\documentclass[12pt,reqno]{article}
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\newcommand{\seqnum}[1]{\href{https://oeis.org/#1}{\rm \underline{#1}}}
\usepackage{tikz}
\usetikzlibrary{patterns}
\newcommand{\FindStat}[1]{%
\ifx&%
\url{www.findstat.org}% #1 is empty
\else \url{www.findstat.org/#1}% #1 is nonempty
\fi}%
\newcommand{\Dfn}[1]{\emph{#1}}%
\begin{document}
\begin{center}
\epsfxsize=4in
\leavevmode\epsffile{logo129.eps}
\end{center}
\theoremstyle{plain}
\newtheorem{thm}{Theorem}
\newtheorem{lem}[thm]{Lemma}
\newtheorem{proposition}[thm]{Proposition}
\theoremstyle{definition}
\newtheorem{dfn}[thm]{Definition}
\begin{center}
\vskip 1cm{\LARGE\bf
Alternating Sign Matrices through X-Rays\\
\vskip .1in } \vskip 1cm \large
Martin Rubey\footnote{Supported by the Austrian Science Fund (FWF): P 29275.}\\
Fakult\"at f\"ur Mathematik und Geoinformation\\
TU Wien\\
Austria\\
\href{mailto:martin.rubey@tuwien.ac.at}{\tt martin.rubey@tuwien.ac.at}
\end{center}
\vskip .2in
\begin{abstract}
We exhibit a bijection between Dyck paths and alternating sign
matrices which are determined by their antidiagonal sums.
\end{abstract}
\section{Introduction}
A fundamental question in discrete tomography is whether a binary
image can be reconstructed from a small number of projections. As a
special case, one might restrict attention to permutation matrices
and try to determine which vectors of antidiagonal sums appear only
once. This problem, considered by Bebeacua, Mansour, Postnikov, and
Severini~\cite{MR2301096}, is apparently still open. The number of
such vectors is recorded in the sequence \seqnum{A289971} in the
{\it On-Line Encyclopedia of Integer Sequences} \cite{sloane}.
In this note, we consider the analogous problem for alternating sign
matrices. An alternating sign matrix is a square matrix of $0$s,
$1$s and $-1$s such that the sum of the entries in each row and each column is $1$,
and the nonzero entries in each row and in each column alternate in
sign. For an $n\times n$-alternating sign matrix $A$, the $k$-th
(antidiagonal) sum is $x_k = \sum_{i+j=k+1} A_{i,j}$ and the
(antidiagonal) \Dfn{X-ray} is the vector $(x_1,\dots,x_{2n-1})$. For
example, the alternating sign matrices of size three together with
their X-rays are as follows:
\begin{center}
\scalebox{0.65}{\mbox{%
$\begin{array}{ccccccc}
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}& \begin{pmatrix}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1
\end{pmatrix}& \begin{pmatrix}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0
\end{pmatrix}& \begin{pmatrix}
0 & 1 & 0 \\
1 & -1 & 1 \\
0 & 1 & 0
\end{pmatrix}& \begin{pmatrix}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0
\end{pmatrix}& \begin{pmatrix}
0 & 0 & 1 \\
1 & 0 & 0 \\
0 & 1 & 0
\end{pmatrix}& \begin{pmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
1 & 0 & 0
\end{pmatrix}\\[16pt]
(1, 0, 1, 0, 1)&
(0, 2, 0, 0, 1)&
(1, 0, 0, 2, 0)&
(0, 2,-1, 2, 0)&
(0, 0, 3, 0, 0)&
(0, 1, 1, 1, 0)&
(0, 1, 1, 1, 0)
\end{array}$
}}
\end{center}
Note that all X-rays except $(0, 1, 1, 1, 0)$ occur precisely once.
Thus, there are five alternating sign matrices determined by their
X-rays. We can now state our main result:
\begin{thm}\label{thm:main}
There is an explicit bijection between Dyck paths of semilength $n$
and $n\times n$-alternating sign matrices which are determined by
their X-rays.
\end{thm}
We recall that the number of Dyck paths of semilength $n$ equals the
$n$-th Catalan number $\frac{1}{n+1}\binom{2n}{n}$, recorded as
\seqnum{A000108}. We note that, although there is also a nice formula
for the total number of size $n$ alternating sign matrices,
$\prod_{k=0}^{n-1}\frac{(3k+1)!}{(n+k)!}$, recorded as
\seqnum{A005130}, it is much harder to prove this.
