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\vskip 1cm{\LARGE\bf On the dominance partial ordering
\vskip .1in
of Dyck paths}
\vskip 1cm
\large
A. Sapounakis, I. Tasoulas, and P. Tsikouras \\
Department of Informatics\\
University of Piraeus\\
18534 Piraeus \\
GREECE\\
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\htmladdnormallink{\texttt{jtas}}{jtas@unipi.gr}, \htmladdnormallink{\texttt{pgtsik}}{pgtsik@unipi.gr}\}\texttt{@unipi.gr}
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\vskip .2 in
\begin{abstract}
The lattice of Dyck paths with the dominance partial order is studied. The
notions of filling and degree of a Dyck path are introduced, studied and used
for the evaluation of the M\"obius function and its powers.
The relation between the symmetric group endowed with the weak
Bruhat order and the set of Dyck paths is studied.
\end{abstract}
\section{Introduction}\label{section1}
A wide range of articles dealing with lattices of combinatorial
objects appear frequently in the literature, e.g., lattices of
integer partitions, \cite{Brylawski, Stanley1}, permutations
\cite{GuilbaudRosenstiehl, Tamari} and noncrossing partitions
\cite{EdelmanSimion, SimionUllman}.
In this paper, the lattice of Dyck paths with the dominance partial order and some
of its relations to other combinatorial structures are studied.
In Section \ref{section2}, some basic definitions and notations referring to the
set $\mathcal{D}_n$ of Dyck paths of semilength $n$, are given.
In Section \ref{section3}, the rank of a Dyck path is studied and the
rank-generating function is exhibited.
In Section \ref{section4}, the notions of filling and degree of a
Dyck path are introduced and studied. Various enumeration results
are presented. These notions are used in Section \ref{section5} for
the evaluation of the M\"obius function and its powers. In this
context, a new appearance of the Fibonacci numbers (\seqnum{A000045}
of \cite{Sloane}) occurs.
Finally, in Section \ref{section6} the relation between the
symmetric group $S_n$ endowed with the weak Bruhat order and
$\mathcal{D}_n$ is studied. More precisely, a partition of $S_n$
into $C_n$ (\mbox{Catalan,} (\seqnum{A000108}) of \cite{Sloane})
classes is constructed, satisfying an ordering condition, and the
cardinal numbers of its members are evaluated.
\section{Preliminaries}\label{section2}
A \textit{Dyck path} of semilength $n$ is a path in the first quadrant, which
begins at the origin, ends at $(2n,0)$, and consists of steps $(1,1)$ and
$(1,-1)$, called \textit{rise} and \textit{fall} respectively.
It is clear that each Dyck path is coded by a word $u \in \{a, \bar{a}\}^*$,
called \textit{Dyck word}, so that every rise (resp. fall) corresponds to the
letter $a$ (resp. $\bar{a}$); see Fig. \ref{fig1}.
\begin{figure}[ht]
\centerline{\includegraphics[width=4in]{fig0.eps}}
\centerline{$a$ $a$ $\bar{a}$ $\bar{a}$ $a$ $a$ $a$ $a$ $a$ $a$ $\bar{a}$
$\bar{a}$ $a$ $\bar{a}$ $\bar{a}$ $\bar{a}$ $a$ $\bar{a}$ $\bar{a}$ $\bar{a}$}
\caption{A Dyck path and its corresponding word}\label{fig1}
\end{figure}
Throughout this paper we will denote with $D$ the set of all Dyck
paths (or equivalently Dyck words). Furthermore, the subset of $D$
that contains the paths $u$ of semilength $l(u) = n$ is denoted by
$D_n$.
We denote with $\epsilon$ the empty path. Every $u \in D \setminus \{\epsilon\}$
can be uniquely decomposed in the form $u = a w \bar{a} v$, $w, v \in D$, which
is called the \textit{first return decomposition} of $u$.
Every Dyck path $u$ that meets the $x$-axis only at its endpoints
(i.e., $u = a w \bar{a}$, with $w \in D$) is called \textit{prime}.
Every Dyck path can be uniquely decomposed into prime paths
\cite{PS}.
It is well known that Dyck paths are enumerated by the Catalan numbers, with
generating function $C(x)$, which satisfies the relation $xC^2(x) = C(x) - 1$.
\medskip
For a parameter $q$ defined on $D$, we will denote with $F_q$ the
generating function of $D$ according to the parameters $l$, $q$,
i.e.,
\[ F_q (x,y) = \sum\limits_{u \in D} x^{l(u)} y^{q(u)}. \]
A point of a Dyck path is called \textit{peak} (resp. \textit{valley}) if it
is preceded by a rise (resp. fall) and followed by a fall (resp. rise). A point
of a Dyck path is called \textit{double rise} (resp. \textit{double fall}) if it
is preceded and followed by a rise (resp. fall).
A convenient way to represent a Dyck word is by using dominating sequences.
A sequence $d = (d_i)_{i \in [n]}$ of non-negative integers is called
\textit{dominating} if it satisfies the following two conditions:
\begin{itemize}
\item[(i)] $\sum\limits_{i=1}^{n} d_i = n$,
\item[(ii)] $\sum\limits_{i=1}^{\nu} d_i \ge \nu$,
for every $\nu \in [n]$.
