Let $[n]$ denote the finite $n-$chain $\{1, 2, \dots, n\}$. A function $\rho$ with  domain and range both being subsets of $[n]$ is termed a \emph{partial transformation} of $[n]$, and it is said to be \emph{full} (or \emph{total}) if its domain is the whole of $[n]$. The set consisting of all partial transformations on $[n]$ is usually denoted by $\mathcal{P}_{n}$ and is known as \emph{the semigroup of all partial transformations} on $[n]$, more commonly referred to as \emph{the partial symmetric monoid}. A transformation $\rho \in \mathcal{P}_{n}$ is called an \emph{isotone} (or order preserving) function (resp., an \emph{antitone} (order reversing) function) if (for all $x,y \in \mathop{\rm Dom}\nolimits\,\rho$) $x \leq y$ implies $x\rho \leq y\rho$ (resp., $x\rho \geq y\rho$); it is called \emph{monotone} if it is either isotone or antitone or both; and it is called \emph{order decreasing} if (for all $a \in \mathop{\rm Dom}\nolimits \, \rho$) $a\rho \leq a$. Let $\mathcal{POD}_{n}$ be \emph{the monoid of all monotone partial transformations} on $[n]$. This monoid first appeared in Fernandes \cite{vm} and subsequently in Umar \cite{auc} and East \emph{et. al.}, \cite{east}, where its congruence/rank properties and combinatorial properties, and its maximal subsemigroups were studied, respectively. The algebraic properties of various subsemigroups of monotone maps in various classes have been investigated over the years, for example see \cite{bugay, bugay1, il}.

In line with \cite{auc},  $\mathcal{LS}_{n}$ (\emph{the semigroup of all isotone and order-decreasing partial transformations} on $[n]$) is  refer to as the \emph{large} \emph{Schr\"{o}der} monoid, which is defined as:

\begin{equation}\label{qn111}
\mathcal{LS}_{n} = \mathcal{OP}_n \cap \mathcal{DP}_n,
\end{equation}

\noindent where  $\mathcal{OP}_n$ and $\mathcal{DP}_n$ denote  the \emph{semigroup of all isotone partial transformations} on $[n]$  and \emph{the semigroup of all order-decreasing partial transformations} on $[n]$, respectively. The order of this monoid, obtained in \cite{al3}, corresponds to the \emph{double} (or \emph{large}) \emph{Schr\"{o}der} number:

\begin{equation}\label{nums}
s_{0}=1, \quad s_{n}= \frac{1}{n+1} \sum\limits_{r=0}^{n}\binom{n+1}{n-r}\binom{n+r}{r}, \quad n\geq 1.
\end{equation}

 Now, let
\begin{equation}\label{qn1}
 	\mathcal{DORP}_{n} = \{\rho \in \mathcal{POD}_{n} : \,   \rho \text{ is \it decreasing} \}
\end{equation}
\noindent be the set of all decreasing maps in $\mathcal{POD}_{n}$. This set can equivalently be expressed as:
\begin{align*}
\mathcal{DORP}_{n} &= \{\rho \in \mathcal{P}_{n} : \,   \rho \text{ is {\it monotone} and \it decreasing} \} \\
&= \{\rho \in \mathcal{P}_{n} : \,   \rho \text{ is {\it isotone} and \it decreasing} \} \cup \{\rho \in \mathcal{P}_{n} : \,   \rho \text{ is {\it antitone} and \it decreasing} \}.
\end{align*}
\noindent Therefore, if we let $DRP_{n} = \{\rho \in \mathcal{P}_{n} : \,   \rho  \text{ is {\it antitone} and \it decreasing}\},$ then
\begin{equation}\label{qn2}
	\mathcal{DORP}_{n} = \mathcal{LS}_{n} \cup DRP_{n}.
\end{equation}
Moreover, the set $\mathcal{DORP}_{n}$ can be expressed as:

$$
\mathcal{DORP}_{n} = \mathcal{POD}_{n} \cap \mathcal{DP}_{n}.
$$
\noindent It is a routine matter to show that $\mathcal{DORP}_{n}$ is a monoid. It shall be refer to  as the \emph{monoid of all monotone and decreasing partial transformations} on $[n]$. There seems to be no prior discussion of the monoid \(\mathcal{DORP}_{n}\) in the literature. This paper investigates certain algebraic features of the monoid \(\mathcal{DORP}_{n}\) and its rank properties.

For  map $\rho\in \mathcal{DORP}_{n}$, we shall adopt the notations $\mathop{\rm Dom}\nolimits \rho$, $1_{[n]}$, $b(\rho) = |\mathop{\rm Dom}\nolimits \, \rho|$, $\mathop{\rm Im}\nolimits \rho$,  $h(\rho) = |\mathop{\rm Im}\nolimits \, \rho|$, and $F(\rho) = \{x \in \mathop{\rm Dom}\nolimits \, \rho : x\rho = x\}$ to denote the domain set of  $\rho$, the identity mapping on $[n]$, the width of $\rho$, the image set of $\rho$, the height of $\rho$,  and the set of fixed points of $\rho$, respectively. Additionally, we will let $f(\rho) = |F(\rho)|$ denote the number of fixed points of $\rho$.  A subset $X$ of $[n]$ is said to be \emph{convex} if, for any $ x, y\in X $ such that $x \leq y$, and for any $c \in [n]$, if $x < c < y$, then $c \in X$. We shall be using the notations $E(S)$ and $\textbf{0}$ to denote the set of idempotents  and the zero element of a semigroup $S$, respectively.

Furthermore, we shall adopt the right-hand composition of two transformations, say $\rho$ and $\sigma$ in $\mathcal{P}_{n}$, defined as
$$
x(\rho \circ \sigma) = ((x)\rho) \sigma
$$
\noindent for all $x \in \mathop{\rm Dom}\nolimits\, \rho$. To be concise, we will denote \(\rho \sigma\) as \(\rho \circ \sigma\).

  Moreover, for $0\le p\le n-1$,  let \begin{equation} \label{kn} I(n, \,p)=\{\rho\in  \mathcal{DORP}_{n}: \, |\mathop{\rm Im}\nolimits \, \rho|\le p\}\end{equation}
  \noindent be the two-sided ideal of $\mathcal{DORP}_{n}$, which consist all decreasing and monotone transformations in  $\mathcal{DORP}_{n}$, each with a height not more than $p$.

  Additionally, given that $p\geq 1$,  denote \begin{equation}\label{knn} {RQ}_{p}(n)= I(n, \,p)/ I(n, \, p-1)  \end{equation}
  \noindent to be the Rees quotient semigroup of $I(n, \,p)$, where the elements in ${RQ}_{p}(n)$ can be regarded as the elements of $\mathcal{DORP}_{n}$ that possess a height of precisely $p$. If the product of two elements from ${RQ}_{p}(n)$ has a height that does not reach $p$, then the result is $\textbf{0}$; otherwise, it is an element of ${RQ}_{p}(n) \setminus \{\textbf{0}\}$.

\indent According to \cite{HRS}, every $\rho \in \mathcal{LS}_{n}$ can be represented in two-line notation as

\begin{equation}
\label{1}
\rho = \begin{pmatrix} A_1 & \dots & A_p \\ a_1 & \dots & a_p \end{pmatrix} \quad (1 \leq p \leq n),
\end{equation}

\noindent where $a_{i} \leq \min A_{i}$ for all $i\in\{1,\ldots, p\}$ since $\rho$ is a decreasing transformation, and each of the sets $A_i$ $(1 \leq i \leq p)$ denotes an equivalence class determined by the following relation:

$$\textnormal{ker } \rho = \{(x, y) \in \mathop{\rm Dom}\nolimits \, \rho \times \mathop{\rm Dom}\nolimits \, \rho : x\rho = y\rho\}.$$

\noindent The collection of these equivalence classes shall be denoted as $\textnormal{\bf Ker } \rho = \{A_1,  \dots, A_p\}$. In addition, the kernel $\textnormal{\bf Ker } \rho$ is linearly ordered (meaning that for any indices $i < j$, the relation $A_{i} < A_{j}$ is equivalent to $x < y$ for all $x \in A_{i}$ and $y \in A_{j}$). Furthermore, we can assume without loss of generality that $1 \leq a_{1} < \dots < a_{p} \leq n$. For a comprehensive introduction to the basic ideas of semigroup theory, we recommend the texts by Howie \cite{howi} and Higgins \cite{ph}.

This paragraph offers a concise explanation of the framework of the paper. Section 1 provide definitions of some basic terms and compute the order of the monoid \( \mathcal{DORP}_{n} \). In Section 2, we characterize of all the Green's relations along with their starred counterparts in the monoid \( \mathcal{DORP}_{n} \) and its two-sided ideal \( I(n, \,p) \). Moreover, we demonstrate that the monoid \( \mathcal{DORP}_{n} \) and its two-sided ideals are abundant semigroups. Finally, in Section 3, we calculate the rank of the Rees quotient semigroup \( RQ_{n}(p) \), the two-sided ideal \( I(n, \,p) \), and the monoid \( \mathcal{DORP}_{n} \). The computation of the rank of this monoid and its two-sided ideals differs from the conventional understanding associated with certain known monoids, where most arguments or generating sets are derive from elements of height \(n-1\). However, in the monoid \( \mathcal{DORP}_{n} \), elements of height \(n-1\) are insufficient to generate the structure; additional elements below this height are necessary, as will be discussed in Section 3.