The coincidence described by the theorem was observed when submitting
the statistic~\FindStat{St000889}, which counts the number of
alternating sign matrices with the same X-rays, to the online database
of combinatorial statistics FindStat~\cite{FindStat2017}, and looking
at the linear terms of the first few generating functions
automatically produced there. We currently have no explanation for
any of the other terms in the distribution.
\section{The bijection}
We define an injective map $\mathcal A$ from Dyck paths to
alternating sign matrices as follows, see Figure~\ref{fig:exampleA}
for an example. For more visual clarity in the pictures, we
sometimes use ($+$)s and ($-$)s instead of $1$s and $-1$s and omit
$0$s.
\begin{itemize}
\item Draw the Dyck path in an $n\times n$ square, beginning in the
top left corner, taking unit east and unit south steps and
terminating in the bottom right corner, never going below the main
diagonal of the matrix.
\item Add the reflection through the main diagonal of the Dyck path
to the picture.
\item For each peak of the Dyck path, fill the cells lying between
the peak and its mirror image on the antidiagonal with $1$s.
\item For each valley of the Dyck path, fill the cells lying between
the valley and its mirror image on the antidiagonal with $-1$s.
\item Fill the remaining cells with $0$s.
\end{itemize}
\begin{figure}[h]
\centering
\begin{tikzpicture}[scale=0.5]
\draw [line width=0.5] (0,0) -- (0,8) -- (8,8) -- (8,0) -- cycle;
\draw[dotted] (0,8)--(8,0);
\draw[rounded corners=1, line width=1]%
(0,8)--(4,8)--%
(4,7)--(6,7)--%
(6,5)--(7,5)--%
(7,4)--(8,4)--%
(8,0);%
\draw[rounded corners=1, line width=1, dotted]%
(0,8)--(0,4)--%
(1,4)--(1,2)--%
(3,2)--(3,1)--%
(4,1)--(4,0)--%
(8,0);%
\node[below left] at (4,8) {$1$};
\node[below left] at (3,7) {$1$};
\node[below left] at (2,6) {$1$};
\node[below left] at (1,5) {$1$};
%
\node[below left] at (6,7) {$1$};
\node[below left] at (5,6) {$1$};
\node[below left] at (4,5) {$1$};
\node[below left] at (3,4) {$1$};
\node[below left] at (2,3) {$1$};
%
\node[below left] at (7,5) {$1$};
\node[below left] at (6,4) {$1$};
\node[below left] at (5,3) {$1$};
\node[below left] at (4,2) {$1$};
%
\node[below left] at (8,4) {$1$};
\node[below left] at (7,3) {$1$};
\node[below left] at (6,2) {$1$};
\node[below left] at (5,1) {$1$};
%%
\node[below left] at (4,7) {$-1$};
\node[below left] at (3,6) {$-1$};
\node[below left] at (2,5) {$-1$};
%
\node[below left] at (6,5) {$-1$};
\node[below left] at (5,4) {$-1$};
\node[below left] at (4,3) {$-1$};
%
\node[below left] at (7,4) {$-1$};
\node[below left] at (6,3) {$-1$};
\node[below left] at (5,2) {$-1$};
%
\foreach\x in {1,2,3,5,6,7,8}
\node[below left] at (\x,8) {$0$};
\foreach\x in {1,2,5,7,8}
\node[below left] at (\x,7) {$0$};
\foreach\x in {1,4,6,7,8}
\node[below left] at (\x,6) {$0$};
\foreach\x in {3,5,8}
\node[below left] at (\x,5) {$0$};
\foreach\x in {1,2,4}
\node[below left] at (\x,4) {$0$};
\foreach\x in {1,3,8}
\node[below left] at (\x,3) {$0$};
\foreach\x in {1,2,3,7,8}
\node[below left] at (\x,2) {$0$};
\foreach\x in {1,2,3,4,6,7,8}
\node[below left] at (\x,1) {$0$};
\end{tikzpicture}