\end{itemize}
It is well known that every non-empty Dyck word $u$ is uniquely
represented by a dominating sequence $d(u) = (d_i)_{i \in l(u)}$
where $d_1$ is the number of $a$'s before the $1^{\rm st}$
occurrence of $\bar{a}$ in $u$, and $d_i$ is the number of $a$'s
between the $(i-1)^{\rm th}$ and the $i^{\rm th}$ occurrence of
$\bar{a}$ in $u$, where $i \in [2, l(u)]$. For example, the word $u =
a a \bar{a} \bar{a} a \bar{a} a a \bar{a} a \bar{a} \bar{a}$ is
represented by the sequence $d(u) = 2, 0, 1, 2, 1, 0$.
\medskip
A sequence $d = (d_i)_{i \in [n]}$ \textit{dominates} another sequence
$d' = (d'_i)_{i \in [n]}$ iff $\sum\limits_{i = 1}^{\nu} d_i \ge
\sum\limits_{i=1}^{\nu} d'_i$ for every $\nu \in [n]$. In this case we say
that $d'$ is \textit{dominated} by $d$. Notice that every dominating sequence
dominates the constant sequence with elements equal to 1.
\medskip
The \textit{dominance partial order ``$\preceq$''} on $D$ is defined as follows:
\smallskip
\centerline{$u \preceq w$ iff $l(u) = l(w)$ and the sequence $d(u)$ is dominated
by the sequence $d(w)$,}
\smallskip
\noindent i.e., the path $w$ lies above (in the broad sense) the
path $u$. If $u \preceq w$ and $u \ne w$, we will write $u \prec w$.
In the sequel, we will denote $(D, \preceq)$ (resp. $(D_n, \preceq)$) with
$\mathcal{D}$ (resp. $\mathcal{D}_n$).
Narayana and Fulton \cite{NarayanaFulton}, using an equivalent language, proved
that the set of Grand-Dyck paths of fixed length $n$ (i.e., paths
defined similarly to Dyck paths but allowed to go below the
$x$-axis, e.g., see \cite{Pergola}) endowed with the dominance order
is a distributive lattice. Kreweras \cite{Kreweras2, Kreweras1} has done
further work in this context. Ferrari and Pinzani \cite{FerrariPinzani}
have recently presented a more general approach to this subject.
It is easy to see that $\mathcal{D}_n$ is a sublattice of the
lattice of all Grand-Dyck paths of semilength $n$, such that if
$d(w) = (d'_i)_{i \in [n]}$, $d(v) = (d''_i)_{i \in [n]}$ we have:
\noindent $d(w \vee v) = (d_i)_{i \in [n]}$, where
\centerline{ $d_1 = \max \{d'_1, d''_1 \}$ and $d_i = \max \{ \sum\limits_{j=1}^{i} d'_j, \sum\limits_{j=1}^{i} d''_j \} -
\max \{ \sum\limits_{j=1}^{i-1} d'_j, \sum\limits_{j=1}^{i-1} d''_j\}$ for
every $i \in [2,n]$\phantom{.}}
\noindent and $d(w \wedge v) = (d_i)_{i \in [n]}$, where
\centerline{$d_1 = \min \{d'_1, d''_1 \}$ and $d_i = \min \{ \sum\limits_{j=1}^{i} d'_j, \sum\limits_{j=1}^{i} d''_j \} -
\min \{ \sum\limits_{j=1}^{i-1} d'_j, \sum\limits_{j=1}^{i-1} d''_j \}$ for every $i \in [2,n]$.}
\bigskip
We note that the least (resp. greatest) element of $\mathcal{D}_n$ is
$0_n = (a\bar{a})^n$ (resp. $1_n = a^n \bar{a}^n$). Finally, for every
$n \ge 2$, $1_n$ covers one and only one element of $\mathcal{D}_n$, namely
$a^{n-1} \bar{a} a \bar{a}^{n-1}$.
The Hasse diagram of $\mathcal{D}_4$ coded by dominating sequences is given
in Fig. \ref{fig2}.
\begin{figure}[ht]
\centerline{\includegraphics[height=3in]{fig1.eps}} \caption{The
Hasse diagram of $\mathcal{D}_4$}\label{fig2}
\end{figure}
All unexplained notations and definitions for posets and Dyck paths can be found
in \cite{Stanley1} and \cite{Deutsch}, respectively.
\section{Chains and Ranks}\label{section3}
In this section we first evaluate the length of maximal chains in intervals
of $\mathcal{D}_n$. For this, notice first that for $u, w \in \mathcal{D}_n$
with $d(u) = (d_i)_{i \in [n]}$ and $d(w) = (d'_i)_{i \in [n]}$, $w$ covers
$u$ iff there exists $j \in [n-1]$ such that $d'_i = d_i$ for every $i \ne
j,j+1$ and $d'_j = d_j+1$, $d'_{j+1} = d_{j+1}-1$.
