boldsymbol{\pi}_{2}=21$ and
consider $\lambda=\frac{5}{2}$. Note that

$\left|\frac{5}{2}(\triangle_{3}-\boldsymbol{\pi}_{3})\cap\mathbb{Z}^{3}\right|=12,$

which we can directly count from the schematic in Figure 4. Observe that
$\triangle_{3}-\boldsymbol{\pi}_{3}$ lives in $\mathbb{R}^{3}$, but is a
2-dimensional simplex. By Theorem 5.9, we can alternatively count the number
of lattice points of $\frac{5}{2}(\triangle_{3}-\boldsymbol{\pi}_{3})$ by
counting the number of lattice points in the five dilates of
$\frac{k}{2}(\triangle_{2}-\boldsymbol{\pi}_{2})$ for $k=1,2,3,4,$ and $5$. In
the schematic, consider the polytopes on their affine span and we can treat
the polytopes as being projected onto the hyperplane $x_{3}=0$ (hence, the
quotations in some of the figure labels). Counting the number of lattice
points in the five dilates of
$\frac{k}{2}(\triangle_{2}-\boldsymbol{\pi}_{2})$ for $k=1,2,3,4,$ and $5$ as
shown in orange, we also obtain 12 lattice points.

###### Proof of Theorem 5.9.

Let $\triangle_{n}^{\prime}=\operatorname{proj}_{(x_{n}=0)}(\triangle_{n})$,
and $\boldsymbol{\pi}_{n}^{\prime}=({\pi}_{1},\ldots,{\pi}_{n-1},0)$ from
$\boldsymbol{\pi}_{n}=(\pi_{1},\ldots,\pi_{n})=L^{\prime}n1$. By Lemma 5.7,

(3)
$\left|\lambda(\triangle_{n+1}-\boldsymbol{\pi}_{n+1})\cap\mathbb{Z}^{n+1}\right|=\left|\lambda(\triangle_{n+1}^{\prime}-\boldsymbol{\pi}_{n+1}^{\prime})\cap\mathbb{Z}^{n+1}\right|.$

Obtain $\hat{\triangle}_{n}$ by lifting $\triangle_{n}$ to $\mathbb{R}^{n+1}$
by appending $0$ to the end of each lattice point of $\triangle_{n}$. By Claim
6.3, for all $\lambda\in\mathbb{R}_{\geq 0}$,

$\displaystyle\lambda(\triangle_{n+1}^{\prime}-\boldsymbol{\pi}_{n+1}^{\prime})\cap\mathbb{Z}^{n+1}=\biguplus_{k=0}^{\lfloor\lambda
n\rfloor}\left(\frac{k}{n}(\hat{\triangle}_{n}-\boldsymbol{\pi}_{n+1}^{\prime})\cap\mathbb{Z}^{n+1}\right),$

that is, the lattice points in the real dilates of
$\triangle^{\prime}_{n+1}-\boldsymbol{\pi}_{n+1}^{\prime}$ correspond the
disjoint union of rational dilates of
$\hat{\triangle}_{n}-\boldsymbol{\pi}_{n+1}^{\prime}$. Thus,

$\displaystyle\left|\lambda(\triangle_{n+1}^{\prime}-\boldsymbol{\pi}_{n+1}^{\prime})\cap\mathbb{Z}^{n+1}\right|$
$\displaystyle=\left|\biguplus_{k=0}^{\lfloor\lambda
n\rfloor}\left(\frac{k}{n}(\hat{\triangle}_{n}-\boldsymbol{\pi}_{n+1}^{\prime})\cap\mathbb{Z}^{n+1}\right)\right|$
$\displaystyle=\sum_{k=0}^{\lfloor\lambda
n\rfloor}\left|\frac{k}{n}(\hat{\triangle}_{n}-\boldsymbol{\pi}_{n+1}^{\prime})\cap\mathbb{Z}^{n+1}\right|.$

Further, we have that

$\displaystyle\left|\lambda(\triangle_{n+1}-\boldsymbol{\pi}_{n+1})\cap\mathbb{Z}^{n+1}\right|$
$\displaystyle=\sum_{k=0}^{\lfloor
n\lambda\rfloor}\left|\frac{k}{n}(\triangle_{n}-\overline{\boldsymbol{\pi}_{n+1}})\cap\mathbb{Z}^{n}\right|,$
and by Lemma 5.8, $\displaystyle=\sum_{k=0}^{\lfloor
n\lambda\rfloor}\left|\frac{k}{n}(\triangle_{n}-\boldsymbol{\pi}_{n})\cap\mathbb{Z}^{n}\right|.$

∎

We continue by presenting a recursive relationship that deals with the
enumeration of relative interior lattice points of real dilates of
$\triangle_{n+1}$ from _rational_ dilates of $\triangle_{n}$.