\quad
\begin{tikzpicture}[scale=0.5]
\draw [line width=0.5] (0,0) -- (0,8) -- (8,8) -- (8,0) -- cycle;
\draw[dotted] (0,8)--(8,0);
\draw[rounded corners=1, line width=1]%
(0,8)--(4,8)--%
(4,7)--(6,7)--%
(6,5)--(7,5)--%
(7,4)--(8,4)--%
(8,0);%
\draw[rounded corners=1, line width=1, dotted]%
(0,8)--(0,4)--%
(1,4)--(1,2)--%
(3,2)--(3,1)--%
(4,1)--(4,0)--%
(8,0);%
\node[below left] at (4,8) {$+$};
\node[below left] at (3,7) {$+$};
\node[below left] at (2,6) {$+$};
\node[below left] at (1,5) {$+$};
%
\node[below left] at (6,7) {$+$};
\node[below left] at (5,6) {$+$};
\node[below left] at (4,5) {$+$};
\node[below left] at (3,4) {$+$};
\node[below left] at (2,3) {$+$};
%
\node[below left] at (7,5) {$+$};
\node[below left] at (6,4) {$+$};
\node[below left] at (5,3) {$+$};
\node[below left] at (4,2) {$+$};
%
\node[below left] at (8,4) {$+$};
\node[below left] at (7,3) {$+$};
\node[below left] at (6,2) {$+$};
\node[below left] at (5,1) {$+$};
%%
\node[below left] at (4,7) {$-$};
\node[below left] at (3,6) {$-$};
\node[below left] at (2,5) {$-$};
%
\node[below left] at (6,5) {$-$};
\node[below left] at (5,4) {$-$};
\node[below left] at (4,3) {$-$};
%
\node[below left] at (7,4) {$-$};
\node[below left] at (6,3) {$-$};
\node[below left] at (5,2) {$-$};
\end{tikzpicture}
\caption{The image of a Dyck path}
\label{fig:exampleA}
\end{figure}
In the remaining two sections we show that the image of the map
$\mathcal A$ consists precisely of those alternating sign matrices
which are determined by their X-ray, thus proving
Theorem~\ref{thm:main}.
\section{A map on diagonally symmetric alternating sign matrices}
In this section we show that any alternating sign matrix which is
uniquely determined by its X-ray is in the image of $\mathcal A$.
Because transposing a matrix preserves the X-ray, only diagonally
symmetric alternating sign matrices may be reconstructible from their
X-ray. We now present a map $\mathcal M$ on diagonally symmetric
alternating sign matrices that preserves the X-ray and is the
identity precisely on the matrices in the image of the map
$\mathcal A$ from the previous section.
These two properties lead to the desired result: suppose that $A$ is
determined by its X-ray. Since the X-rays of $A$ and $\mathcal M(A)$
are the same, $\mathcal M(A)$ is also determined by its X-ray, and
equals $A$.
Let $A$ be a diagonally symmetric alternating sign matrix. Then the
matrix $\mathcal M(A)$ is obtained as follows. See
Figure~\ref{fig:exampleM} for an example.
\begin{itemize}
\item Imagine a sun in the north-east, such that the $1$s in $A$ cast
shadows, and trace out a Dyck path by following the shadow line.
This path is indicated in solid black in Figure~\ref{fig:exampleM}.
\item Consider the entries of $A$ which are strictly south-west of
the entries just below the Dyck path. In
Figure~\ref{fig:exampleM}, these are the entries in the shaded
region. Reflect these entries through the subdiagonal of the
matrix.
\item Into each cell just south-west of a valley of the Dyck path
which is not on the subdiagonal and which contains a $0$, place a
$-1$, and place a $1$ in the cell reflected through the
subdiagonal.