From the above observation, we deduce that every path $w$ which covers
the path $u$ is obtained by turning a valley $(x,y)$ of $u$ into the peak
$(x,y+2)$.
\begin{prop}\label{prop1}
Let $u,w \in \mathcal{D}_n$ with $u \preceq w$ and $d(u) = (d_i)_{i \in [n]}$,
$d(w) = (d'_i)_{i \in [n]}$. Then, the length of every maximal chain $\mathcal{C}$
of $[u,w]$ is equal to
\[\sum\limits_{i=1}^{n} i (d_i - d'_i).\]
\end{prop}
\noindent \textit{Proof}. We will use induction with respect to the length $k$
of $\mathcal{C}$. If $k=1$ then $w$ covers $u$; so there exists $j \in [n-1]$
with $d_i = d'_i$ for every $i \in [n] \setminus \{j, j+1\}$ and $d'_j = d_j +
1$, $d'_{j+1} = d_{j+1} - 1$.
Hence
\begin{align*}\sum\limits_{i = 1}^{n} i (d_i - d'_i)
& = j (d_j - d'_j) + (j + 1) (d_{j+1} - d'_{j+1}) \\
& = j (-1) + (j + 1) 1 = 1.
\end{align*}
Since in this case the only maximal chain of $[u,w]$ is $\{u,w\}$ and has
length 1, the result holds when $k=1$.
Suppose now that the result holds for every maximal chain of $[u,w]$ of
length $k$, for every $u, w \in \mathcal{D}_n$ with $u \prec w$. We will
prove that the result also holds for any maximal chain $\mathcal{C}$ of
$[u,w]$ of length $k+1$.
If $v$ is the predecessor of $w$, then
$C \setminus \{w\}$ is a maximal chain of $[u, v]$ of length $k$; so
by the induction hypothesis we have
\[k = \sum\limits_{i=1}^n i (d_i - d''_i)\]
\noindent where $d(v) = (d''_i)_{i \in [n]}$.
Since $\sum\limits_{i=1}^n i (d''_i - d'_i) = 1$, we obtain that
$\sum\limits_{i=1}^n i (d_i - d'_i) = k+1$. \hfill{$\Box$}
\bigskip
If we apply the previous proposition for $u = 0_n$ and $w = 1_n$ we obtain that
every maximal chain in $\mathcal{D}_n$ has length equal to ${n (n-1)/2}$.
Thus, the lattice $\mathcal{D}_n$ is graded of rank ${n \choose 2}$.
In the sequel, we investigate the parameter ``rank'' of $\mathcal{D}$ defined as
follows : $\rho(\epsilon) = 0$ and for $u \ne \epsilon$, $\rho(u)$ is the rank
of $u$ in $\mathcal{D}_{l(u)}$.
By Proposition \ref{prop1}, it follows easily that if $u \ne \epsilon$ with
$d(u) = (d_i)_{i \in l(u)}$, then
\begin{equation}\label{eq0}
\rho(u) = {l(u) (l(u) +1)}/{2} - \sum\limits_{i=1}^{l(u)} i d_i.
\end{equation}
Ferrari and Pinzani \cite{FerrariPinzani} give an equivalent expression of
the above relation through the notion of the area of a Dyck path.
Using relation \eqref{eq0} (or alternatively its equivalent relation in
\cite{FerrariPinzani}), we can deduce that the parameter ``rank'' satisfies the
following properties:
\begin{itemize}
\item[(i)] $\rho(wv) = \rho(w) + \rho(v)$,
\item[(ii)] $\rho(aw\bar{a}) = \rho(w) + l(w)$,
for every $w, v \in \mathcal{D}$.
\end{itemize}
\medskip
Taking into account the above properties and the fact that every Dyck path is either empty or of the form
$a w \bar{a} v$, where $w, v \in D$, we can easily deduce the following relation.
\begin{equation}\label{relation2} F_{\rho} (x, y) = 1 + x F_{\rho} (xy, y) F_{\rho} (x,y). \end{equation}
Furthermore, if $f_n (y) = \sum\limits_{k=0}^{n \choose 2} a_{n,k} y^k$,
where $a_{n,k}$ denotes the number of elements of $\mathcal{D}_n$ of rank $k$,
then by relation~\eqref{relation2} we have
\begin{align*}
& \sum\limits_{n=0}^{\infty} f_n (y) x^n = 1 + x\left ( \sum\limits_{n=0}^{\infty} f_n
(y) y^n x^n \right ) \left (\sum\limits_{n=0}^{\infty} f_n (y) x^n \right )
\end{align*}
\noindent or, equivalently,
\begin{align*}
& \sum\limits_{n=0}^{\infty} f_{n+1}(y) x^n = \sum\limits_{n=0}^{\infty}
\left (\sum\limits_{\nu=0}^{n} f_{\nu} (y) f_{n - \nu}(y) y^{\nu} \right ) x^n.
\end{align*}
Hence we obtain the following result.
\begin{prop}
The rank-generating function of $\mathcal{D}_n$ is given by the following
recursive formula
\[f_{n+1} (y) = \sum\limits_{\nu=0}^n f_{\nu}(y) f_{n - \nu} (y) y^{\nu},\]
\noindent where $f_0 (y) = 1$.
\end{prop}
\section{Fillings and Degrees}\label{section4}
In this section we first introduce and study the notion of the filling of a
Dyck path.
The \textit{filling} $\widetilde{u}$ of $u \in \mathcal{D} \setminus \{ \epsilon
\}$ is defined to be the Dyck path obtained by turning each valley $(x,y)$ of
$u$ into the peak $(x,y+2)$. The filling of the empty path is assumed to be the
empty path.