######  Proposition 5.11.

For all $\lambda\in\mathbb{R}_{\geq 0}$,

$\displaystyle
L_{\mathbb{R}}((\triangle_{n+1}-\boldsymbol{\pi}_{n+1})^{\circ};\lambda)$
$\displaystyle=\left|(\lambda(\triangle_{n+1}-\boldsymbol{\pi}_{n+1}))^{\circ}\cap\mathbb{Z}^{n+1}\right|$
$\displaystyle=\sum_{k=1}^{\lceil
n\lambda\rceil-1}\left|\left(\frac{k}{n}(\triangle_{n}-\boldsymbol{\pi}_{n})\right)^{\circ}\cap\mathbb{Z}^{n}\right|=\sum_{k=1}^{\lceil
n\lambda\rceil-1}L_{\mathbb{R}}\left((\triangle_{n}-\boldsymbol{\pi}_{n})^{\circ};\frac{k}{n}\right).$

###### Proof.

The proof follows similar arguments to that of the proof of Theorem 5.9;
modifying that proof the claim holds. ∎

We conclude this section by making use of the similarities in the recurrence
relations of Theorem 5.9 and Proposition 5.11 to prove the following result by
induction. The result shows that the real lattice-point count for the
$\lambda$-th dilate of $\triangle_{n}-\boldsymbol{\pi}_{n}$ coincides with the
relative interior lattice-point count for the $(\lambda+2)$-th dilate of
$\triangle_{n}-\boldsymbol{\pi}_{n}$ for every real number $\lambda$.
Furthermore, the result shows that $\triangle_{n}$ is Gorenstein of index $2$.
This alternative proof is included because it allows for the Gorenstein result
without knowledge of the entire lattice-point enumerator.

######  Theorem 5.12.

For $\lambda\in\mathbb{R}$,

$L_{\mathbb{R}}(\triangle_{n}-\boldsymbol{\pi}_{n};\lambda)=L_{\mathbb{R}}((\triangle_{n}-\boldsymbol{\pi}_{n})^{\circ};\lambda+2).$

Hence, any integer translate of $\triangle_{n}-\boldsymbol{\pi}_{n}$, in
particular $\triangle_{n}$, is Gorenstein of index $2$.

###### Proof.

First, we prove the claim: for all $t\in\mathbb{Z}$,

$\left|\left(\left(\frac{t}{n}+\frac{2n-1}{n}\right)(\triangle_{n}-\boldsymbol{\pi}_{n})\right)^{\circ}\cap\mathbb{Z}^{n}\right|=\left|\frac{t}{n}(\triangle_{n}-\boldsymbol{\pi}_{n})\cap\mathbb{Z}^{n}\right|.$

In words, the relative interior lattice-point count for the
$\left(\frac{t}{n}+\frac{2n-1}{n}\right)$-dilate of
$\triangle_{n}-\boldsymbol{\pi}_{n}$ matches the lattice-point count of the
$\frac{t}{n}$-dilate of $\triangle_{n}-\boldsymbol{\pi}_{n}$. We proceed by
induction on $n$, which relies on manipulations of summations.

Consider $n=2$. Observe that $\triangle_{2}-\boldsymbol{\pi}_{2}$ is simply a
line segment from $(0,0)$ to $(-1,1)$,
$(\frac{t}{2}+\frac{3}{2})(\triangle_{2}-\boldsymbol{\pi}_{2})$ is a line
segment from $(0,0)$ to $(-\frac{t}{2}-\frac{3}{2},\frac{t}{2}+\frac{3}{2})$,
and $\frac{t}{2}(\triangle_{2}-\boldsymbol{\pi}_{2})$ is a line segment from
$(0,0)$ to $(-\frac{t}{2},\frac{t}{2})$. Notice that all the interior lattice
points of the line segment from $(0,0)$ to
$(-\frac{t}{2}-\frac{3}{2},\frac{t}{2}+\frac{3}{2})$ lie on the line segment
from $(0,0)$ to $(-\frac{t}{2},\frac{t}{2})$ giving the following relation:

$L_{\mathbb{R}}\left((\triangle_{2}-\boldsymbol{\pi}_{2})^{\circ};\frac{t}{2}+\frac{3}{2}\right)=L_{\mathbb{R}}\left(\triangle_{2}-\boldsymbol{\pi}_{2};\frac{t}{2}\right)$

for all $t\in\mathbb{Z}$.

Assume the statement holds for some $m\in\mathbb{Z}_{\geq 2}$. We then have:

$\displaystyle\left|\frac{t}{m+1}(\triangle_{m+1}-\boldsymbol{\pi}_{m+1})\cap\mathbb{Z}^{m+1}\right|$
$\displaystyle=\sum_{k=0}^{\lfloor
m\cdot\frac{t}{m+1}\rfloor}\left|\frac{k}{m}(\triangle_{m}-\boldsymbol{\pi}_{m})\cap\mathbb{Z}^{m}\right|$
$\displaystyle=\underbrace{\sum_{k=0}^{\lfloor
m\cdot\frac{t}{m+1}\rfloor}\left|\left(\frac{k}{m}+\frac{2m-1}{m}(\triangle_{m}-\boldsymbol{\pi}_{m})\right)^{\circ}\cap\mathbb{Z}^{m}\right|}_{(\bigstar)}.$

By Claim 6.4, we can derive the following:

$\displaystyle(\bigstar)$ $\displaystyle=\sum_{k=2m-1}^{\lceil
m\cdot\frac{t}{m+1}+\frac{2m+1}{m+1}\rceil-1}\left|\left(\frac{k}{m}(\triangle_{m}-\boldsymbol{\pi}_{m})\right)^{\circ}\cap\mathbb{Z}^{m}\right|$
$\displaystyle=\sum_{k=0}^{\lceil
m\cdot\frac{t}{m+1}+\frac{2m+1}{m+1}\rceil-1}\left|\left(\frac{k}{m}(\triangle_{m}-\boldsymbol{\pi}_{m})\right)^{\circ}\cap\mathbb{Z}^{m}\right|-\sum_{k=0}^{2m-2}\left|\left(\frac{k}{m}(\triangle_{m}-\boldsymbol{\pi}_{m})\right)^{\circ}\cap\mathbb{Z}^{m}\right|$
$\displaystyle=\left|\left(\left(\frac{t}{m+1}+\frac{2m+1}{m+1}\right)(\triangle_{m+1}-\boldsymbol{\pi}_{m+1})\right)^{\circ}\cap\mathbb{Z}^{m+1}\right|-\sum_{k=0}^{2m-2}\left|\left(\frac{k}{m}(\triangle_{m}-\boldsymbol{\pi}_{m})\right)^{\circ}\cap\mathbb{Z}^{m}\right|.$

Here, note that by Proposition 5.11,

$\displaystyle\sum_{k=0}^{2m-2}\left|\left(\frac{k}{m}(\triangle_{m}-\boldsymbol{\pi}_{m})\right)^{\circ}\cap\mathbb{Z}^{m}\right|$
$\displaystyle=\left|\left(\frac{2m-1}{m}(\triangle_{m+1}-\boldsymbol{\pi}_{m+1})\right)^{\circ}\cap\mathbb{Z}^{m+1}\right|$
$\displaystyle=\left|\left(\triangle_{m+1}-\boldsymbol{\pi}_{m+1}\right)^{\circ}\cap\mathbb{Z}^{m+1}\right|,$
$\displaystyle=0,$

where the last equality holds by Theorem 4.3. Hence,

$(\bigstar)=\left|\left(\left(\frac{t}{m+1}+\frac{2m+1}{m+1}\right)(\triangle_{m+1}-\boldsymbol{\pi}_{m+1})\right)^{\circ}\cap\mathbb{Z}^{m+1}\right|.$

Therefore, by induction, we have proved that for all $t\in\mathbb{Z}_{\geq
0}$, $n\in\mathbb{Z}_{\geq 2}$,

$\left|\left(\left(\frac{t}{n}+\frac{2n-1}{n}\right)(\triangle_{n}-\boldsymbol{\pi}_{n})\right)^{\circ}\cap\mathbb{Z}^{n}\right|=\left|\frac{t}{n}(\triangle_{n}-\boldsymbol{\pi}_{n})\cap\mathbb{Z}^{n}\right|.$

Now, by Theorem 5.9, for all $n\in\mathbb{Z}_{\geq 3}$

$\displaystyle L_{\mathbb{R}}(\triangle_{n}-\boldsymbol{\pi}_{n};\lambda)$
$\displaystyle=\sum_{k=0}^{\lfloor(n-1)\lambda\rfloor}L_{\mathbb{R}}\left(\triangle_{n-1}-\boldsymbol{\pi}_{n-1};\frac{k}{n}\right).$
By our recently proven claim,
$\displaystyle=\sum_{k=0}^{\lfloor(n-1)\lambda\rfloor}\left|\left(\left(\frac{k}{n-1}+\frac{2n-3}{n-1}\right)(\triangle_{n-1}-\boldsymbol{\pi}_{n-1})\right)^{\circ}\cap\mathbb{Z}^{n-1}\right|$
$\displaystyle=\sum_{k=0}^{\lceil(n-1)(\lambda+2)\rceil-1}\left|\left(\left(\frac{k}{n-1}\right)(\triangle_{n-1}-\boldsymbol{\pi}_{n-1})\right)^{\circ}\cap\mathbb{Z}^{n-1}\right|$
$\displaystyle\text{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}-\sum_{k=0}^{2n-4}\left|\left(\left(\frac{k}{n-1}\right)(\triangle_{n-1}-\boldsymbol{\pi}_{n-1})\right)^{\circ}\cap\mathbb{Z}^{n-1}\right|,$
and by Theorem 4.3,
$\displaystyle=\left|((\lambda+2)(\triangle_{n}-\boldsymbol{\pi}_{n}))^{\circ}\cap\mathbb{Z}^{n}\right|-0$
$\displaystyle=L_{\mathbb{R}}((\triangle_{n}-\boldsymbol{\pi}_{n})^{\circ};\lambda+2).$

One can also show by solving inequalities that for $n=2$ the result holds, but
we omit the technicalities here. Therefore, for all $n\in\mathbb{Z}_{\geq 2}$,

$L_{\mathbb{R}}(\triangle_{n}-\boldsymbol{\pi}_{n};\lambda)=L_{\mathbb{R}}((\triangle_{n}-\boldsymbol{\pi}_{n})^{\circ};\lambda+2).$

In particular, for all $n\in\mathbb{Z}_{\geq 2}$ and $t\in\mathbb{Z}_{\geq
0}$, $\triangle_{n}$ is Gorenstein of index $2$, i.e.,

$L_{\mathbb{Z}}(\triangle_{n};t)=L_{\mathbb{Z}}(\triangle_{n}^{\circ};t+2).$

∎

######  Remark 5.13.

If we extend the previous definition and define a polytope to be _real
Gorenstein of index $k$_ for an integer $k$, if the relations from Equation 2
hold for all real dilates $t>k$, we believe this matches the rational/real
Gorenstein definitions given in [1] and we can say that
$\triangle_{n}-\boldsymbol{\pi}_{n}$ is “real Gorenstein” of index $2$.