\end{itemize}
\begin{figure}[h]
\centering
\begin{tikzpicture}[scale=0.5]
\draw [line width=0.5] (0,0) -- (0,8) -- (8,8) -- (8,0) -- cycle;
\draw [fill=gray, opacity=0.3]
(0,7) -- (3,7) -- (3,6) -- (5,6) -- (5,4) -- (6,4) -- (6,3) --
(7,3) -- (7,0) -- (4,0) -- (4,1) -- (3,1) -- (3,2) -- (1,2) --
(1,4) -- (0,4) -- cycle;
\draw[dotted] (0,7)--(7,0);
\draw[rounded corners=1, line width=1]%
(0,8)--(4,8)--%
(4,7)--(6,7)--%
(6,5)--(7,5)--%
(7,4)--(8,4)--%
(8,0);%
\draw[rounded corners=1, line width=1, dotted]%
(0,8)--(0,4)--%
(1,4)--(1,2)--%
(3,2)--(3,1)--%
(4,1)--(4,0)--%
(8,0);%
%
\node[below left] at (2,7) {$+$};
%
\node[below left] at (4,8) {$+$};
\node[below left] at (1,5) {$+$};
%
\node[below left] at (6,7) {$+$};
\node[below left] at (5,6) {$+$};
\node[below left] at (3,4) {$+$};
\node[below left] at (2,3) {$+$};
%
\node[below left] at (7,5) {$+$};
\node[below left] at (4,2) {$+$};
%
\node[below left] at (8,4) {$+$};
\node[below left] at (7,3) {$+$};
\node[below left] at (6,2) {$+$};
\node[below left] at (5,1) {$+$};
%%
\node[below left] at (4,7) {$-$};
\node[below left] at (2,5) {$-$};
%
\node[below left] at (7,4) {$-$};
\node[below left] at (6,3) {$-$};
\node[below left] at (5,2) {$-$};
%
\end{tikzpicture}
\quad\raisebox{50pt}{$\mapsto$}\quad
\begin{tikzpicture}[scale=0.5]
\draw [line width=0.5] (0,0) -- (0,8) -- (8,8) -- (8,0) -- cycle;
\draw [fill=gray, opacity=0.3]
(0,7) -- (3,7) -- (3,6) -- (5,6) -- (5,4) -- (6,4) -- (6,3) --
(7,3) -- (7,0) -- (4,0) -- (4,1) -- (3,1) -- (3,2) -- (1,2) --
(1,4) -- (0,4) -- (0,7);
\draw[dotted] (0,7)--(7,0);
\draw[rounded corners=1, line width=1]%
(0,8)--(4,8)--%
(4,7)--(6,7)--%
(6,5)--(7,5)--%
(7,4)--(8,4)--%
(8,0);%
\draw[rounded corners=1, line width=1, dotted]%
(0,8)--(0,4)--%
(1,4)--(1,2)--%
(3,2)--(3,1)--%
(4,1)--(4,0)--%
(8,0);%
%
\node[below left] at (1,6) {$+$};
%
\node[below left] at (4,8) {$+$};
\node[below left] at (3,7) {$+$};
%
\node[below left] at (6,7) {$+$};
\node[below left] at (5,6) {$+$};
\node[below left] at (4,5) {$+$};
\node[below left] at (2,3) {$+$};
%
\node[below left] at (7,5) {$+$};
\node[below left] at (6,4) {$+$};
%
\node[below left] at (8,4) {$+$};
\node[below left] at (7,3) {$+$};
\node[below left] at (6,2) {$+$};
\node[below left] at (5,1) {$+$};
%%
\node[below left] at (4,7) {$-$};
\node[below left] at (3,6) {$-$};
%
\node[below left] at (7,4) {$-$};
\node[below left] at (6,3) {$-$};
\node[below left] at (5,2) {$-$};
\end{tikzpicture}
\quad\raisebox{50pt}{$\mapsto$}\quad
\begin{tikzpicture}[scale=0.5]
\draw [line width=0.5] (0,0) -- (0,8) -- (8,8) -- (8,0) -- cycle;
\draw[dotted] (0,7)--(7,0);
\draw[rounded corners=1, line width=1]%
(0,8)--(4,8)--%
(4,7)--(6,7)--%
(6,5)--(7,5)--%
(7,4)--(8,4)--%
(8,0);%
\draw[rounded corners=1, line width=1, dotted]%
(0,8)--(0,4)--%
(1,4)--(1,2)--%
(3,2)--(3,1)--%
(4,1)--(4,0)--%
(8,0);%
%
\node[below left] at (1,6) {$+$};
\node[below left, red] (A) at (6,5) {$-$};
\node[draw=black, circle, minimum size=12pt] at (A) {};
\node[below left, red] (B) at (3,2) {$+$};
\node[draw=black, circle, minimum size=12pt] at (B) {};
%
\node[below left] at (4,8) {$+$};
\node[below left] at (3,7) {$+$};
%
\node[below left] at (6,7) {$+$};
\node[below left] at (5,6) {$+$};
\node[below left] at (4,5) {$+$};
\node[below left] at (2,3) {$+$};
%
\node[below left] at (7,5) {$+$};
\node[below left] at (6,4) {$+$};
%
\node[below left] at (8,4) {$+$};
\node[below left] at (7,3) {$+$};
\node[below left] at (6,2) {$+$};
\node[below left] at (5,1) {$+$};
%%
\node[below left] at (4,7) {$-$};
\node[below left] at (3,6) {$-$};
%
\node[below left] at (7,4) {$-$};
\node[below left] at (6,3) {$-$};
\node[below left] at (5,2) {$-$};
\end{tikzpicture}
\caption{A diagonally symmetric alternating sign matrix and its
image}
\label{fig:exampleM}
\end{figure}
Although not used subsequently, we note that the Dyck path
constructed in the first step returns to the main diagonal exactly
once for each direct summand of $A$, when regarding $A$ as a block
diagonal matrix. Thus, the map $\mathcal M$ is such that it can be
applied to each direct summand of $A$ individually.