For example, if $u = a a \bar{a} \bar{a} a a \bar{a} a \bar{a} \bar{a}$ then
$\widetilde{u} = a a \bar{a} a \bar{a} a a \bar{a} \bar{a} \bar{a}$.
The main properties of the filling are given in the following proposition, which
is easy to prove.
\begin{prop}
For a non-empty Dyck path $u$ we have : \\
\rm{i}) $l(\widetilde{u}) = l(u)$, $u \preceq \widetilde{u}$ and
$u = \widetilde{u}$ if $u = 1_{l(u)}$. \\
\rm{ii}) The length $l(u,\widetilde{u})$ of the interval $[u, \widetilde{u}]$ is
equal to the number $\rm{val}(u)$ of all valleys of $u$.\\
\rm{iii}) The cardinal number of the interval $[u,\widetilde{u}]$ is equal to
$2^{\rm{val}(u)}$. \\
\rm{iv}) If $u \preceq v$, then $\widetilde{u} \preceq \widetilde{v}$.
\end{prop}
We note that since the parameter $\rm{val}$ defined by the number of valleys
follows the Narayana distribution \cite{Deutsch}, we deduce that the number
$a_{n,k}$ of all $u \in \mathcal{D}_n$ such that the interval $[u, \widetilde{u}]$
contains exactly $k$ elements is equal to
\[a_{n,k} = \begin{cases} \frac{1}{n}{n \choose \lambda}{n \choose
\lambda - 1}, & \textrm{if $k = 2^{\lambda}$, $\lambda \in
\mathbb{N}^*$}; \\
0, & \textrm{if $k \ne 2^{\lambda}$, $\lambda \in \mathbb{N}^*$}.
\end{cases}\]
\medskip
Next, we consider the Dyck paths which are fillings of Dyck paths. For this, we
need the following characterization.
\begin{prop}
A Dyck path $u$ is a filling of a Dyck path iff $u$ is prime and avoids
$\bar{a}\bar{a} a a$.
\end{prop}
\noindent \textit{Proof}. If $u = \widetilde{v}$ for some $v \in \mathcal{D}$, then
clearly $u$ has no valleys on level zero and so $u$ is prime. Furthermore, if
$u$ contains a $\bar{a} \bar{a} a a$, then there exists a valley $(x,y)$ of $u$
such that the points $(x-1,y+1)$ and $(x+1,y+1)$ are a double fall and a double
rise of $u$, respectively. It follows that the points $(x,y)$, $(x-1,y+1)$ and
$(x+1,y+1)$ remain unchanged during the generation of $u$ from $v$, so that
$(x,y)$ is also a valley of $v$. This contradicts the definition of the filling,
since $(x,y+2)$ is not a peak of $u$.
Conversely, assume that $u$ is prime and avoids $\bar{a} \bar{a} aa$.
Let $v$ be the path that we obtain from $u$ by turning each peak $(x,y)$ of $u$
into the valley $(x,y-2)$. Clearly, since $u$ is prime it has no low peaks,
hence $v$ never crosses the $x$-axis and so $v \in \mathcal{D}$. In order to
show that $\widetilde{v} = u$, it is enough to show that every valley of $v$ is
generated by a peak of $u$ according to the above procedure. Indeed, if this
is not true and $(x,y)$ is a valley of $v$ that is not generated in such a way,
then $(x,y)$ must be a valley of $u$, too.
Since $u$ avoids $\bar{a} \bar{a} a a$, it follows that at least one of the
points $(x-1,y+1)$ and $(x+1,y+1)$ is a peak of $u$ and hence a valley of $v$,
which contradicts the fact that $(x,y)$ is a valley of $v$. \hfill{$\Box$}
\bigskip
We remark that the Dyck path $v$ constructed by turning each peak of a Dyck
path $u$ into a valley as in the converse of the above proof, is the least Dyck
path with the property $\widetilde{v} = u$. In this case, we say that $v$ is the
\textit{antifilling} of $u$.
In the following, we study the set $\mathcal{\widetilde{D}}$ of all Dyck paths
that are fillings of Dyck paths.
\begin{prop}
The generating function $F$ of $\mathcal{\widetilde{D}}$ according to
semilength satisfies the equation
\begin{equation}\label{eq1} F(x) = 1 + x F^2(x) - x^2 F(x).\end{equation}
\noindent Furthermore, the number $a_n$ of all Dyck paths of semilength $n$
that are fillings of Dyck paths is given by the formula
\begin{equation}\label{equation4} a_n = \sum\limits_{k=[n/2]}^n \frac{(-1)^{n-k}}{k} {k \choose n-k}{3k-n \choose k-1},
\textit{ for } n \ge 2
\end{equation}
\noindent whereas $a_0 = a_1 = 1$.
\end{prop}
\noindent \textit{Proof}. Let $\mathcal{A}$ be the set of all Dyck paths that
avoid $\bar{a} \bar{a} a a$ and $A(x)$ its generating function according to
semilength.
Clearly, by the previous proposition, every non-empty element of
$\mathcal{\widetilde{D}}$ can be written uniquely in the form $a w \bar{a}$
where $w \in \mathcal{A}$, so that $F(x) = 1 + xA(x)$.
Furthermore, we can easily check that every non-empty element $u \in \mathcal{A}$
can be written uniquely in one of the following forms: $u = a \bar{a} v$,
$u = a w \bar{a}$ or $u = a w \bar{a} a \bar{a} v$ where $w, v \in \mathcal{A}$
and $w \ne \epsilon$.