## Dedication & Acknowledgments

In loving memory of a dear friend, Luca Elghanayan.

The authors thank Matthias Beck, Simon Rubenstein-Salzedo, and Julie Vega for
fruitful conversation. Andrés R. Vindas-Meléndez also thanks Cameron Lowe,
Justin McClung, Emily Muller-Foster, and Heather Willis for conversations at
the start of this project. Andrés R. Vindas-Meléndez is partially supported by
the National Science Foundation under Award DMS-$2102921$.

## References

  * [1] Matthias Beck, Sophia Elia, and Sophie Rehberg, _Rational Ehrhart theory_ , preprint (arXiv:2110.10204).
  * [2] Matthias Beck and Sinai Robins, _Computing the continuous discretely_ , second ed., Undergraduate Texts in Mathematics, Springer, New York, 2015, Integer-point enumeration in polyhedra.
  * [3] Miklós Bóna, _A survey of stack-sorting disciplines_ , Electron. J. Combin. 9 (2002/03), no. 2, Article 1, 16.
  * [4] Katie L. Bright and Carla D. Savage, _The geometry of lecture hall partitions and quadratic permutation statistics_ , 22nd International Conference on Formal Power Series and Algebraic Combinatorics (FPSAC 2010), Discrete Math. Theor. Comput. Sci. Proc., vol. AN, Assoc. Discrete Math. Theor. Comput. Sci., Nancy, 2010, pp. 569–580.
  * [5] Sylvie Corteel, Sunyoung Lee, and Carla D. Savage, _Enumeration of sequences constrained by the ratio of consecutive parts_ , Sém. Lothar. Combin. 54A (2005/07), Art. B54Aa, 12. MR 2180869
  * [6] Alan J. Hoffman and Joseph B. Kruskal, _Integral boundary points of convex polyhedra_ , pp. 49–76, Springer Berlin Heidelberg, Berlin, Heidelberg, 2010.
  * [7] Donald E. Knuth, _The art of computer programming._ , vol. 1, Addison-Wesley, 1968.
  * [8] Günter M. Ziegler, _Lectures on polytopes_ , Graduate Texts in Mathematics, vol. 152, Springer-Verlag, New York, 1995.

##  6\. Appendix

Due to the length and technicalities of the proofs of the following claims, we
have decided to include them in an appendix. Before beginning, it is necessary
to outline a few more terms:

Given a collection of points $C=\\{\mathbf{x}_{1},\dots,\mathbf{x}_{k}\\}$ in
$\mathbb{R}^{n}$, we define a _pointed cone_ with apex
$\mathbf{v}\in\mathbb{R}^{n}$ as a set of the form

(4)
$\operatorname{cone}_{\mathbf{v}}({C}):=\left\\{\mathbf{v}+\sum_{i=1}^{k}\lambda_{i}(\mathbf{x}_{i}-\mathbf{v}):\lambda_{i}\in\mathbb{R}_{\geq
0}\right\\}.$

This creates a cone that starts from the point $\mathbf{v}$ and consists of
the infinite region bounded by the $k$ rays formed from $\mathbf{v}$ through a
given point $\mathbf{x}_{i}$.

Similarly, we define _the pyramid over $P$ from $\mathbf{v}$_ as

$\operatorname{pyr}_{v}(P):=\left\\{\mathbf{v}+\sum_{i=1}^{k}\lambda_{i}(\mathbf{x}_{i}-\mathbf{v}):\lambda_{i}\in\mathbb{R}_{\geq
0}\text{ and }\sum_{i=1}^{k}\lambda_{i}=1\right\\}.$

Note that the choice of point $\mathbf{v}$ will affect the information that is
encoded in the cone/pyramid. For a polytope $P$ that is not full-dimensional,
it is a careful selection of $\mathbf{v}\not\in\operatorname{aff}(P)$ that
allows dilates of $P$ to perfectly encode themselves in ‘slices’ of the cone
or pyramid over $\mathbf{v}$.

######  Definition 6.1.

Let $S\subseteq\mathbb{R}^{n}$ be an $(n-1)$-dimensional affine subspace and
take $\mathbf{x}\in\mathbb{R}^{n}-S$. Then the _$S$ -halfspace with
$\mathbf{x}$_ denoted as $\mathcal{H}_{\mathbf{x}}(S)$ is defined as the
unique halfspace in $\mathbb{R}^{n}$ formed from the hyperplane $S$ that
includes the point $\mathbf{x}$.

######  Remark 6.2.

Note that
$\operatorname{pyr}_{\mathbf{v}}(P)=\operatorname{cone}_{\mathbf{v}}(P)\cap\mathcal{H}_{\mathbf{v}}(P)$
in our case where $P$ is $(n-1)$-dimensional in $n$-space and
$\mathbf{v}\not\in\operatorname{aff}(P)$.

The first claim shows that the lattice points in the real dilates of
$\triangle^{\prime}_{n+1}-\boldsymbol{\pi}_{n+1}^{\prime}$ correspond to the
disjoint union of rational dilates of
$\hat{\triangle}_{n}-\boldsymbol{\pi}_{n+1}^{\prime}$.