\begin{lem}
The map $\mathcal M$, applied to a diagonally symmetric alternating
sign matrix, produces an alternating sign matrix.
\end{lem}
\begin{proof}
Let $A$ be a diagonally symmetric alternating sign matrix. Let us
call the region in $A$ which is symmetric with respect to the subdiagonal
and whose south-west border is the reflected Dyck path, the \Dfn{shade}
of $A$. This is the shaded region in Figure~\ref{fig:exampleM}.
Consider a column $c$ of $A$, and its reflection $r$ through the
subdiagonal. Thus, when $c$ is the first column, $r$ is the second
row of $A$.
Suppose first that the topmost nonzero entry of $c$ and the
rightmost nonzero entry of $r$ within the shade of $A$ are both
$1$. This is the case when $c$ (or $r$) does not contain a peak of
the Dyck path, or the valley below (or to the left of) the peak
contains a $-1$. In this case, column $c$ and row $r$ of
$\mathcal M(A)$ satisfy the alternating sign matrix conditions,
since the bottommost nonzero entry of any column and the leftmost
nonzero entry of any row of $A$ is within the shade and equals $1$.
Let us now consider the second scenario, where the topmost nonzero
entry of $c$ within the shade of the original matrix $A$ is $-1$.
In the example of Figure~\ref{fig:exampleM} this happens in the
sixth column.
In this case, the Dyck path must have a peak in this column. Let
$v$ be the cell just south-west of the valley below the peak. Note
that $v$ must contain a $0$.
The cell $v$ must be strictly above the diagonal, because otherwise
the reflection of the $-1$ below it through the main diagonal would
lie on or above the Dyck path. Thus, by definition of
$\mathcal M$, we place a $-1$ into the cell $v$. The effect of
this is that column $c$ of $\mathcal M(A)$ is alternating.
Furthermore, we place a $1$ in the cell $v'$ corresponding to $v$
reflected through the subdiagonal. This satisfies the alternating
sign matrix conditions, because after reflecting through the
subdiagonal the row containing $v'$ has a $-1$ as its leftmost
nonzero entry.
\end{proof}
\begin{lem}
The map $\mathcal M$, applied to a diagonally symmetric alternating
sign matrix $A$, is the identity if and only if $A$ is in the image
of $\mathcal A$.
\end{lem}
\begin{proof}
If $A$ is in the image of $\mathcal A$, the shade of $A$ is
symmetric. Moreover, the cells just south-west of the valleys
which are above the subdiagonal all contain $-1$s. Thus,
$\mathcal M(A) = A$.
Otherwise, since $A$ is symmetric, and the shade of $A$ is
reflected through the subdiagonal, $\mathcal M(A)$ cannot be
symmetric.
\end{proof}
\section{Reconstructing the alternating sign matrix}
To complete the proof of Theorem~\ref{thm:main}, we have to show the following:
\begin{lem}
The X-ray corresponding to an alternating sign matrix in the image
of $\mathcal A$ determines the matrix unambiguously.
\end{lem}
\begin{proof}
Consider the antidiagonal sums beginning at the north-west corner.
Suppose that the entries of the first $k$ antidiagonals are
uniquely determined by the first $k$ entries $x_1,x_2,\dots,x_k$ of
the X-ray, and suppose that $x_k\neq 0$,
$x_{k+1}=\dots=x_{\ell-1}=0$ and $x_{\ell}\neq 0$.