Thus, we have:
\begin{align*}
A(x) & = 1 + x A(x) + x (A(x)-1) + x^2 (A(x) -1)A(x).
\end{align*}
So
\[x^2 A^2(x) - (x-1)^2 A(x) + 1 - x = 0\]
\noindent and hence
\[ xF^2(x) - (1+x^2) F(x) + 1 = 0 \]
\noindent which gives formula \eqref{eq1}.
For the proof of formula \eqref{equation4} we consider the generating
function $F(x,y)$ satisfying the equation
\begin{equation}\label{equation5} F(x,y) = 1 + y (x F^2(x,y) - x^2 F(x,y)). \end{equation}
If we set $P(\lambda) = x \lambda^2 - x^2 \lambda$, then $F(x,y) = 1 + y P(F(x,y))$; so
applying the Lagrange inversion formula with respect to the variable $y$, in the form
given by Deutsch \cite{Deutsch}, we obtain that
\begin{equation}\label{equation6} [y^k] F = \frac{1}{k} [\lambda^{k-1}](P(1+\lambda))^k \end{equation}
\noindent for every $k \in \mathbb{N}^*$.
Furthermore, we have
\begin{align*}
(P(1+\lambda))^k & = x^k \sum\limits_{\nu=0}^k \sum\limits_{\rho=0}^k {k \choose \nu}{k \choose \rho} \lambda^{\nu} (\lambda
-x)^{\rho} \\
& = \sum\limits_{\nu=0}^k \sum\limits_{\rho=0}^k \sum\limits_{i=0}^{\rho} (-1)^{\rho-i} {k \choose \nu}
{k \choose \rho} {\rho \choose i} \lambda^{\nu +i} x^{k + \rho - i}.
\end{align*}
Then, for $i = k - \nu -1$, from relation \eqref{equation6} we obtain that
\[ [y^k] F = \frac{1}{k} \sum\limits_{\rho = 0}^k \sum\limits_{\nu=0}^{k-1} (-1)^{\rho-k+\nu+1} {k \choose \nu}{k \choose
\rho} {\rho \choose k - \nu -1} x^{\rho + \nu +1}. \]
Thus,
\begin{align*}
F(x,y) & = 1 + \sum\limits_{k=1}^{\infty} \sum\limits_{\rho=0}^{k} \sum\limits_{\nu =0}^{k-1} (-1)^{\rho-k+\nu+1}
\frac{1}{k} {k \choose \nu} {k \choose \rho} {\rho \choose k - \nu -1} x^{\rho+\nu +1} y^k \\
& = 1 + xy + \sum\limits_{n=2}^{\infty} \sum\limits_{k = [n / 2]}^n \sum\limits_{\nu =0}^{k-1} (-1)^{n-k} \frac{1}{k}
{k \choose \nu} {k \choose n - \nu - 1}{n - \nu - 1 \choose k - \nu -1} x^n y^k \\
& = 1 + xy + \sum\limits_{n=2}^{\infty} \sum\limits_{k = [n / 2]}^n (-1)^{n-k} \frac{1}{k} {k \choose n -k}{3k -n
\choose k-1} x^n y^k.
\end{align*}
Thus, since by relations \eqref{eq1} and \eqref{equation5} we have $F(x) = F(x,1)$, it follows that
\[ a_0 = a_1 = 1 \textrm{ and } a_n = \sum\limits_{k = [n/2]}^n (-1)^{n-k} \frac{1}{k} {k \choose n -k}{3k - n \choose k-1},
\textrm{ for $n \ge 2$. } \]
$\enspace$\\[-48pt]
\hfill{$\Box$}
\bigskip
By formula \eqref{equation4} by obtain the sequence
1,1,1,2,5,13,35,97,275,\ldots, which also counts the number of Dyck paths
that avoid $a a \bar{a} \bar{a}$
(\seqnum{A086581} of \cite{Sloane}).
This can also be seen by proving that the set of all antifillings coincides with
the set of all Dyck paths that avoid $a a \bar{a} \bar{a}$.
\medskip
The rest of this section deals with the notion of the degree of a Dyck path.
For this, we define the $i^{\rm th}$ filling $u^{(i)}$ of a non-empty Dyck path $u$
recursively, as follows:
\centerline{$u^{(0)} = u$ and $u^{(i)} = \widetilde{u^{(i-1)}}$, for $i \ge 1$.}
We define the \textit{degree} $\delta(u)$ of $u \in \mathcal{D} \setminus \{
\epsilon \}$ to be the least non-negative integer such that
$u^{(\delta(u))} = 1_{l(u)}$.
The degree of the empty path is assumed to be equal to zero.
For example, if $u = a a \bar{a} \bar{a} a a \bar{a} a \bar{a} \bar{a}$ then
$\delta(u) = 4$.
\medskip
The main properties of the degree are given in the following
result.
\begin{prop}\label{prop4.5}
For every non-empty Dyck path $u$ we have: \\
\rm{i}) $\delta(\widetilde{u}) = \delta(u) - 1$, for every $u \ne 1_{l(u)}$. \\
\rm{ii}) $\delta(au\bar{a}) = \delta(u)$. \\
\rm{iii}) $0 \le \delta(u) \le l(u) - 1$, and $\delta(u) = l(u) -1$ iff $u$ is non-prime. \\
\rm{iv}) If $u \prec v$, then $\delta(v) \le \delta(u)$.