$\operatorname{aff}(\frac{5}{2}(\hat{\triangle}_{2}-\boldsymbol{\pi}_{3}^{\prime}))$$\operatorname{aff}(\hat{\triangle}_{2}-\boldsymbol{\pi}_{3}^{\prime})$$\triangle^{\prime}_{3}$$\hat{\triangle}_{2}$(1,2,0)(2,3,0)(2,2,0)(2,1,0)(0,0,0)$\triangle^{\prime}_{3}-\boldsymbol{\pi}_{3}$(0,-5,0)$\frac{5}{2}\left(\triangle^{\prime}_{3}-\boldsymbol{\pi}_{3}\right)$$\operatorname{cone}_{\mathbf{0_{3}}}(\hat{\triangle}_{2}-\boldsymbol{\pi}_{3}^{\prime})$

Figure 5. This figure corresponds to the set-up of the proof of Claim 6.3,
where $\lambda=\frac{5}{2}$, $n=2$, and distance $d=\sqrt{2}$.

######  Claim 6.3.

The following equality holds for all $\lambda\in\mathbb{R}_{\geq 0}$:

$\displaystyle\lambda(\triangle_{n+1}^{\prime}-\boldsymbol{\pi}_{n+1}^{\prime})\cap\mathbb{Z}^{n+1}=\biguplus_{k=0}^{\lfloor\lambda
n\rfloor}\left(\frac{k}{n}(\hat{\triangle}_{n}-\boldsymbol{\pi}_{n+1}^{\prime})\cap\mathbb{Z}^{n+1}\right).$

###### Proof.

Let $\mathbf{0_{n}}$ be the origin in $\mathbb{R}^{n}$, $d$ be the distance
between $\mathbf{0_{n+1}}$ and
$\operatorname{aff}(\hat{\triangle}_{n}-\boldsymbol{\pi}_{n+1}^{\prime})$, and
$\mathbf{v}$ be any fixed vector on the plane $x_{n+1}=0$ that is orthogonal
to $\operatorname{aff}(\hat{\triangle}_{n}-\boldsymbol{\pi}_{n+1}^{\prime})$
such that $\mathbf{v}\cdot\boldsymbol{\pi}_{n+1}^{\prime}\geq 0$ (see Figure
5, for a concrete visual example).

We will show that the lattice points in $\lambda$-th dilate of
$\triangle_{n+1}^{\prime}-\boldsymbol{\pi}_{n+1}^{\prime}$ corresponds to the
lattice points in the union of slices of
$\lambda(\triangle_{n+1}^{\prime}-\boldsymbol{\pi}_{n+1}^{\prime})$ obtained
by intersecting the polytope with a infinite set of hyperplanes
$\operatorname{aff}(\lambda(\hat{\triangle}_{n}-\boldsymbol{\pi}_{n+1}^{\prime}))+k\mathbf{v}$
for all real numbers $k\in[0,\frac{\lambda d}{|\mathbf{v}|}]$.

Take
$\mathcal{H}_{\mathbf{0_{n+1}}}(\lambda(\hat{\triangle}_{n}-\boldsymbol{\pi}_{n+1}^{\prime}))$
as defined using Definition 6.1. Then

$\displaystyle\lambda(\triangle_{n+1}^{\prime}-\boldsymbol{\pi}_{n+1}^{\prime})\cap\mathbb{Z}^{n+1}$
$\displaystyle=\lambda(\triangle_{n+1}^{\prime}-\boldsymbol{\pi}_{n+1}^{\prime})\cap\mathcal{H}_{\mathbf{0_{n+1}}}(\lambda(\hat{\triangle}_{n}-\boldsymbol{\pi}_{n+1}^{\prime}))\cap\mathbb{Z}^{n+1}$
$\displaystyle=\lambda(\triangle_{n+1}^{\prime}-\boldsymbol{\pi}_{n+1}^{\prime})\cap\left(\bigcup_{0\leq
k\leq\frac{\lambda
d}{|\mathbf{v}|}}(\operatorname{aff}(\lambda(\hat{\triangle}_{n}-\boldsymbol{\pi}_{n+1}^{\prime}))+k\mathbf{v})\right)\cap\mathbb{Z}^{n+1}$
(5) $\displaystyle=\bigcup_{0\leq k\leq\frac{\lambda
d}{|\mathbf{v}|}}\left(\lambda(\triangle_{n+1}^{\prime}-\boldsymbol{\pi}_{n+1}^{\prime})\cap(\operatorname{aff}(\lambda(\hat{\triangle}_{n}-\boldsymbol{\pi}_{n+1}^{\prime}))+k\mathbf{v})\cap\mathbb{Z}^{n+1}\right).$

Since,

$\operatorname{aff}(\hat{\triangle}_{n}-\boldsymbol{\pi}_{n+1}^{\prime})=\\{\mathbf{x}\in\mathbb{R}^{n+1}|x_{1}+\cdots+x_{n}=-n,x_{n+1}=0\\},$

note that $\mathbf{v}=c(1,\ldots,1,0)\in\mathbb{R}^{n+1}$ for some
$c\in\mathbb{R}$.