Now let us briefly examine the elements in $DRP_{n}$. First, notice that every element of height 1 in $\mathcal{DORP}_{n}$ is both isotone and antitone; that is, it belongs to both $\mathcal{LS}_{n}$ and $DRP_{n}$. However, antitones of height greater than or equal to 2 cannot be isotones. Therefore, we have the following remark.

\begin{remark}\label{rmk1}
Every element in $DRP_{n}$ of height $1 < p < n$ of the form
\begin{equation}\label{2}
\rho = \begin{pmatrix} A_1 & \dots & A_p \\ a_p & \dots & a_1 \end{pmatrix},
\end{equation}
 has the following properties:
\begin{itemize}
  \item[(i)] $1 \leq a_{1} < \cdots < a_{p} \leq \min A_{1}$;
  \item[(ii)] Note that when multiplying two isotone maps or two antitone maps, the result remains isotone. In contrast, the product of an isotone map with an antitone map yields an antitone map. Similarly, when an antitone map is multiplied by an isotone map, the outcome is antitone as well.

  \item[(iii)] It should also be emphasized that all elements in $\mathcal{DORP}_{n}$ of height greater than $\lceil \frac{n}{2} \rceil$ is necessarily isotone.
\end{itemize}
\end{remark}

Consequently, every element in $DRP_{n}$ has height $p$ that ranges from $1 \leq p \leq \lceil \frac{n}{2} \rceil$; thus, we have the following lemma.

\begin{lemma}\label{rev}
An element $\rho \in \mathcal{LS}_{n}$, as expressed in \eqref{1}, is reversible; that is, $\begin{pmatrix} A_1 & \dots & A_p \\ a_p & \dots & a_1 \end{pmatrix} \in \mathcal{DORP}_{n}$ if and only if $a_{p} \leq \min A_{1}$ and $1 \leq p \leq \lceil \frac{n}{2} \rceil$.
\end{lemma}
\begin{proof} Let $w=\min (\mathop{\rm Dom}\nolimits \, \rho) =\min A_{1}\geq a_{p}$, by the order decreasing property. However, notice that $n-p+1\geq w $ and $a_{p}\geq p$,  and so,  $n-p+1\geq w\geq a_{p}\geq p$, which implies $n+1\geq 2p$. Hence $p\leq \lfloor \frac{n+1}{2}\rfloor=\lceil\frac{n}{2}\rceil$. The converse is trivial.

\end{proof}
We note the following well known combinatorial identity, which is useful in our subsequent discussions.

 \begin{lemma}[\cite{44}, (3b), p.8]\label{id}
For all n \( m, n, k\in \mathbb{N} \), it follows that
\[
\sum_{j=k}^{m} \binom{j}{k} \binom{m+n-j}{n} = \binom{m+n+1}{n+k+1}.
\]
\end{lemma}

 It is now clear that to obtain the size of $\mathcal{DORP}_{n}$, it is sufficient to compute the size of the set $DRP_{n}$. For a transformation $\rho$ on a finite chain, let $b(\rho) = r$ and $h(\rho) = p$. Then, $p \leq r \leq n$. Define the combinatorial functions:

\[F(n,r,p)=|\{\rho\in {DRP}_n : \, b(\rho)=r \text{ and } h( \rho)=p\}|\]
\noindent and
\[F(n,p)=|\{\rho\in {DRP}_n : \,  h(\rho)=p\}|.\]

Then we present the following theorem.

For map $\rho\in \mathcal{DORP}_{n}$, we shall adopt the notations $\mathop{\rm Dom}\nolimits \rho$, $1_{[n]}$, $b(\rho) = |\mathop{\rm Dom}\nolimits \, \rho|$, $\mathop{\rm Im}\nolimits \rho$, $h(\rho) = |\mathop{\rm Im}\nolimits \, \rho|$, and $F(\rho) = \{x \in \mathop{\rm Dom}\nolimits \, \rho : x\rho = x\}$ to denote the domain set of $\rho$, the identity mapping on $[n]$, the width of $\rho$, the image set of $\rho$, the height of $\rho$, and the set of fixed points of $\rho$, respectively. Additionally, we will let $f(\rho) = |F(\rho)|$ denote the number of fixed points of $\rho$. A subset $X$ of $[n]$ is said to be \emph{convex} if, for any $ x, y\in X $ such that $x \leq y$, and for any $c \in [n]$, if $x < c < y$, then $c \in X$. We shall be using the notations $E(S)$ and $\textbf{0}$ to denote the set of idempotents and the zero element of a semigroup $S$, respectively.

Moreover, for $0\le p\le n-1$, let \begin{equation} \label{kn} I(n, \,p)=\{\rho\in \mathcal{DORP}_{n}: \, |\mathop{\rm Im}\nolimits \, \rho|\le p\}\end{equation}
 \noindent be the two-sided ideal of $\mathcal{DORP}_{n}$, which consist all decreasing and monotone transformations in $\mathcal{DORP}_{n}$, each with a height not more than $p$.

Additionally, given that $p\geq 1$, denote \begin{equation}\label{knn} {RQ}_{p}(n)= I(n, \,p)/ I(n, \, p-1) \end{equation}
 \noindent to be the Rees quotient semigroup of $I(n, \,p)$, where the elements in ${RQ}_{p}(n)$ can be regarded as the elements of $\mathcal{DORP}_{n}$ that possess a height of precisely $p$. If the product of two elements from ${RQ}_{p}(n)$ has a height that does not reach $p$, then the result is $\textbf{0}$; otherwise, it is an element of ${RQ}_{p}(n) \setminus \{\textbf{0}\}$.

Now let us briefly examine the elements in $DRP_{n}$. First, notice that every element of height 1 in $\mathcal{DORP}_{n}$ is both isotone and antitone; that is, it belongs to both $\mathcal{LS}_{n}$ and $DRP_{n}$. However, antitones of height greater than or equal to 2 cannot be isotones. Therefore, we have the following remark.

\[F(n,r,p)=|\{\rho\in {DRP}_n : \, b(\rho)=r \text{ and } h( \rho)=p\}|\]
\noindent and
\[F(n,p)=|\{\rho\in {DRP}_n : \, h(\rho)=p\}|.\]

Then we present the following theorem.

For map $\rho\in \mathcal{DORP}_{n}$, we shall adopt the notations $\dom \rho$, $1_{[n]}$, $b(\rho) = |\dom \, \rho|$, $\im \rho$, $h(\rho) = |\im \, \rho|$, and $F(\rho) = \{x \in \dom \, \rho : x\rho = x\}$ to denote the domain set of $\rho$, the identity mapping on $[n]$, the width of $\rho$, the image set of $\rho$, the height of $\rho$, and the set of fixed points of $\rho$, respectively. Additionally, we will let $f(\rho) = |F(\rho)|$ denote the number of fixed points of $\rho$. A subset $X$ of $[n]$ is said to be \emph{convex} if, for any $ x, y\in X $ such that $x \leq y$, and for any $c \in [n]$, if $x < c < y$, then $c \in X$. We shall be using the notations $E(S)$ and $\textbf{0}$ to denote the set of idempotents and the zero element of a semigroup $S$, respectively.

Additionally, given that $p\geq 1$, denote \begin{equation}\label{knn} {RQ}_{p}(n)= I(n, \,p)/ I(n, \, p-1) \end{equation}
 \noindent to be the Rees quotient semigroup of $I(n, \,p)$, where the elements in ${RQ}_{p}(n)$ can be regarded as the elements of $\mathcal{DORP}_{n}$ that possess a height of precisely $p$. If the product of two elements from ${RQ}_{p}(n)$ has a height that does not reach $p$, then the result is $\textbf{0}$; otherwise, it is an element of ${RQ}_{p}(n) \setminus \{\textbf{0}\}$.

\[F(n,r,p)=|\{\rho\in {DRP}_n : \, b(\rho)=r \text{ and } h( \rho)=p\}|\]
\noindent and
\[F(n,p)=|\{\rho\in {DRP}_n : \, h(\rho)=p\}|.\]

Let $\rho\in{DRP}_n$ be as expressed in \eqref{2}, where $b(\rho)=r$ and $h(\rho)=p$. Moreover, let $w=\min(\dom \, \rho)$. Note that $p\leq w\leq n - p + 1$, from the proof of Lemma \ref{rev}, and for all $x_i \in \dom \, \rho$ and $a_j \in \im \, \rho$, we have $ a_j\leq x_i$. To count the number of $\rho\in{DRP}_n$, we first choose the domain elements. Now, since $w\in \dom \, \rho$, we can choose the remaining $r-1$ elements from $[n]\setminus\{1,\ldots,w\}$, i.e., in ${n - w \choose r - 1}$ ways. Next, we partition the $r$ elements in the domain into $p$ convex (modulo $\dom \, \rho$) blocks in ${r - 1 \choose p - 1}$ ways. Then, we can choose the $p$ images from the set $\{1,\ldots,w\}$ in ${w \choose p}$ ways. Finally, taking the sum of the product: ${n-w\choose r-1} {r-1\choose p-1} {w\choose p}$, from $w=p$ to $w=n-p+1$ produces
\begin{align*} F(n,r,p)&=\sum_{w=p}^{n-p+1} {n-w\choose r-1} {r-1\choose p-1} {w\choose p}\\&=
{r-1\choose p-1}\sum_{w=p}^{n-p+1} {n-w\choose r-1} {w\choose p}
\\&=
{r-1 \choose p-1} {n+1 \choose r+p}, \quad (\text{by Lemma \ref{id}})
\end{align*}
\noindent as postulated.