\end{prop}
\noindent \textit{Proof}. i) is obvious, whereas ii) is based on the
equality $(a w \bar{a})^{(i)} = a w^{(i)} \bar{a}$, for every $i \in
\mathbb{N}$.
For the proof of iii) we first show by induction with respect to the semilength of $u$ that if
$u$ is non-prime then $\delta(u) = l(u) -1$.
Indeed, if $l(u) =2$, then $u = a \bar{a} a \bar{a}$ and $\delta(u) = 1 = l(u) - 1$.
Assume that the result holds for every non-prime Dyck path with semilength equal
to $n-1$, where $n \ge 3$, and let $u$ be a non-prime path of semilength $n$.
Then, using the prime decomposition of $u$, there exists a finite sequence
$(w_i)_{i \in [k]}$, $k \ge 2$, of Dyck words such that
\centerline{$u = a w_1 \bar{a} a w_2 \bar{a} a \cdots \bar{a} a w_{k-1} \bar{a} a w_k \bar{a}$.}
It follows easily that
\centerline{$\widetilde{u} = a \widetilde{w}_i a \bar{a}
\widetilde{w}_2 a \bar{a} \cdots a \bar{a} \widetilde{w}_{k-1} a \bar{a}
\widetilde{w}_k \bar{a}$.}
If we set
\centerline{$z = \widetilde{w}_i a \bar{a} \widetilde{w}_2 a \bar{a} \cdots a \bar{a}
\widetilde{w}_{k-1} a \bar{a} \widetilde{w}_k$,}
\noindent then $z \in \mathcal{D}$,
$l(z) = n-1$, $\widetilde{u} = a z \bar{a}$ and z is non-prime. Thus, by the
induction hypothesis we have $\delta(z) = l(z) - 1$ and so
\centerline{$\delta(u) = \delta(\widetilde{u}) + 1 = \delta(z) + 1 = l(z) =
l(u) -1$.}
Next, we note that $0 \le \delta(u)$ and $\delta(u) = 0$ iff $u =
1_{l(u)}$. Furthermore, if $u \ne 1_{l(u)}$, there exists $\nu \in
\mathbb{N}$ such that $u = a^{\nu} w \bar{a}^{\nu}$, where w is a
non-prime Dyck path.
Then, by ii) we deduce that
\centerline{$\delta(u) = \delta(w) = l(w) - 1 \le l(u) - 1$.}
It remains to check that $\delta(u) < l(u) -1$ for every prime Dyck path $u$.
Indeed, $u = aw\bar{a}$ with $w \in \mathcal{D}$, so that
$\delta(u) = \delta(w) \le l(w) - 1 = l(u) - 2$.
Finally, we show iv) using induction with respect to the semilength of $u$.
It is clear that the result is true when $l(u) = 1$. Assuming that the result
holds for Dyck paths of semilength $n-1$, where $n \ge 2$, we will show that
if $u, v \in \mathcal{D}_n$ with $u \prec v$ then $\delta(v) \le \delta(u)$.
Clearly, by iii), it is enough to restrict ourselves to the case where
$u$ is a prime word. Thus, if $u = a u' \bar{a}$ where $u' \in \mathcal{D}_{n-1}$,
it follows easily that $v = a v' \bar{a}$, where $v' \in \mathcal{D}_{n-1}$
and $u' \prec v'$. From the induction hypothesis it follows that $\delta(v) =
\delta(v') \le \delta(u') = \delta(u)$. \hfill{$\Box$}
\bigskip
We conclude this section with the following result.
\begin{prop}
The number of all $u \in \mathcal{D}_{n}$, $n \ge 2$, with degree equal to $k$,
where \mbox{$k \in [n - 1]$}, is equal to $C_{k+1} - C_k$.
\end{prop}
\noindent \textit{Proof}. Clearly, since every non-empty Dyck path $u$ can be
written uniquely in either of the forms $u = a w \bar{a}$ or
$u = a w \bar{a} v$, where $w, v \in \mathcal{D}$ and $v \ne \epsilon$, applying
Proposition \ref{prop4.5} we obtain that
\begin{align*}
F_{\delta}(x,y) & = 1 + x F_{\delta}(x,y) + x C(xy) (C(xy) - 1).