Without loss of generality, take $c=1$ and consider
$\mathbf{v}=(1,\ldots,1,0)\in\mathbb{R}^{n+1}$. Note that $d=|\mathbf{v}|$ and
thus, we can proceed with (5) as follows:

$\lambda(\triangle_{n+1}^{\prime}-\boldsymbol{\pi}_{n+1}^{\prime})\cap\mathbb{Z}^{n+1}=\operatorname{cone}_{\mathbf{0_{n+1}}}(\lambda(\hat{\triangle}_{n}-\boldsymbol{\pi}_{n+1}^{\prime}))\cap\bigcup_{0\leq
k\leq\lambda}\left((\operatorname{aff}(\lambda(\hat{\triangle}_{n}-\boldsymbol{\pi}_{n+1}^{\prime}))+k\mathbf{v})\cap\mathbb{Z}^{n+1}\right),$

with $\mathbf{v}=(1,\ldots,1,0)\in\mathbb{R}^{n+1}$.

Now, we show that lattice points can only exist on the slices of
$\lambda(\triangle_{n+1}^{\prime}-\boldsymbol{\pi}_{n+1}^{\prime})$ obtained
from the hyperplanes
$\operatorname{aff}(\lambda(\hat{\triangle}_{n}-\boldsymbol{\pi}_{n+1}^{\prime}))+k\mathbf{v}$
for $k\in[0,\lambda]$ such that $kn$ is a integer. Equivalently, we show that
if $kn\notin[0,\lambda]\cap\mathbb{Z}$, then
$\lambda(\triangle_{n+1}^{\prime}-\boldsymbol{\pi}_{n+1}^{\prime})\cap(\operatorname{aff}(\lambda(\hat{\triangle}_{n}-\boldsymbol{\pi}_{n+1}^{\prime}))+k\mathbf{v})=\emptyset$.
So, first we consider $\lambda(\triangle_{n+1}-\boldsymbol{\pi})$, which is
$\lambda(\triangle_{n+1}^{\prime}-\boldsymbol{\pi}_{n+1}^{\prime})$ before the
projection to $x_{n+1}=0$. Notice that the slice of
$\lambda(\triangle_{n+1}-\boldsymbol{\pi})$ obtained from intersecting it with
the hyperplane $H_{h}:=\\{\mathbf{x}\in\mathbb{R}^{n+1}|x_{n+1}=h\\}$ only
contains lattice points if $h$ is an integer. Thus,

$h\not\in\mathbb{Z}\cap[-\lambda
n,0]\Longrightarrow(\lambda(\triangle_{n+1}-\boldsymbol{\pi})\cap\mathbb{Z}^{n+1})\cap
H_{h}=\emptyset.$

Next, we show that the slice of $\lambda(\triangle_{n+1}-\boldsymbol{\pi})$
obtained from intersecting it with $H_{h}$ is equivalent to the slice of
$\lambda(\triangle_{n+1}-\boldsymbol{\pi})$ obtained from intersecting it with
$H^{\text{vert}}+(1-\frac{h}{n})\mathbf{v}$, where

$H^{\text{vert}}:=\\{\mathbf{x}\in\mathbb{R}^{n+1}|x_{1}+\cdots+x_{n}=-n\\}.$

Note that

$\operatorname{aff}(\lambda(\triangle_{n+1}-\boldsymbol{\pi}_{n+1}))=\\{\mathbf{x}\in\mathbb{R}^{n+1}|x_{1}+\cdots+x_{n+1}=0\\}.$

Now, we can derive the following:

$\displaystyle
H_{h}\cap\operatorname{aff}(\lambda(\triangle_{n+1}-\boldsymbol{\pi}_{n+1}))$
$\displaystyle=\\{\mathbf{x}\in\mathbb{R}^{n+1}|x_{n+1}=h\\}\cap\\{\mathbf{x}\in\mathbb{R}^{n+1}|x_{1}+\cdots+x_{n+1}=0\\}$
$\displaystyle=\\{\mathbf{x}\in\mathbb{R}^{n+1}|x_{1}+\cdots+x_{n}=-h,x_{n+1}=h\\}$
$\displaystyle=\\{\mathbf{x}\in\mathbb{R}^{n+1}|x_{1}+\cdots+x_{n}=-h\\}$
$\displaystyle\text{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\cap\\{\mathbf{x}\in\mathbb{R}^{n+1}|x_{1}+\cdots+x_{n+1}=0\\}$
$\displaystyle=\left\\{\left(x_{1}+\left(1-\frac{h}{n}\right),\ldots,x_{n}+\left(1-\frac{h}{n}\right),x_{n+1}\right)\Bigg{|}x_{1}+\cdots+x_{n}=-n\right\\}$
$\displaystyle\text{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\cap\\{\mathbf{x}\in\mathbb{R}^{n+1}|x_{1}+\cdots+x_{n+1}=0\\}$
$\displaystyle=\left(\\{\mathbf{x}\in\mathbb{R}^{n+1}|x_{1}+\cdots+x_{n}=-n\\}+\left(1-\frac{h}{n}\right)\mathbf{v}\right)\cap\operatorname{aff}(\lambda(\triangle_{n+1}-\boldsymbol{\pi}))$
$\displaystyle=\left(H^{\text{vert}}+\left(1-\frac{h}{n}\right)\mathbf{v}\right)\cap\operatorname{aff}(\lambda(\triangle_{n+1}-\boldsymbol{\pi})).$

Hence,

$(\lambda(\triangle_{n+1}-\boldsymbol{\pi}_{n+1})\cap\mathbb{Z}^{n+1})\cap
H_{k}=(\lambda(\triangle_{n+1}-\boldsymbol{\pi}_{n+1})\cap\mathbb{Z}^{n+1})\cap\left(H^{\text{vert}}+\left(1-\frac{h}{n}\right)\mathbf{v}\right).$