The number of elements in ${DRP}_n$ of a fixed height $1\le p\le \lceil\frac{n}{2}\rceil$ is
\begin{equation}\label{coo} F(n,p)= \sum_{r=p}^{n-p+1}{r-1 \choose p-1} {n+1 \choose r+p}.\end{equation}
\end{corollary}
\begin{proof}
 We get the number by varying the width within the possible range which is $p\leq r\leq n-p+1$.
\end{proof}

We will now state the following lemma.
\begin{lemma} Let $F(n,p)$ be as in \eqref{coo} and let $a_{n} = \sum\limits_{p=2}^{\lceil\frac{n}{2} \rceil} F(n, p)$. Then
\[
a_{n} = \sum\limits_{p=2}^{\lceil\frac{n}{2} \rceil} \sum_{r=p}^{n-p+1} {r-1 \choose p-1} {n+1 \choose r+p}.
\]
\end{lemma}

\noindent\textbf{(ii.)} The result follows from Remark \ref{remm}.
\end{proof}
Next, we have the following theorem.
\begin{theorem} Let $\mathcal{DORP}_{n}$ be as defined in \eqref{qn1}. Then the number of convex vital elements of height $2 \leq p\leq \lceil \frac{n}{2} \rceil$ in $\mathcal{DORP}_{n}$ is \[(n+2)\left(\left\lceil \frac{n}{2} \right\rceil-1\right)-\left(\left\lceil \frac{n}{2} \right\rceil\left(\left\lceil \frac{n}{2} \right\rceil+1\right)-2\right).\]
\end{theorem}
\begin{proof} By summing the convex vital elements in Lemma \ref{nid}(i) over $2 \leq p \leq \lceil \frac{n}{2} \rceil,$ together with some algebraic manipulations, we obtain \[\sum_{p=2}^{\left\lceil \frac{n}{2} \right\rceil} (n - 2p + 2) = (n + 2)\left(\left\lceil \frac{n}{2} \right\rceil - 1\right) - \left(\left\lceil \frac{n}{2} \right\rceil\left(\left\lceil \frac{n}{2} \right\rceil + 1\right) - 2\right).\]
\end{proof}

\begin{lemma}\label{mminknp} For $0\leq p\le n-1$,
$W(p)$ is the minimum generating set of $I(n, \,p)$.
\end{lemma}
\begin{proof} The result follows from the fact that $G(p)$ is the minimum generating set of $\langle J^{*}_{p} \rangle$ by Lemma \ref{minnn}, and also from the fact that each minimum generating set of the ideal $I(n, \,p)$ must contain all the extreme convex vital elements below height $p$, as stated in Remark \ref{extr}.
\end{proof}
\begin{lemma}\label{ooknp} Let $W(p)$ be as defined in \eqref{w}. Then \[|W(p)|=\left\{
 \begin{array}{lll}
 1, & \hbox{if $p=0$;} \\
 2^{n}-1, & \hbox{if $p=1$;} \\
 (n-2)+\sum\limits_{r=p}^{n}{\binom{n}{r}}{\binom{r-1}{p-1}}, & \hbox{if $p \in \{2, \ldots, n-1\}$.}
 \end{array}
 \right.\]
\end{lemma}
\begin{proof} For the cases $p=0$ and $p=1$, the result is trivial. Next, if $2 \leq p \leq \lceil \frac{n}{2} \rceil$, then $|W(p)|=|G(p)|+|\{\delta_{p, \, i}, \, \, \delta_{p, \, n-i+1}: \, 2\leq i\le p-1\}|$. Thus, by Lemma \ref{nid2}, we see that $$|W(p)|=(n-2p+2)+\sum\limits_{r=p}^{n}{\binom{n}{r}}{\binom{r-1}{p-1}}+2(p-2)=(n-2)+\sum\limits_{r=p}^{n}{\binom{n}{r}}{\binom{r-1}{p-1}}.$$

\begin{equation}\label{2}
\rho = \begin{pmatrix} A_1 & \dots & A_p \\ a_p & \dots & a_1 \end{pmatrix},
\end{equation}

\begin{lemma}[\cite{44}, (3b), p.8]\label{id}
For all n \( m, n, k\in \mathbb{N} \), it follows that
\[
\sum_{j=k}^{m} \binom{j}{k} \binom{m+n-j}{n} = \binom{m+n+1}{n+k+1}.
\]
\end{lemma}

\begin{lemma}\label{rev}
An element $\rho \in \mathcal{LS}_{n}$, as expressed in \eqref{1}, is reversible; that is, $\begin{pmatrix} A_1 & \dots & A_p \\ a_p & \dots & a_1 \end{pmatrix} \in \mathcal{DORP}_{n}$ if and only if $a_{p} \leq \min A_{1}$ and $1 \leq p \leq \lceil \frac{n}{2} \rceil$.
\end{lemma}

\begin{proof}

 Let $\rho\in{DRP}_n$ be as expressed in \eqref{2}, where $b(\rho)=r$ and $h(\rho)=p$. Moreover, let $w=\min(\mathop{\rm Dom}\nolimits \, \rho)$. Note that $p\leq w\leq n - p + 1$, from the proof of Lemma \ref{rev},   and for all $x_i \in \mathop{\rm Dom}\nolimits \, \rho$ and $a_j \in \mathop{\rm Im}\nolimits \, \rho$, we have $ a_j\leq x_i$. To count the number of $\rho\in{DRP}_n$, we first choose the domain elements. Now, since $w\in \mathop{\rm Dom}\nolimits \, \rho$, we can choose the remaining $r-1$ elements from $[n]\setminus\{1,\ldots,w\}$, i.e., in ${n - w \choose r - 1}$ ways. Next, we partition the $r$ elements in the domain into $p$  convex (modulo $\mathop{\rm Dom}\nolimits \, \rho$) blocks in ${r - 1 \choose p - 1}$ ways. Then, we can choose the $p$ images from the set $\{1,\ldots,w\}$  in ${w \choose p}$ ways. Finally, taking the sum of the product: ${n-w\choose r-1} {r-1\choose p-1} {w\choose p}$, from $w=p$ to $w=n-p+1$ produces
\begin{align*}  F(n,r,p)&=\sum_{w=p}^{n-p+1} {n-w\choose r-1} {r-1\choose p-1} {w\choose p}\\&=
{r-1\choose p-1}\sum_{w=p}^{n-p+1} {n-w\choose r-1}  {w\choose p}
\\&=
{r-1 \choose p-1} {n+1 \choose r+p}, \quad (\text{by Lemma \ref{id}})
\end{align*}
\noindent as postulated.

\end{proof}
\bigskip

\begin{corollary}\label{co}

The number of elements in ${DRP}_n$ of a fixed height $1\le p\le \lceil\frac{n}{2}\rceil$ is
\begin{equation}\label{coo} F(n,p)= \sum_{r=p}^{n-p+1}{r-1 \choose p-1} {n+1 \choose r+p}.\end{equation}
\end{corollary}
\begin{proof}
 We get the number by varying the width within the possible range which is $p\leq r\leq n-p+1$.
\end{proof}

We will now state the following lemma.
\begin{lemma} Let $F(n,p)$ be as in  \eqref{coo} and let $a_{n} = \sum\limits_{p=2}^{\lceil\frac{n}{2} \rceil} F(n, p)$. Then
\[
a_{n} =  \sum\limits_{p=2}^{\lceil\frac{n}{2} \rceil}  \sum_{r=p}^{n-p+1}  {r-1 \choose p-1} {n+1 \choose r+p}.
\]
\end{lemma}

At this point, we present the following result.

\begin{theorem} Let $s_{n}$ be as defined in \eqref{nums}. Then $|\mathcal{DORP}_{n}|=s_{n}+a_{n}$.
\end{theorem}
\begin{proof} Notice that, as in \eqref{qn2}, $\mathcal{DORP}_{n} = \mathcal{LS}_{n} \cup DRP_{n}$. However, elements of exactly height $1$ are both isotone and antitone maps, so  $$|\mathcal{LS}_{n} \cap DRP_{n}|=F(n,1).$$ \noindent Thus,  $$|\mathcal{DORP}_{n}| =|\mathcal{LS}_{n} \cup DRP_{n}|= |\mathcal{LS}_{n}|+| DRP_{n}|-| \mathcal{LS}_{n} \cap DRP_{n}|,$$ \noindent and therefore
\[|\mathcal{DORP}_{n}| =s_{n}+\sum\limits_{p=1}^{\lceil\frac{n}{2} \rceil} {F(n, p)}-F(n,1)=s_{n}+ F(n,1)+\sum\limits_{p=2}^{\lceil\frac{n}{2} \rceil} {F(n, p)}-F(n,1)=s_{n}+a_{n},\] as required.
\end{proof}

\begin{remark}
 It is  significant to highlight that the triangle of numbers $F(n,\, p)$ and the sequences $a_{n}$ and $s_{n}+a_{n}$ are not yet recorded in The Online Encyclopedia of Integer Sequences \cite{slone}.
\end{remark}

Let $\rho\in{DRP}_n$ be as expressed in \eqref{2}, with $b(\rho)=r$, $h(\rho)=p$, and let $w=\min(\mathop{\rm Dom}\nolimits \, \rho)$. From the proof of Lemma \ref{rev}, $p\leq w\leq n-p+1$, and for all $x_i\in\mathop{\rm Dom}\nolimits\,\rho$ and $a_j\in\mathop{\rm Im}\nolimits\,\rho$ we have $a_j\le x_i$. To count such $\rho$, first choose the domain: since $w\in\mathop{\rm Dom}\nolimits\,\rho$, choose the remaining $r-1$ domain elements from $[n]\setminus\{1,\ldots,w\}$ in ${n-w\choose r-1}$ ways. Next, partition the $r$ domain elements into $p$ convex (modulo $\mathop{\rm Dom}\nolimits\,\rho$) blocks in ${r-1\choose p-1}$ ways. Then choose the $p$ images from $\{1,\ldots,w\}$ in ${w\choose p}$ ways. Summing over $w$ gives
\[
F(n,r,p)=\sum_{w=p}^{n-p+1}{n-w\choose r-1}{r-1\choose p-1}{w\choose p}={r-1\choose p-1}\sum_{w=p}^{n-p+1}{n-w\choose r-1}{w\choose p}={r-1\choose p-1}{n+1\choose r+p}\quad(\text{by Lemma \ref{id}}),
\]
as claimed.