For simplicity, assume that $x_k > 0$. By hypothesis, the
alternating sign matrix then has the following form:
\begin{center}
\begin{tikzpicture}[scale=0.5]
\draw[dotted] (0,8)--(8,0);
\draw[rounded corners=1, line width=1]%
(0,8)--(4,8)--%
(4,7)--(6,7);%
\draw[rounded corners=1, line width=1, dotted]%
(0,8)--(0,4)--%
(1,4)--(1,2);%
\draw [fill=gray, opacity=0.3] (6,6)--(6,2)--(2,2)--cycle;
\draw [preaction={fill,red,opacity=0.3}, pattern=dots] (6,2)--(9,2)--(6,-1)--cycle;
\node[below left] at (4,8) {$1$};
\node[below left] at (3,7) {$1$};
\node[below left] at (2,6) {$1$};
\node[below left] at (1,5) {$1$};
\node[below left] at (4,7) {$-1$};
\node[below left] at (3,6) {$-1$};
\node[below left] at (2,5) {$-1$};
\node[below left] at (6,7) {$1$};
\node[below left] at (5,6) {$1$};
\node[below left] at (4,5) {$1$};
\node[below left] at (3,4) {$1$};
\node[below left] at (2,3) {$1$};
%
\foreach\x in {1,2,3,5,6,7}
\node[below left] at (\x,8) {$0$};
\foreach\x in {1,2,5}
\node[below left] at (\x,7) {$0$};
\foreach\x in {1,4}
\node[below left] at (\x,6) {$0$};
\foreach\x in {3}
\node[below left] at (\x,5) {$0$};
\foreach\x in {1,2}
\node[below left] at (\x,4) {$0$};
\foreach\x in {1}
\node[below left] at (\x,3) {$0$};
\foreach\x in {1}
\node[below left] at (\x,2) {$0$};
\end{tikzpicture}
\end{center}
Let us first note that there cannot be any nonzero entries on the
antidiagonals $k+1,\dots,\ell-1$. According to our assumption, all
these have sum zero. Suppose for the sake of contradiction that
there is such an antidiagonal and consider the first of these.
Because every row and every column of an alternating sign matrix
must begin with a $1$, this antidiagonal can have $-1$s only in the
triangular region south-east of the sequence of $1$s in the $k$-th
antidiagonal - shaded grey in the example above. However, there
cannot be any $1$s on the same antidiagonal, necessarily outside of
this triangular region: any such $1$ below the main diagonal would
be followed by another $1$ in the same column above it.
We now distinguish two cases: if $x_\ell < 0$, since the matrix is
in the image of $\mathcal A$, the $\ell$-th antidiagonal intersects
the triangular region defined above and all cells of the
antidiagonal within this region are filled with $-1$s. Thus, in
this case the entries on the $\ell$-th antidiagonal are also
uniquely determined by the antidiagonal sum.
On the other hand, if $x_\ell > 0$, since the matrix is in the
image of $\mathcal A$, the $\ell$-th antidiagonal intersects the
triangular region which is dotted and shaded red in the example.
In this case, all cells of the antidiagonal within this region are
filled with $1$s. Since there cannot be any $1$s on the same
antidiagonal outside of the red triangular region, also in this
case the entries of the $\ell$-th antidiagonal are uniquely
determined by their sum.
\end{proof}
\begin{thebibliography}{1}
\bibitem{MR2301096}
C.~Bebeacua, T.~Mansour, A.~Postnikov, and S.~Severini,
\newblock On the {X}-rays of permutations, in
\newblock {\it Proceedings of the {W}orkshop on {D}iscrete
{T}omography and its {A}pplications, Electron. Notes Discrete Math.} {\bf 20} (2005), 193--203. Available at \url{https://arxiv.org/abs/math/0506334}.
\bibitem{FindStat2017}
M.~Rubey, C.~Stump, et~al.,
\newblock {FindStat}---{T}he combinatorial statistics database,
\newblock \url{https://www.FindStat.org}.
\bibitem{sloane}
N. J. A. Sloane et al.
\newblock {\it The On-Line Encyclopedia of Integer Sequences}, 2021.
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\end{thebibliography}
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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 05A19.
\noindent \emph{Keywords: }
bijective combinatorics, combinatorial statistic, Dyck path, alternating sign matrix.
\bigskip
\hrule
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\noindent (Concerned with sequences
\seqnum{A000108}, \seqnum{A005130}, and \seqnum{A289971}.)
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\vspace*{+.1in}
\noindent
Received December 14 2020;
revised versions received April 22 2021; April 28 2021.
Published in {\it Journal of Integer Sequences}, May 19 2021.
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\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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