\end{align*}
It follows that
\begin{align*}
F_{\delta}(x,y)
& = \frac{1 + y^{-1} \sum\limits_{n=1}^{\infty} C_n (xy)^n - x
\sum\limits_{n=0}^{\infty} C_n (xy)^n}{1 - x} \\
& = \frac{1 + \sum\limits_{n=1}^{\infty} \left (C_n - C_{n-1} \right ) y^{n-1}
x^n}{1 - x} \\
& = \left (\sum\limits_{n=0}^{\infty} g_n (y) x^n \right ) \left (\sum\limits_{n=0}^{\infty}
x^n\right),
\end{align*}
\noindent where \[g_n(y) = \begin{cases}
\left(C_{n} - C_{n-1} \right ) y^{n-1}, & \textrm{if $n \ge 1$}; \\
1, & \textrm{if $n = 0$.}
\end{cases}\]
Therefore, we have
\begin{align*} F_{\delta}(x,y) & = \sum\limits_{n=0}^{\infty} \sum\limits_{k=0}^n
g_k(y) x^n \\
& = \sum\limits_{n=0}^{\infty} x^n + \sum\limits_{n=2}^{\infty}
\sum\limits_{k=1}^{n-1} (C_{k+1} - C_k) y^k x^n,
\end{align*}
\noindent which gives the required result. \hfill{$\Box$}
\section{The M\"obius function}\label{section5}
In this section we study the M\"obius function of $\mathcal{D}$ and its
powers. We recall \cite{Stanley1} that the M\"obius function $\mu$ of a poset $(P,
\preceq)$ is defined by
\[\mu(x,y) = - \sum\limits_{x \preceq z \prec y} \mu(x,z) \textrm{ for $x \prec
y$ and $\mu(x,x) = 1$}.\]
Furthermore, the $k$-th power of $\mu$, for $k \ge 2$, is defined by
\[\mu^k(x,y) = \sum\limits_{x=x_0 \preceq x_1 \preceq \cdots \preceq
x_k =y} \mu(x_0,x_1) \mu(x_1,x_2) \cdots \mu(x_{k-1},x_k).\]
It is known \cite{BarnabeiPezzoli} that if $(P, \preceq)$ is a locally finite distributive lattice,
then its M\"obius function is given by the formula
\[
\mu(x,y) = \begin{cases} (-1)^{\nu}, & \textrm{if $y$ is a join of $\nu$ elements covering $x$}; \\
0, & \textrm{if $y$ is not a join of elements covering $x$.}
\end{cases}
\]
For the lattice of Dyck paths, we have that $v \in (u, \widetilde{u}]$ with
$l(u,v) = \nu$ iff $v$ is obtained by turning $\nu$ of the valleys of $u$ into
peaks or, equivalently, iff $v$ is a join of $\nu$ elements of $D$ covering $u$.
Thus, from the above formula we obtain the following result.
\begin{prop}\label{prop5.1}
The M\"obius function of $\mathcal{D}$ is given by the formula
\[\mu(u,v) = \begin{cases}
(-1)^{l(u,v)}, & \textrm{if $u \preceq v \preceq \widetilde{u}$}; \\
0, & \textrm{otherwise}
\end{cases}\]
\noindent for every $u, v \in \mathcal{D}$, where $l(u,v)$
denotes the length of the interval $[u,v]$.
\end{prop}
In the sequel we study the powers of the M\"obius function of $\mathcal{D}_n$.
For this, we need the following result, which is an easy consequence of
Proposition \ref{prop5.1}.
\begin{cor}\label{lemma5.2}
For every multichain $u_0 \preceq u_1 \preceq \cdots \preceq u_k$ in
$\mathcal{D}_n$ with
\centerline{$\mu(u_0, u_1) \mu(u_1, u_2) \cdots \mu(u_{k-1},u_k) \ne 0$}
\noindent we have $u_i \preceq \widetilde{u_{i-1}}$ and
$u_i \preceq u_0^i$ for every $i \in [k]$.
\end{cor}
Clearly, if $u \in \mathcal{D}_n$ and $k < \delta(u)$, for every multichain
$u = u_0 \preceq u_1 \preceq \cdots \preceq u_k = 1_n$ we have that
$u^{(k)} \prec u_k$ so that, by the above Corollary, we deduce that
\centerline{$\mu(u_0, u_1) \mu(u_1, u_2) \cdots \mu(u_{k-1},u_k) = 0$.}
This shows that $\mu^{k}(u,1_n) = 0$ for every $k < \delta(u)$.
\medskip
In the following result we consider the case where $k = \delta(u)$.
\begin{prop}
Let $u \in \mathcal{D}_n$ and $j, \nu \in \mathbb{N}^*$ such that
$u^{(j)} = 0_{l(u)}^{\nu}$. Then, we have that
\[\mu^{\delta(u)}(u,1_{n}) = (-1)^{l(u^{(j)}, 1_{n})} \mu^j(u, u^{(j)}).\]
\end{prop}
\noindent \textit{Proof}.
Let $u = u_0 \preceq u_1 \preceq \cdots \preceq u_{\delta(u)} = 1_{n}$ be a
multichain of $\mathcal{D}_n$ with
\centerline{$\mu(u_0, u_1) \mu(u_1, u_2) \cdots \mu(u_{\delta(u)-1}, u_{\delta(u)}) \ne 0$;}
\noindent then, by Corollary \ref{lemma5.2}
it follows that $u_i \preceq \widetilde{u_{i-1}}$ and $u_i \preceq u^{(i)}$ for
every $i \in [\delta(u)]$. We show that $u_i = u^{(i)}$ for every $i \ge j$.
Indeed, if this is not true, let $\xi$ be the greatest element of $[j, \delta(u)]$
such that $u_{\xi} \prec u^{(\xi)}$.
Clearly, since $u_{\delta(u)} = 1_{l(u)} = u^{(\delta(u))}$, we have that
$\xi < \delta(u)$.
Since $u^{(\xi+1)} = u_{\xi+1} \preceq \widetilde{u_{\xi}} \preceq u^{(\xi+1)}$, we
obtain that $u^{(\xi+1)} = \widetilde{u_{\xi}}$.