Let $\rho\in{DRP}_n$ be as expressed in
\begin{equation}\label{2}
\rho = \begin{pmatrix} A_1 & \dots & A_p \\ a_p & \dots & a_1 \end{pmatrix},
\end{equation}
with $b(\rho)=r$, $h(\rho)=p$, and let $w=\min(\mathop{\rm Dom}\nolimits \, \rho)$. From the proof of the following lemma,
\begin{lemma}\label{rev}
An element $\rho \in \mathcal{LS}_{n}$, as expressed in \eqref{1}, is reversible; that is, $\begin{pmatrix} A_1 & \dots & A_p \\ a_p & \dots & a_1 \end{pmatrix} \in \mathcal{DORP}_{n}$ if and only if $a_{p} \leq \min A_{1}$ and $1 \leq p \leq \lceil \frac{n}{2} \rceil$.
\end{lemma}
we have $p\leq w\leq n-p+1$, and for all $x_i\in\mathop{\rm Dom}\nolimits\,\rho$ and $a_j\in\mathop{\rm Im}\nolimits\,\rho$ we have $a_j\le x_i$. To count such $\rho$, first choose the domain: since $w\in\mathop{\rm Dom}\nolimits\,\rho$, choose the remaining $r-1$ domain elements from $[n]\setminus\{1,\ldots,w\}$ in ${n-w\choose r-1}$ ways. Next, partition the $r$ domain elements into $p$ convex (modulo $\mathop{\rm Dom}\nolimits\,\rho$) blocks in ${r-1\choose p-1}$ ways. Then choose the $p$ images from $\{1,\ldots,w\}$ in ${w\choose p}$ ways. Summing over $w$ gives
\[
F(n,r,p)=\sum_{w=p}^{n-p+1}{n-w\choose r-1}{r-1\choose p-1}{w\choose p}={r-1\choose p-1}\sum_{w=p}^{n-p+1}{n-w\choose r-1}{w\choose p}={r-1\choose p-1}{n+1\choose r+p}\quad(\text{by the following lemma}),
\]
\begin{lemma}[\cite{44}, (3b), p.8]\label{id}
For all n \( m, n, k\in \mathbb{N} \), it follows that
\[
\sum_{j=k}^{m} \binom{j}{k} \binom{m+n-j}{n} = \binom{m+n+1}{n+k+1}.
\]
\end{lemma}
as claimed.

The number of elements in ${DRP}_n$ of a fixed width $r$ and height $p$ is $F(n,r,p)= {r-1 \choose p-1} {n+1 \choose r+p} $,

Classification or Bijection
Inequality or Bound

Question:
Let $[n]=\{1,2,\dots,n\}$, and let ${DRP}_n$ denote the set of order-reversing partial transformations of $[n]$. For $\rho\in {DRP}_n$, define its width by $b(\rho)=|\operatorname{dom}\rho|$ and its height by $h(\rho)=|\operatorname{im}\rho|$. Also define
\[
F(n,r,p)=\left|\{\rho\in {DRP}_n: b(\rho)=r \text{ and } h(\rho)=p\}\right|.
\]
Which explicit formula gives the number of elements of ${DRP}_n$ with fixed width $r$ and height $p$?

\[
F(n,r,p)=\binom{r-1}{p-1}\binom{n+1}{r+p}.
\]

Choices:
B.
\[
F(n,r,p)=\binom{r-1}{p-1}\binom{n}{r+p-1}.
\]
C.
\[
F(n,r,p)=\binom{r-1}{p-1}\sum_{w=p}^{\,n-p+1}\binom{n-w}{r-1}\binom{w}{p}.
\]
D.
\[
F(n,r,p)=\binom{r}{p}\binom{n+1}{r+p}.
\]
E.
\[
F(n,r,p)=\binom{r-1}{p-1}\binom{n+1}{r+p-1}.
\]

Vandermonde-index shift after summing over w

closed-form binomial evaluation by Lemma \ref{id}

number of convex block partitions of the domain

final upper binomial parameter r+p from convolution identity

The stem defines the objects and asks for the counting formula, but it does not reveal or strongly hint at the specific closed form. There is no explicit answer leakage.

The item is essentially a direct request for the known formula for F(n,r,p). It tests recall/recognition of the theorem rather than applying it in a new context.

Some reasoning is needed to distinguish the exact closed form from nearby off-by-one variants and from the weaker true summation formula, but the task is still mainly formula recognition rather than substantial derivation.

The distractors are mathematically plausible and well targeted: several encode common indexing/convolution mistakes, and one option is a weaker true expression that increases subtlety.

A solid multiple-choice item with no answer leakage and strong distractors, but it is largely theorem-recall and therefore weak on tautology avoidance and deep generative reasoning.


\label{sec.Intro}
Aperiodic tilings of Euclidean space, a topic once regarded as an esoteric pursuit, are now widely recognized as a subject of central interest. Beginning with work of Hao Wang and Robert Berger in the 1960's studying a relationship between aperiodic tilings and the Halting problem for Turing machines, results surrounding aperiodic tilings have had substantial impacts in the last century within mathematics, computer science, the natural sciences, and even the liberal arts. Excellent summaries of many of these results, up until about 2010, can be found in \cite[Chapter 1]{Gard1997}, \cite[Chapters 1, 2]{Sene1995}, \cite[Forward by Roger Penrose]{BaakGrim2013}, and \cite[Chapter 1]{BaakGrim2013}. There has also been significant and highly publicized progress in this subject in the last decade.

One landmark, which has also become an inspiration in art and architecture, was Roger Penrose's discovery of a pair of polygonal tiles which could tile the plane, but could only do so aperiodically. Penrose published a first version of his tilings in 1974 \cite{Penr1974} (see also \cite{Penr1979}). Soon after, they were introduced to the general public in an article by Martin Gardner in the Scientific American \cite{Gard1977}.

\begin{figure}[h]
 \caption{Tiles used to construct rhombic Penrose tilings.\\}\label{fig.PenTiles}
 \centering
 \vspace*{-30bp}
 \includegraphics[width=0.75\textwidth]{PenTiles1mod}
\end{figure}

There are several related collections of tilings which are referred to as Penrose tilings. In this article we will consider \textit{rhombic Penrose tilings} (RPTs), which are tilings of the plane by copies of two rhombuses with angles $72^\circ$ and $36^\circ$, pictured in Figure \ref{fig.PenTiles}. The rhombuses are allowed to be rotated, but they must fit together edge to edge, and the matching rules indicated by the arrows must also be enforced. A small patch of an RPT is pictured in Figure \ref{fig.PenPatch} (see also \cite[Figures 1.2, 6.44]{BaakGrim2013}). When the rhombuses are chosen to have side length 1, we call the resulting tilings \textit{unit length RPTs}.

The main problem that we wish to investigate in this paper is that of estimating the number of vertices of an RPT which lie in $B_R$, the closed Euclidean ball of radius $R$, centered at some point in the plane which we fix to be the origin. RPTs can be defined using primitive inflation rules (see \cite[Section 6.2]{BaakGrim2013}), which guarantee existence of limiting frequencies of tile types in large regions. Although it is not obvious, it is not too difficult to deduce (see for example \cite[Remark 5]{SchmTrev2018}) that if $V$ is the vertex set of a unit length RPT then, for $R\ge 1$,
\begin{equation}\label{eqn.PenVertTrivEst}
 \#(V\cap B_R)=\pi C_P R^2 + O(R),
\end{equation}
where
\[C_P=\frac{\phi}{5}\sqrt{10+2\sqrt{5}}\approx 1.231,\]
with $\phi=(1+\sqrt{5})/2$. The error term on the right hand side of \eqref{eqn.PenVertTrivEst} comes from counting tiles which intersect the boundary of $B_R$.
Since there are in fact $\gg R$ such tiles, it may appear that this error term is essentially best possible (cf. results of Solomon from \cite{Solo2011} on counting tiles in \textit{supertiles} of primitive substitutions). However, we will show that for this special class of tilings, one can actually do better.

\begin{equation}\label{eqn.PenVertTrivEst}
	\#(V\cap B_R)=\pi C_P R^2 + O(R),
\end{equation}

\begin{figure}[h]
	\caption{A patch of an RPT constructed using the cut and project method detailed in Section \ref{sec.ModelSets}. The 4 colors of vertices are translates of the cut and project sets produced using the different windows.\\}\label{fig.PenPatch}
	\centering
	\includegraphics[width=0.75\textwidth]{PenPatch2}
\end{figure}

\begin{figure}[h]
	\caption{Tiles used to construct rhombic Penrose tilings.\\}\label{fig.PenTiles}
	\centering
	\vspace*{-30bp}
	\includegraphics[width=0.75\textwidth]{PenTiles1mod}
\end{figure}