Further, note that the slice of
$\lambda(\triangle_{n+1}-\boldsymbol{\pi}_{n+1})$ obtained from intersecting
it with $H^{\text{vert}}+(1-\frac{h}{n})\mathbf{v}$ has no lattice point on it
if $h$ is not an integer, i.e.,

$h\not\in\mathbb{Z}\cap[-\lambda
n,0]\Longrightarrow(\lambda(\triangle_{n+1}-\boldsymbol{\pi}_{n+1})\cap\mathbb{Z}^{n+1})\cap\left(H^{\text{vert}}+\left(1-\frac{h}{n}\right)\mathbf{v}\right)=\emptyset.$

Lastly, we show that the projection onto $x_{n+1}=0$ of the slice of
$\lambda(\triangle_{n+1}-\boldsymbol{\pi}_{n+1})$ obtained from intersecting
it with $H^{\text{vert}}+(1-\frac{h}{n})\mathbf{v}$ is equivalent to the slice
of $\lambda(\hat{\triangle}_{n}-\boldsymbol{\pi}_{n+1}^{\prime})$ obtained
from intersecting it with
$\operatorname{aff}(\lambda(\hat{\triangle}_{n}-\boldsymbol{\pi}_{n+1}^{\prime}))+(1-\frac{h}{n})\mathbf{v}$.
So, consider

$\operatorname{proj}_{x_{n+1}=0}\left((\lambda(\triangle_{n+1}-\boldsymbol{\pi}_{n+1})\cap\mathbb{Z}^{n+1})\cap\left(H^{\text{vert}}+\left(1-\frac{h}{n}\right)\mathbf{v}\right)\right).$

Then

$\displaystyle\operatorname{proj}_{x_{n+1}=0}$
$\displaystyle\left((\lambda(\triangle_{n+1}-\boldsymbol{\pi}_{n+1})\cap\mathbb{Z}^{n+1})\cap\left(H^{\text{vert}}+\left(1-\frac{h}{n}\right)\mathbf{v}\right)\right)$
$\displaystyle=\operatorname{proj}_{x_{n+1}=0}\left(\lambda(\triangle_{n+1}-\boldsymbol{\pi}_{n+1})\cap\mathbb{Z}^{n+1}\right)\cap\operatorname{proj}_{x_{n+1}=0}\left(H^{\text{vert}}+\left(1-\frac{h}{n}\right)\mathbf{v}\right)$
$\displaystyle=(\lambda(\triangle_{n+1}^{\prime}-\boldsymbol{\pi}_{n+1}^{\prime})\cap\mathbb{Z}^{n+1})\cap\left(\\{\mathbf{x}\in\mathbb{R}^{n+1}|x_{1}+\cdots+x_{n}=-n,x_{n+1}=0\\}+\left(1-\frac{h}{n}\right)\mathbf{v}\right)$
$\displaystyle=(\lambda(\triangle_{n+1}^{\prime}-\boldsymbol{\pi}_{n+1}^{\prime})\cap\mathbb{Z}^{n+1})\cap\left(\operatorname{aff}(\lambda(\hat{\triangle}_{n}-\boldsymbol{\pi}_{n+1}^{\prime}))+\left(1-\frac{h}{n}\right)\mathbf{v}\right).$

Thus,

$h\not\in\mathbb{Z}\cap[-\lambda
n,0]\Longrightarrow(\lambda(\triangle_{n+1}^{\prime}-\boldsymbol{\pi}_{n+1}^{\prime})\cap\mathbb{Z}^{n+1})\cap\left(\operatorname{aff}(\lambda(\hat{\triangle}_{n}-\boldsymbol{\pi}_{n+1}^{\prime}))+\left(1-\frac{h}{n}\right)\mathbf{v}\right)=\emptyset.$

Therefore,

$kn\not\in\mathbb{Z}\cap[0,\lambda
n]\Longrightarrow(\lambda(\triangle_{n+1}^{\prime}-\boldsymbol{\pi}_{n+1}^{\prime})\cap\mathbb{Z}^{n+1})\cap\left(\operatorname{aff}(\lambda(\hat{\triangle}_{n}-\boldsymbol{\pi}_{n+1}^{\prime}))+k\mathbf{v}\right)=\emptyset.$

To conclude, by “flossing” out all the slices of
$\lambda(\triangle_{n+1}^{\prime}-\boldsymbol{\pi}_{n+1}^{\prime})$ that don’t
contain lattice points, we can derive the following:

$\displaystyle\bigcup_{0\leq k\leq\lambda}$
$\displaystyle\left(\lambda(\triangle_{n+1}^{\prime}-\boldsymbol{\pi}_{n+1}^{\prime})\cap(\operatorname{aff}(\lambda(\hat{\triangle}_{n}-\boldsymbol{\pi}_{n+1}^{\prime}))+k\mathbf{v})\cap\mathbb{Z}^{n+1}\right)$
$\displaystyle=\biguplus_{\begin{subarray}{c}0\leq k\leq\lambda:\\\
kn\in\mathbb{Z}\end{subarray}}\left(\lambda(\triangle_{n+1}^{\prime}-\boldsymbol{\pi}_{n+1}^{\prime})\cap(\operatorname{aff}(\lambda(\hat{\triangle}_{n}-\boldsymbol{\pi}_{n+1}^{\prime}))+k\mathbf{v})\cap\mathbb{Z}^{n+1}\right)$
$\displaystyle=\biguplus_{k=0}^{\lfloor\lambda
n\rfloor}\left(\lambda(\triangle_{n+1}^{\prime}-\boldsymbol{\pi}_{n+1}^{\prime})\cap(\operatorname{aff}(\lambda(\hat{\triangle}_{n}-\boldsymbol{\pi}_{n+1}^{\prime}))+\frac{k}{n}\mathbf{v})\cap\mathbb{Z}^{n+1}\right)$
$\displaystyle=\biguplus_{k=0}^{\lfloor\lambda
n\rfloor}\left(\operatorname{pyr}_{\mathbf{0_{n+1}}}(\lambda(\hat{\triangle}_{n}-\boldsymbol{\pi}_{n+1}^{\prime}))\cap(\operatorname{aff}(\lambda(\hat{\triangle}_{n}-\boldsymbol{\pi}_{n+1}^{\prime}))+\frac{k}{n}\mathbf{v})\cap\mathbb{Z}^{n+1}\right)$
$\displaystyle=\biguplus_{k=0}^{\lfloor\lambda
n\rfloor}\left(\operatorname{cone}_{\mathbf{0_{n+1}}}(\hat{\triangle}_{n}-\boldsymbol{\pi}_{n+1}^{\prime})\cap(\operatorname{aff}(\lambda(\hat{\triangle}_{n}-\boldsymbol{\pi}_{n+1}^{\prime}))+\frac{k}{n}\mathbf{v})\cap\mathbb{Z}^{n+1}\right)$
$\displaystyle=\biguplus_{k=0}^{\lfloor\lambda
n\rfloor}\left(\frac{k}{n}(\hat{\triangle}_{n}-\boldsymbol{\pi}_{n+1}^{\prime})\cap\mathbb{Z}^{n+1}\right).$

Hence, the claim holds. ∎

######  Claim 6.4.

For $\boldsymbol{\pi}_{n}:=L^{\prime}n1$, we have

$\sum_{k=0}^{\lfloor
n\cdot\frac{t}{n+1}\rfloor}\left|\left(\frac{k}{n}+\frac{2n-1}{n}(\triangle_{n}-\boldsymbol{\pi}_{n})\right)^{\circ}\cap\mathbb{Z}^{n}\right|=\sum_{k=2n-1}^{\lceil
n\cdot\frac{t}{n+1}+\frac{2n+1}{n+1}\rceil-1}\left|\left(\frac{k}{n}(\triangle_{n}-\boldsymbol{\pi}_{n})\right)^{\circ}\cap\mathbb{Z}^{n}\right|.$

###### Proof.

By dividing the cases for when $n+1|t$ and $n+1\nmid t$, we can derive the
following:

$\displaystyle\sum_{k=0}^{\lfloor n\cdot\frac{t}{n+1}\rfloor}$
$\displaystyle\left|\left(\frac{k}{n}+\frac{2n-1}{n}(\triangle_{n}-\boldsymbol{\pi}_{n})\right)^{\circ}\cap\mathbb{Z}^{n}\right|$
$\displaystyle=\begin{cases}\displaystyle\sum_{k=2n-1}^{\lceil
n\cdot\frac{t}{n+1}+2n\rceil-1}\left|\left(\frac{k}{n}(\triangle_{n}-\boldsymbol{\pi}_{n})\right)^{\circ}\cap\mathbb{Z}^{n}\right|&\text{if
}n+1|t\\\ \displaystyle\sum_{k=2n-1}^{\lceil
n\cdot\frac{t}{n+1}+2n-1\rceil-1}\left|\left(\frac{k}{n}(\triangle_{n}-\boldsymbol{\pi}_{n})\right)^{\circ}\cap\mathbb{Z}^{n}\right|&\text{if
}n+1\nmid t.\end{cases}$

Note that if $n+1\mid t$, then $n\cdot\frac{t}{n+1}+2n\in\mathbb{Z}$, thus

$\left\lceil n\cdot\frac{t}{n+1}+2n\right\rceil=\left\lceil
n\cdot\frac{t}{n+1}+2n-\frac{n}{n+1}\right\rceil.$

Alternatively, if $n+1\nmid t$, then $0<n\cdot\frac{t}{n+1}+2n-1-\lceil
n\cdot\frac{t}{n+1}+2n-1\rceil\leq\frac{n}{n+1}$, thus

$\left\lceil n\cdot\frac{t}{n+1}+2n-1\right\rceil=\left\lceil
n\cdot\frac{t}{n+1}+2n-1+\frac{1}{n+1}\right\rceil.$

Therefore,

$\sum_{k=0}^{\lfloor
n\cdot\frac{t}{n+1}\rfloor}\left|\left(\frac{k}{n}+\frac{2n-1}{n}(\triangle_{n}-\boldsymbol{\pi}_{n})\right)^{\circ}\cap\mathbb{Z}^{n}\right|=\sum_{k=2n-1}^{\lceil
n\cdot\frac{t}{n+1}+\frac{2n+1}{n+1}\rceil-1}\left|\left(\frac{k}{n}(\triangle_{n}-\boldsymbol{\pi}_{n})\right)^{\circ}\cap\mathbb{Z}^{n}\right|.$

∎