There are several related collections of tilings which are referred to as Penrose tilings. In this article we will consider \textit{rhombic Penrose tilings} (RPTs), which are tilings of the plane by copies of two rhombuses with angles $72^\circ$ and $36^\circ$, pictured in Figure \ref{fig.PenTiles}. The rhombuses are allowed to be rotated, but they must fit together edge to edge, and the matching rules indicated by the arrows must also be enforced. A small patch of an RPT is pictured in Figure \ref{fig.PenPatch} (see also \cite[Figures 1.2, 6.44]{BaakGrim2013}). When the rhombuses are chosen to have side length 1, we call the resulting tilings \textit{unit length RPTs}.

\begin{figure}[h]
 \caption{A patch of an RPT constructed using the cut and project method detailed in Section \ref{sec.ModelSets}. The 4 colors of vertices are translates of the cut and project sets produced using the different windows.\\}\label{fig.PenPatch}
 \centering
 \includegraphics[width=0.75\textwidth]{PenPatch2}
\end{figure}

The work of de Bruijn, et al., mentioned at the beginning of this section (see references listed above), implies the following result.
\begin{theorem}\label{thm.PenVerts}
Every non-singular point set $\Lambda_\omega$, constructed as above, is the set of vertices of a unit length rhombic Penrose tiling. Furthermore, the vertex set of any unit length rhombic Penrose tiling can be obtained, after rotation, as the limit in an appropriate topology (e.g. the Chabauty-Fell topology) of a sequence of non-singular point sets $\Lambda_\omega$.
\end{theorem}
An example of a RPT obtained using the cut and project method described above is shown in Figure \ref{fig.PenPatch}. Theorem \ref{thm.PenVerts} will allow us to derive Theorem \ref{thm.Pen} from the following result.
\begin{theorem}\label{thm.PenCPVersion}
For the cut and project sets $\Lambda_{m,\omega}$ defined above, there is a constant $C>0$ with the property that, for any $\omega\in\C$, any integer $1\le m\le 4$, and any $R\ge 2$,
\[\left|\#\left(\Lambda_{m,\omega}\cap B_R\right)-\pi R^2\mr{dens}(\cL)\left|W_{m,\omega}\right|\right|\le CR^{2/3}(\log R)^{2/3},\]
where $|W_{m,\omega}|$ denotes the area of $W_{m,\omega}$, and $\mr{dens}(\cL)$ is the reciprocal of the volume of a fundamental domain for $\R^4/\cL$.
\end{theorem}
An important point is that the error term in this theorem is uniform in $\omega$, which allows us to take limits of non-singular point sets, to account for all possible vertex sets of RPTs, in order to establish Theorem \ref{thm.Pen}. There are a few other small details that have to be ironed out, but they are not difficult to justify, so we postpone their discussion until the end of Section \ref{sec.PenProof}. The majority of the remainder of the paper will be focused on proving Theorem \ref{thm.PenCPVersion}.

If $W\subseteq\R^2$ then we write $\chi_W$ for the indicator function of $W$. The similarity in notation to $\chi_R$, as defined above, will not cause confusion in what follows. Observe that
\[\widehat{\chi}_W(y)=\int_We(-x\cdot y)\mr{d}x=\frac{-1}{(2\pi|y|)^2}\int_W\nabla_x^2\left( e(-x\cdot y)\right)\mr{d}x,\]
where $\nabla_x^2=\nabla_x\cdot\nabla_x$ is the Laplacian. By the Divergence Theorem, under reasonable hypotheses on $W$, this allows us to express the Fourier transform as a line integral over the boundary of $W$. Applying this argument in the case when $W$ is a polygon leads to a particularly nice formula for the Fourier transform, which we reproduce here in the form given in \cite{BranColzTrav1997}.
\begin{proposition}\label{prop.FourTranPoly}
 Let $P\subseteq\R^2$ be a polygon with counterclockwise oriented vertices $\{a_j\}_{j=1}^m$. Denote by $\sigma_j$ the unit vector parallel to the side $[a_j,a_{j+1}]$ and by $v_j$ the outward unit normal to this side. Then, with $a_{m+1}:=a_1$, we have
 \[\widehat{\chi}_P(y)=\frac{-1}{(2\pi|y|)^2}\sum_{j=1}^m\left(\frac{e(-y\cdot a_{j+1})-e(-y\cdot a_{j})}{y\cdot \sigma_j}\right)y\cdot v_j.\]
\end{proposition}
This proposition is proved as \cite[Lemma 2.2]{BranColzTrav1997}. The quantity in parenthesis in the sum is interpreted to equal
\[-\left(2\pi i |a_{j+1}-a_j|\right)e\left(\frac{-y\cdot(a_j+a_{j+1})}{2}\right),\]
when $y\cdot\sigma_j=0$. Proposition \ref{prop.FourTranPoly} implies that
\begin{equation}\label{eqn.PolyFourEst}
 \widehat{\chi}_P(y)\ll_P \frac{1}{|y|}\sum_{j=1}^m\min\left\{1,\frac{1}{|y\cdot\sigma_j|}\right\}.
\end{equation}
One thing to note is that these Fourier coefficients decay in most directions like $|y|^{-2}$. However, in directions perpendicular to the sides of $P$, they decay only like $|y|^{-1}$. This turns out to be a source of some difficulty, which is given special attention in the proofs below.

\noindent\emph{Bound for $E_2$:} To estimate $E_2$, first write
\[\mc{S}_{m,n}=\left\{\theta\in L^\circledast : \frac{2^{m-1}}{R}<|\theta|\le\frac{2^m}{R},~\frac{nR}{M2^m}\le|\theta^\star|<\frac{(n+1)R}{M2^m}\right\},\]
with $M$ chosen as in \eqref{eqn.MChoice}. Then we have that
\begin{align}
 E_2&\ll R^{1/2}\sum_{m=1}^{\lceil \log_2R\rceil}\sum_{n=1}^\infty\sum_{\theta\in\mc{S}_{m,n}}\frac{1}{|\theta|^{3/2}}|\widehat{\chi}_W(\theta^\star)|\left(1+\delta^2|\theta^\star|^2\right)^{-N}\nonumber\\
 &\ll R^2 \sum_{m=1}^{\lceil\log_2R\rceil}2^{-3m/2}\sum_{n=1}^\infty\left(1+\frac{\delta^2n^2R^2}{2^{2m}}\right)^{-N}\sum_{\theta\in\mc{S}_{m,n}}|\widehat{\chi}_W(\theta^\star)|.\label{eqn.E2Bound1}
\end{align}
Here we will use the following lemma.
\begin{lemma}\label{lem.ChiWEst2}
 With notation as above, for $m,n\in\N$ with $m\le\lceil \log_2R\rceil$, we have that
 \[\sum_{\theta\in\mc{S}_{m,n}}|\widehat{\chi}_W(\theta^\star)|\ll\frac{2^{2m}\log n}{nR^2}.\]
\end{lemma}
\begin{proof}
 Let us write
 \[\tilde{\mc{S}}=\left\{\lambda\in \frac{1}{25}L : \frac{2^{m-1}}{R}<|\lambda|\le\frac{2^m}{R},~\frac{nR}{M2^m}\le|\lambda^\star|<\frac{(n+1)R}{M2^m},~|\mr{arg}(\lambda^\star)|\le \frac{\pi}{5}\right\}.\]
 By the argument used in the proof of Lemma \ref{lem.ChiWEst1} leading up to equation \eqref{eqn.ChiWEst1}, it is sufficient for this lemma to prove that
 \begin{equation*}
 \sum_{\lambda\in\tilde{\mc{S}}}\frac{1}{|\lambda^\star|^2|\sin(\mathrm{arg}(\lambda^\star))|}\ll\frac{2^{2m}\log n}{nR^2}.
 \end{equation*}
 Label the elements of $\star(\tilde{\mc{S}})$ as
 \[\lambda_j^\star=\rho_je^{it_j},\]
 for $1\le j\le J:=|\tilde{\mc{S}}|$, with each $\rho_j\in\R^+$ and with $0< |t_1|\le |t_2|\le\cdots \le |t_J|\le\pi/5$. Also, label the element of $\tilde{\mc{S}}$ corresponding to each $\lambda_j^\star$ as $\lambda_j$.

\noindent\emph{Proof of Theorem \ref{thm.Pen}:} Let $V$ be the vertex set of a unit length RPT. If $V$ can be constructed as a cut and project set, as $V=\Lambda_\omega$ for some $\omega\in\C$, then the result of Theorem \ref{thm.Pen} follows immediately from Theorem \ref{thm.PenCPVersion}. Otherwise, by Theorem \ref{thm.PenVerts}, there is a sequence of complex numbers $\{\omega_j\}_{j=1}^\infty$ with the property that, after a possible rotation,
\[V=\lim_{j\rar\infty}\Lambda_{\omega_j},\]
where the limit is taken in the Chabauty-Fell topology. For $R\ge 2$, the number of points in $\R^2$ which lie on the boundary of $B_R$, and which are the limits of sequences of points $\lambda_j\in\Lambda_{\omega_j}$, as $j\rar\infty,$ must be $\ll R^{2/3}(\log R)^{2/3}$. This follows from the uniformity over $\omega\in \C$ of the error term in Theorem \ref{thm.PenCPVersion}. It then follows, again by uniformity, that
\begin{align}
 \#(V\cap B_R)&=\lim_{j\rar\infty}\#\left(\Lambda_{\omega_j}\cap B_R\right)+O(R^{2/3}(\log R)^{2/3})\nonumber\\
 &=\pi R^2\mr{dens}(\cL)\sum_{m=1}^4|W_{m,\omega+m}|+O( R^{2/3}(\log R)^{2/3}).\label{eqn.ChabLim1}
\end{align}
Finally, we compute that
\[\mr{dens}(\cL)=\mr{disc}(L)^{-1/2}=\frac{4}{25\sqrt{5}}.\]
Substituting this in \eqref{eqn.ChabLim1}, together with the areas of the pentagonal windows, completes the proof of Theorem \ref{thm.Pen}.

Readers who are familiar with the analogous counting problem for integer lattice points in the plane, called the Gauss circle problem, may at this point recognize some similarities. Let us write
\[\#(\mathbb{Z}^2\cap B_R)=\pi R^2 + E(R).\]
Gauss himself observed that $|E(R)|\le 2\sqrt{2}\pi R$, and he conjectured that it should be possible to improve the exponent on $R$  to obtain an error term of $\ll R^{1/2}$. The next substantial result was published in 1906 by Sierpi\'nski \cite{Sier1906}, who showed that $E(R)\ll R^{2/3}$ (see also \cite{Schi1972}). Subsequent improvements have been made by many authors (the introduction of \cite{Huxl2003} contains a list), and currently the best known upper bound, due to Martin Huxley \cite{Huxl2003}, is $E(R)\ll R^{131/208}(\log R)^\tau$, with $\tau=18627/8320$.

The important link which allows us to employ techniques used in the Gauss circle problem in order to study vertices of Penrose tilings is a remarkable observation of de Bruijn, that the vertex set of any RPT can be obtained from a cut and project set, by projecting `slices' of lattice points in a higher dimensional space onto a 2 dimensional subspace. We describe this connection in detail in Section \ref{sec.ModelSets}. With this as a general framework, we are able in Sections \ref{sec.Fourier}-\ref{sec.PenProof} to bring in tools from Fourier analysis and to approach our point counting problem through the lens of the Poisson summation formula (PSF). Results of a similar flavor, estimating discrepancies of points in growing patches of cut and projects sets, have recently been explored by Koivusalo and Lagac{\'e} in \cite{KoivLagacBjorHart2025}, and by R\"uhr, Smilansky, and Weiss in \cite{RuhrSmilWeis2024}.

We remark that our proof, although structurally similar, is not a na\"ive application of known methods (e.g. Huxley's Fourier analytic approach to lattice point counting in regions with smooth boundary, using Van der Corput's method). Here, there are two new difficulties which are not encountered in the classical versions of the Gauss circle problem. The first is that, for the cut and project sets that we consider, the problem amounts to counting points in a lattice in a $4$-dimensional cylinder which is only growing radially. Since the height of the cylinder stays fixed, the smoothing necessary to apply the PSF must shrink faster in the direction of the height than in the radial direction. Secondly, the lattices from which we derive RPTs are irrational. This has the effect that, depending on how well approximable the lattice points are by rational points, smoothing our regions may pick up points which are unusually close to their boundaries. This is a real phenomenon, which can potentially cause a blow up in the Fourier coefficients after using the PSF. It is one of the main issues that we must confront using the specific geometric and algebraic data which define RPTs.

\begin{figure}[h]
	\caption{A patch of an RPT constructed using the cut and project method detailed in Section \ref{sec.ModelSets}. The 4 colors of vertices are translates of the cut and project sets produced using the different windows.\\}\label{fig.PenPatch}
	\centering
	\includegraphics[width=0.75\textwidth]{PenPatch2}
\end{figure}

Finally, there is the question of how these results could be generalized. Several parts of the proofs below can be adapted quite well to deal with much wider classes of cut and project sets. However, there are some crucial details in the proof, in particular those contained in Section \ref{sec.PenProof}, which form a narrow gate that any more general result would need to pass through. The cut and project sets used to define RPTs are extremely special, both in the algebraic properties of the lattice involved and in the exact shape, symmetry, and orientation of the windows involved, with respect to the specific underlying lattice. Any attempt to modify the lattice, or change the windows (even by a small rotation) will destroy important features of the argument and make it difficult or even impossible to control the sizes of the Fourier coefficients involved.

On the other hand, it seems likely that the arguments given here should apply equally well, with the appropriate modifications, to vertex sets of Ammann-Beenker tilings (see \cite[Section 10.4]{GrunShep1987} and \cite{Been1982}). We leave it to the interested reader to work out the details.

To prove Theorem~\ref{thm.Pen}, the authors connect vertex counting in a unit length RPT to the Gauss circle problem by using “a remarkable observation of de Bruijn, that the vertex set of any RPT can be obtained from a cut and project set, by projecting `slices' of lattice points in a higher dimensional space onto a 2 dimensional subspace” (detailed in Section~\ref{sec.ModelSets}). With this framework, they “bring in tools from Fourier analysis” and “approach [the] point counting problem through the lens of the Poisson summation formula (PSF)” (Sections~\ref{sec.Fourier}--\ref{sec.PenProof}).

They emphasize two main technical obstacles (and thus key components of the argument): (1) the counting reduces to “counting points in a lattice in a $4$-dimensional cylinder which is only growing radially,” so because “the height of the cylinder stays fixed, the smoothing necessary to apply the PSF must shrink faster in the direction of the height than in the radial direction”; and (2) the relevant lattices are “irrational,” so “depending on how well approximable the lattice points are by rational points, smoothing our regions may pick up points which are unusually close to their boundaries,” which “can potentially cause a blow up in the Fourier coefficients after using the PSF.” Controlling this requires “the specific geometric and algebraic data which define RPTs,” and the authors note that “crucial details in the proof, in particular those contained in Section~\ref{sec.PenProof}, … make it difficult or even impossible to control the sizes of the Fourier coefficients involved” under general perturbations of lattice/windows.

To prove the main theorem, the authors connect vertex counting in a unit length RPT to the Gauss circle problem by using “a remarkable observation of de Bruijn, that the vertex set of any RPT can be obtained from a cut and project set, by projecting `slices' of lattice points in a higher dimensional space onto a 2 dimensional subspace” (detailed later). With this framework, they “bring in tools from Fourier analysis” and “approach [the] point counting problem through the lens of the Poisson summation formula (PSF)” (proved later).

They emphasize two main technical obstacles (and thus key components of the argument): (1) the counting reduces to “counting points in a lattice in a $4$-dimensional cylinder which is only growing radially,” so because “the height of the cylinder stays fixed, the smoothing necessary to apply the PSF must shrink faster in the direction of the height than in the radial direction”; and (2) the relevant lattices are “irrational,” so “depending on how well approximable the lattice points are by rational points, smoothing our regions may pick up points which are unusually close to their boundaries,” which “can potentially cause a blow up in the Fourier coefficients after using the PSF.” Controlling this requires “the specific geometric and algebraic data which define RPTs,” and the authors note that “crucial details in the proof, in particular those contained in the later argument, … make it difficult or even impossible to control the sizes of the Fourier coefficients involved” under general perturbations of lattice/windows.

\label{thm.Pen}
Let $V$ be the set of vertices of any unit length RPT. Then, for $R\ge 2$,
\[\#(V\cap B_R)=\pi C_P R^2 + O(R^{2/3}(\log R)^{2/3}).\],

Question:
Let $V$ be the vertex set of any unit length rhombic Penrose tiling, meaning a tiling of the plane by copies of the two Penrose rhombi of side length $1$ with angles $72^\circ$ and $36^\circ$, placed edge-to-edge and satisfying the usual matching rules. Let $B_R$ denote the closed Euclidean ball of radius $R$ centered at the origin, and define
\[
C_P=\frac{\phi}{5}\sqrt{10+2\sqrt{5}}, \qquad \phi=\frac{1+\sqrt{5}}{2}.
\]
Which quantitative estimate holds for the number of vertices of the tiling in $B_R$ when $R\ge 2$?

For every such $V$ and all $R\ge 2$,
\[
\#(V\cap B_R)=\pi C_P R^2 + O\!\left(R^{2/3}(\log R)^{2/3}\right).
\]

Choices:
B.
For every such $V$ and all $R\ge 2$,
\[
\#(V\cap B_R)=\pi C_P R^2 + O\!\left(R^{1/2}(\log R)^{2/3}\right).
\]
C.
For every such $V$ and all $R\ge 2$,
\[
\#(V\cap B_R)=\pi C_P R^2 + o(R).
\]
D.
For every such $V$ there exists a constant $K_V>0$ such that for all $R\ge 2$,
\[
\#(V\cap B_R)=\pi C_P R^2 + O\!\left(K_V R^{2/3}(\log R)^{2/3}\right).
\]
E.
For every such $V$ and all $R\ge 2$,
\[
\#(V\cap B_R)=\pi C_P R^2 + O\!\left(R^{2/3}\right).
\]

The stem does not explicitly reveal the correct error term or identify choice A. It only defines the setting and the main-density constant, so the answer is not leaked.

This is essentially a theorem-recall item: it asks which asymptotic statement holds, and the correct option is the theorem statement itself with its precise error term.

There is some reasoning pressure because the options differ by subtle issues of strength, quantifiers, and logarithmic losses. However, the item mainly tests recognition of the exact known result rather than generating a conclusion from supplied mathematical structure.

The distractors are plausible and mathematically meaningful: one is too strong, one is weaker-but-true, one alters uniformity/quantifiers, and one removes the log factor. These align well with common failure modes in asymptotic reasoning.

A solid theorem-discrimination MCQ with strong distractors and no answer leakage, but it is mostly a direct restatement of a known result rather than a genuinely generative reasoning task.


Question:
Let $V$ be the vertex set of any unit length rhombic Penrose tiling, meaning a tiling of the plane by copies of the two Penrose rhombi of side length $1$ with angles $72^\circ$ and $36^\circ$, placed edge-to-edge and satisfying the usual matching rules. Let $B_R$ denote the closed Euclidean ball of radius $R$ centered at the origin, and define
\[
C_P=\frac{\phi}{5}\sqrt{10+2\sqrt{5}}, \qquad \phi=\frac{1+\sqrt{5}}{2}.
\]
Which quantitative estimate holds for the number of vertices of the tiling in $B_R$ when $R\ge 2$?

Choices:
B.
For every such $V$ and all $R\ge 2$,
\[
\#(V\cap B_R)=\pi C_P R^2 + O\!\left(R^{1/2}(\log R)^{2/3}\right).
\]
C.
For every such $V$ and all $R\ge 2$,
\[
\#(V\cap B_R)=\pi C_P R^2 + o(R).
\]
D.
For every such $V$ there exists a constant $K_V>0$ such that for all $R\ge 2$,
\[
\#(V\cap B_R)=\pi C_P R^2 + O\!\left(K_V R^{2/3}(\log R)^{2/3}\right).
\]
E.
For every such $V$ and all $R\ge 2$,
\[
\#(V\cap B_R)=\pi C_P R^2 + O\!\left(R^{2/3}\right).
\]

The stem defines the setting and the constant C_P but does not reveal the asymptotic error term, the logarithmic factor, or the quantifier structure. There is no explicit or trivial cue pointing to choice A.

The item is largely a theorem-identification question: one option states the precise target estimate, and the task is to pick the correct formulation among close variants. This is better than a pure restatement, but still only a mild reformulation rather than a genuinely new conclusion derived from the setup.

There is some reasoning pressure because the choices differ in subtle but meaningful ways: stronger false bounds, a weaker true statement, a quantifier-dependent variant, and omission of a log factor. However, the item primarily tests precise theorem recall/recognition rather than substantial generative mathematical reasoning from first principles.

The distractors are strong: B is an over-strong error term, C is a weaker true statement, D probes uniformity versus tiling-dependent constants, and E tests whether the log factor is necessary. These are plausible and mathematically aligned with common failure modes.

A solid MCQ with no answer leakage and high-quality distractors, but it functions mainly as precise theorem recognition rather than a deeply generative reasoning task.


\begin{thm}
  \label{main1}
  Let $X$ be a tropical variety in $\mathbb R^n$.
  Then $\mathbf E(|X|)$ is finitely generated if and only if $|X|$ is a rational affine linear subspace of $\mathbb R^n$.
\end{thm}

Tropical geometry is a subject of mathematics developed in recent decades, which studies geometric objects called tropical varieties.
Sometimes tropical geometry is called \textquotedblleft the algebraic geometry over $\mathbb T$\textquotedblright, where $\mathbb T$ is a semifield called the tropical semifield.
Since $\mathbb T$ is not a field, tropical geometry is not included in classical algebraic geometry.
However, for many albedo-geometric theories, their tropical versions are studied.
For instance, the tropical intersection theory \cite{allermann2010first,jensen2013computing,jensen2016stable,katz2012tropical,rau2016intersections}, the theory of divisors on tropical curves \cite{amini2013reduced,an2014canonical,haase2012linear,hladky2013rank,len2014brill}, and the theory of tropical homology \cite{amini2020hodge,gross2019sheaf,itenberg2019tropical,ruddat2021homology} have been developed.

One of the basic objects in algebraic geometry is the ideals of rings.
In tropical geometry, there are several analogous objects.
The simplest one is the concept of ideal of a semiring, which is defined as a submodule of the semiring itself.
Another one is the concept of tropical ideal, which is defined by Maclagan and Rinc\'{o}n in \cite{maclagan2018tropical}.
It is an analogue of an ideal of a ring since the quotient semiring $R/E$ is defined.
Moreover, there are several specific class of congruences.
In \cite{giansiracusa2016equations}, J. Giansiracusa and N. Giansiracusa define \textit{bend congruences} to develop the theory of scheme theoretic tropicalization.
Another important class is the congruences of the form $\mathbf{E}(Z)$ for some $Z \subset \mathbb R^n$.
Here,
$$\mathbf{E}(Z) = \{ (f,g) \in \mathbb T[x_1, \ldots, x_n] \ | \ f|_Z = g|_Z \}.$$
The semiring $\mathbb T[x_1, \ldots, x_n]$ is sometimes replaced by another one such as the Laurent polynomial semiring $\mathbb T[x_1^{\pm}, \ldots, x_n^{\pm}]$, the rational function semifield $\mathbb T(x_1, \ldots, x_n)$, the polynomial ring $\mathbb B[x_1, \ldots, x_n]$ over the Boolean semifield, or the polynomial function semiring $\mathbb T[x_1, \ldots, x_n]_{\mathrm{fcn}}$.
This is a simple analogue of the concept of vanishing ideal in algebraic geometry.
The congruences of this form are studied by Bertram and Easton in \cite{bertram2017tropical}, Jo\'{o} and Mincheva in \cite{joo2018prime}, Song in \cite{song2024congruences}, and the author in \cite{ito2022local}.
Also, \textit{prime congruence} are defined in \cite{joo2018prime}, which are of course analogues of the prime ideals of rings.

Almost all rings appeared in classical algebraic geometry are Noetherian, and their ideals are finitely generated.
This fact is used to develop the dimension theory.
Hence it is natural that we consider whether a given ideal (or congruence) is finitely generated.
There are few known results.
In \cite[Example 3.10]{maclagan2018tropical}, an example of tropical ideal which is not finitely generated is given.
Also, it holds by definition that a bend relation defined by a polynomial is finitely generated.
In this paper, we consider the congruences on $\mathbb T[x_1^{\pm}, \ldots, x_n^{\pm}]$ of the form $\mathbf E(Z)$.
In particular, we consider the case that $Z$ is the support of a tropical variety.
The following is our main result.

Almost all rings appeared in classical algebraic geometry are Noetherian, and their ideals are finitely generated.
This fact is used to develop the dimension theory.
Hence it is natural that we consider whether a given ideal (or congruence) is finitely generated.
There are few known results.
In \cite[Example 3.10]{maclagan2018tropical}, an example of tropical ideal which is not finitely generated is given.
Also, it holds by definition that a bend relation defined by a polynomial is finitely generated.
In this paper, we consider the congruences on $\mathbb T[x_1^{\pm}, \ldots, x_n^{\pm}]$ of the form $\mathbf E(Z)$.
In particular, we consider the case that $Z$ is the support of a tropical variety.
The following is our main result.

Thus, for almost all tropical variety $X$, $\mathbf E(X)$ is not finitely generated.
The only if part of this theorem is a consequence of the following more general result.

\begin{thm}
 \label{main}
 Let $X$ be a tropical variety in $\mathbb R^n$.
 Then $\mathbf E(X)$ is finitely generated if and only if $|X|$ is a rational affine linear subspace of $\mathbb R^n$.
\end{thm}

Hence, for a subset $Z \subset W$, in order to show that $\mathbf E(Z)$ is not finitely generated, it is enough to show that $\mathbf E(\iota^{-1}(Z))$ is not finitely generated.
Thus we may assume that $Z$ is not included in any proper rational affine linear subspace of $\mathbb R^n$.
Note that if $Z$ is the support of a tropical variety in $\mathbb R^n$, then so is $\iota^{-1}(Z)$ by Lemma \ref{trop var reduction}.

Let $X$ be a tropical variety in $\mathbb R^n$.
Assume that $W := |X|$ is a $d$-dimensional rational affine linear subspace of $\mathbb R^n$.
By Lemma \ref{translation}, we may assume that $\mathbf 0 \in W$, i.e., $W$ is a linear subspace of $\mathbb R^n$.
Since $W$ is rational, the orthogonal complement $W'$ (with respect to the standard inner product) of $W$ is also rational.
We use the notations $\Lambda(W) = W \cap \mathbb Z^n$ and $\Lambda(W') = W' \cap \mathbb Z^n$.
Let $\{ \bm q_1, \ldots, \bm q_{n-d} \}$ be a basis of $\Lambda(W')$.
We see that the congruence $\mathbf E(W)$ is generated by the finite set $S := \left\{ \left(\overline{\bm x^{\bm q_i}}, 0 \right) \ | \ i=1, \ldots, n-d \right\}$.

\begin{prop}
 \label{trop unbounded}
 Let $X$ be a tropical variety in $\mathbb R^n$.
 Assume that $|X|$ is not included in a proper rational affine linear subspace of $\mathbb R^n$.
 Then $|X|$ is unbounded in all rational directions.
\end{prop}

\begin{proof}[Proof of \textquotedblleft Proposition \ref{trop unbounded in f} $\Longrightarrow$ Proposition \ref{trop unbounded}\textquotedblright]
 Assume that Proposition \ref{trop unbounded in f} holds.
 Let $X$ be a tropical variety in $\mathbb R^n$ whose support is not included in a proper rational affine linear subspace of $\mathbb R^n$.
 Take any $\bm p \in \mathbb Z^n \setminus \{ \mathbf 0 \}$.
 If the function $f: |X| \to \mathbb R, \bm a \mapsto \bm a \cdot \bm p$ is constant with the value $k$, then $|X|$ is included in the set $\{ \bm a \in \mathbb R^n \ | \ \bm a \cdot \bm p = k \}$, which is a proper rational affine linear subspace of $\mathbb R^n$.
 This contradicts to the assumption, hence $f$ is not constant.
 Thus $f$ is upper unbounded.
 Since $\bm p$ is an arbitrary element of $\mathbb Z^n \setminus \{ \mathbf 0 \}$, $|X|$ is unbounded in all rational directions.
\end{proof}

\begin{proof}[Proof of the only if part of Theorem \ref{main}]
 Let $X$ be a tropical variety such that $|X|$ is not a rational affine linear subspace of $\mathbb R^n$.
 We may assume that $|X|$ is not included in any proper rational affine linear subspace of $\mathbb R^n$ by Lemma \ref{fg reduction} and the observation after it.
 Hence, by Proposition \ref{trop unbounded}, $|X|$ is unbounded in all rational directions.
 In order to use Theorem \ref{not fg}, we check that $n \geq 2$.
 If $n =0$, $|X| = \{ \mathbf 0 \}$.
 It contradicts to the assumption that $|X|$ is not an affine linear subspace.
 If $n =1$, $|X|$ is a singleton or the whole space $\mathbb R$ (by Lemma \ref{n in n}).
 It also contradicts to the assumption.
 Thus $n \geq 2$, and hence by Theorem \ref{not fg}, $\mathbf E(X)$ is not finitely generated. \qedhere
\end{proof}

Thus, for almost all tropical variety $X$, $\mathbf E(X)$ is not finitely generated.
The only if part of this theorem is a consequence of the following more general result.

\begin{thm}
  \label{main2}
  Assume that $n \geq 2$.
  Let $Z$ be a proper closed subset of $\mathbb R^n$ that is unbounded in all rational directions.
  Then $\mathbf E(Z)$ is not finitely generated.
\end{thm}

The strategy of the proof of Theorem \ref{main2} is as follows.
First we assume that $\mathbf E(Z)$ is finitely generated and fix a generating set.
Then we construct a pair $(f,g) \in \mathbf E(Z)$ such that $f \neq g$ and the distance of any two lattice points of the Newton polytope of $f$ is sufficiently large.
The existence of such pair causes a contradiction.
See Section 3 for detail.

Note that we cannot apply Theorem \ref{main2} for showing the only if part of Theorem \ref{main1} directly because $|X|$ is not unbounded in all rational directions in general.
In particular, if $|X|$ is included in a proper rational affine linear subspace of $\mathbb R^n$, then $|X|$ is not unbounded in all rational directions.
To overcome this issue, we show that we may replace the ambient space with a lower dimensional one.
See Section 4 for detail.

Finally, we give an example of minimal generating set of $\mathbf E(L)$ for the standard tropical line $L \subset \mathbb R^2$.
See Section \ref{section min gen} for detail.

\begin{thm}
  \label{main3}
  The set
  $$\mathcal S = \left\{ \left( \overline{0 \oplus x^u y^v}, f_{\Delta(\{ (0,0), (u,v) \})} \right) \ \middle| \ \begin{aligned}
  &\text{$u$ and $v$ are integers with $\gcd(u,v) = 1$,}\\
  &\text{ $u > 0$ or $(u,v) = (0,1)$}
  \end{aligned} \right\}$$
  is a minimal generating set of $\mathbf E(L)$.
\end{thm}

This paper is organized as follows.
In Section 2, we recall some basic definitions and facts about semirings, tropical algebra, tropical varieties, and Newton polytopes.
Also we show some lemmas we need later.
In Section 3, we consider a way to prove that a given congruence on $\mathbb T[x_1^{\pm}, \ldots, x_n^{\pm}]$ is not finitely generated, and use it to show Theorem \ref{main2}.
In Section 4, we discuss about the reduction of the ambient spaces.
The result of this section is necessary for showing Theorem \ref{main1}.
In Section 5, we prove Theorem \ref{main1}.
In Section 6, we give an example of minimal generating set of $\mathbf E(L)$ for the standard tropical line $L$.
This section is almost independent of Section 3-5.
In Appendix, we give proofs of some facts about free abelian groups and polytopes, which are used in this paper.

For Theorem~\ref{main1}, the text says: “The only if part of this theorem is a consequence of the following more general result” (Theorem~\ref{main2}), but “we cannot apply Theorem~\ref{main2} for showing the only if part of Theorem~\ref{main1} directly because $|X|$ is not unbounded in all rational directions in general.” “To overcome this issue, we show that we may replace the ambient space with a lower dimensional one.”

For Theorem~\ref{main2}, the stated proof strategy is: “First we assume that $\mathbf E(Z)$ is finitely generated and fix a generating set. Then we construct a pair $(f,g) \in \mathbf E(Z)$ such that $f \neq g$ and the distance of any two lattice points of the Newton polytope of $f$ is sufficiently large. The existence of such pair causes a contradiction.”

Question:
Let $X$ be a tropical variety in $\mathbb R^n$, and let $|X|$ denote its support. For a subset $Z\subset \mathbb R^n$, define the congruence
$$\mathbf E(Z)=\{(f,g)\in \mathbb T[x_1^{\pm},\ldots,x_n^{\pm}]\mid f(\mathbf a)=g(\mathbf a)\text{ for all }\mathbf a\in Z\},$$
where $\mathbb T[x_1^{\pm},\ldots,x_n^{\pm}]$ is the tropical Laurent polynomial semiring. A rational affine linear subspace of $\mathbb R^n$ means an affine translate of a linear subspace defined over $\mathbb Q$ (equivalently, whose direction is spanned by rational vectors). Which statement holds for every such tropical variety $X$?

Choices:
B. $\mathbf E(|X|)$ is finitely generated if and only if $|X|$ is an affine linear subspace of $\mathbb R^n$.
C. If $|X|$ is a rational affine linear subspace of $\mathbb R^n$, then $\mathbf E(|X|)$ is finitely generated.
D. $\mathbf E(|X|)$ is finitely generated whenever $|X|$ is contained in a rational affine linear subspace of $\mathbb R^n$.
E. $\mathbf E(|X|)$ is finitely generated if and only if $|X|$ is a finite union of rational affine linear subspaces of $\mathbb R^n$.

equality with a rational affine linear subspace replaced by mere containment

single rational affine linear subspace replaced by finite union

The stem does not reveal the correct choice explicitly or by a near-verbatim hint. It asks for an equivalent characterization, and the correct option must be identified from several closely related geometric statements.

The item is essentially a theorem-recall question about an equivalence, so it is somewhat close to restating a known result. However, it is not fully tautological because the choices introduce subtly different conditions such as rationality, containment, and finite unions.

Some reasoning is needed to distinguish the exact equivalence from weaker or altered variants, especially between equality, containment, and rationality. Still, the question mainly tests precise recall/recognition of a theorem rather than deeper generative problem solving.

The distractors are mathematically plausible and target common failure modes: dropping rationality, weakening equality to containment, replacing a single subspace by a finite union, or reformulating via congruence equality with a larger subspace.

A solid theorem-recognition MCQ with strong distractors and no answer leakage, but it leans more toward precise recall than genuine generative reasoning.


Question:
Let $X$ be a tropical variety in $\mathbb R^n$, and let $|X|$ denote its support. For a subset $Z\subset \mathbb R^n$, define the congruence
$$\mathbf E(Z)=\{(f,g)\in \mathbb T[x_1^{\pm},\ldots,x_n^{\pm}]\mid f(\mathbf a)=g(\mathbf a)\text{ for all }\mathbf a\in Z\},$$
where $\mathbb T[x_1^{\pm},\ldots,x_n^{\pm}]$ is the tropical Laurent polynomial semiring. A rational affine linear subspace of $\mathbb R^n$ means an affine translate of a linear subspace defined over $\mathbb Q$ (equivalently, whose direction is spanned by rational vectors). Which statement holds for every such tropical variety $X$?

Choices:
B. $\mathbf E(|X|)$ is finitely generated if and only if $|X|$ is an affine linear subspace of $\mathbb R^n$.
C. If $|X|$ is a rational affine linear subspace of $\mathbb R^n$, then $\mathbf E(|X|)$ is finitely generated.
D. $\mathbf E(|X|)$ is finitely generated whenever $|X|$ is contained in a rational affine linear subspace of $\mathbb R^n$.
E. $\mathbf E(|X|)$ is finitely generated if and only if $|X|$ is a finite union of rational affine linear subspaces of $\mathbb R^n$.

The stem defines the objects involved but does not reveal the finite-generation criterion or otherwise point directly to choice A.

The item is essentially an exact theorem-identification question: the correct option states the full characterization directly, so the task is largely to recognize the theorem rather than apply it.

There is some reasoning in comparing 'if and only if' versus weaker, stronger, or tampered variants, but the problem mainly tests recall of the precise statement rather than genuine derivation or synthesis.

The distractors are plausible and mathematically meaningful: they vary rationality, weaken to one direction, replace equality by containment, or broaden to finite unions, all of which reflect realistic failure modes.

Well-constructed in terms of non-leaky wording and strong distractors, but it is primarily a theorem-recall/restatement item rather than a high-generative-reasoning question.