Furthermore, since $u^{(\xi)} = 0_{n}^{(\nu+\xi-j)}$, $u^{(\xi+1)} =
0_{n}^{(\nu+\xi-j+1)}$ and the antifilling of $0^{(k+1)}_n$ is $0^{(k)}_n$ for every
$k \in [n-2]$, we obtain
that $u^{(\xi)}$ is the antifilling of $u^{(\xi+1)}$, though
$\widetilde{u_{\xi}} = u^{(\xi+1)}$ and $u_{\xi} \prec u^{(\xi)}$, which is a
contradiction.
Thus, $u_i = u^{(i)}$ for every $i \ge j$.
It follows that
\bigskip
\centerline{$
\mu^{\delta(u)}(u,1_{n}) = \sum \mu(u_0, u_1) \mu(u_1, u_2) \cdots
\mu(u_{j-1},\! u^{(j)}) \mu(u^{(j)},\! u^{(j+1)}) \mu(u^{(j+1)},\! u^{(j+2)}) \cdots
\mu(u^{(\delta(u)-1)},\! u^{(\delta(u))})
$}
\bigskip
\noindent where the sum is taken over all multichains $u = u_0 \preceq u_1
\preceq \cdots u_j = u^{(j)}$ of $\mathcal{D}_n$.
Since, by Proposition \ref{prop5.1}, we have
\begin{align*}
& \mu(u^{(j)}, u^{(j+1)}) \mu(u^{(j+1)}, u^{(j+2)}) \cdots \mu(u^{(\delta(u)-1)},
u^{(\delta(u))}) \\
& \qquad \qquad \qquad = (-1)^{l(u^{(j)}, u^{(j+1)})} (-1)^{l(u^{(j+1)}, u^{(j+2)})} \cdots
(-1)^{l(u^{(\delta(u)-1)}, u^{(\delta(u))})} \\
& \qquad \qquad \qquad = (-1)^{l(u^{(j)}, 1_{n})},
\end{align*}
\noindent we deduce that
\[\mu^{\delta(u)}(u,1_{n}) = (-1)^{l(u^{(j)}, 1_{n})} \mu^j(u, u^{(j)}).\]
$\enspace$\\[-48pt]
\hfill{$\Box$}
\bigskip
\noindent \textbf{Remark} Let $\mathcal{N}$ be the set of all
non-empty Dyck paths $u$ such that $\widetilde{u} =0^{\nu}_{l(u)}$
for some $\nu \in \mathbb{N}^*$. Then, taking $j =1$ in the previous
proposition, we obtain that
\begin{align*}
\mu^{\delta(u)}(u,1_{l(u)})
& = (-1)^{l(\widetilde{u}, 1_{l(u)})} \mu(u, \widetilde{u}) \\
& = (-1)^{l(\widetilde{u}, 1_{l(u)})} (-1)^{l(u, \widetilde{u})} \\
& = (-1)^{l(u, 1_{l(u)})}
\end{align*}
\noindent for every $u \in \mathcal{N}$.
In particular if $u =0_n$ for some $n \in \mathbb{N}^*$, we have
that $\delta(u) = n-1$ and
\[\mu^{n-1} (0_n, 1_n) = (-1)^{n \choose 2}.\]
Thus, by Lemma 4.1 in \cite{Edelman} we deduce that the zeta polynomial of
$\mathcal{D}_n$ satisfies the following formula
\[Z(\mathcal{D}_n, -k) = \begin{cases} (-1)^{{n \choose 2}}, & \textrm{if $k = n -1$}; \\
0, & \textrm{if $1 \le k < n -1$.} \end{cases} \]
\bigskip
We close this section by enumerating the sets $\mathcal{N} \cap \mathcal{D}_n$.
For this, we consider the set
\[\mathcal{N}_{\nu,n} = \{ u \in \mathcal{D}_n : \widetilde{u} = 0_n^{\nu} \}\]
\noindent where $\nu \in \mathbb{N}^*$ and $\nu \le n -1$.
Clearly, for every $p \in \mathbb{N}^*$, by considering the
bijection $u \to a^p u \bar{a}^p$ we can deduce that
\begin{equation}\label{eq2}|\mathcal{N}_{\nu,n}| = |\mathcal{N}_{\nu+p,n+p}|\end{equation}
\noindent for every $p \in \mathbb{N}^*$.
Furthermore, we have the following result.
\begin{prop}
For every $\nu \in \mathbb{N}^*$, the sequence
$(\mathcal{N}_{\nu,n})$, $n \ge \nu +1$ satisfies the following
relation
\[|\mathcal{N}_{\nu,n}| = F_{n-\nu+2},\]
\noindent where $(F_n)$ denotes the Fibonacci sequence.
\end{prop}
\noindent \textit{Proof}. In view of relation \eqref{eq2} it is enough to show
that
\[|\mathcal{N}_{1,n}| = F_{n+1}\]
\noindent for every $n \ge 2$.
Clearly, since $|\mathcal{N}_{1,2}| = 2$ and $|\mathcal{N}_{1,3}| = 3$, it is
enough to show that
\[|\mathcal{N}_{1,n}| = |\mathcal{N}_{1,n-1}| + |\mathcal{N}_{1,n-2}|\]
\noindent for every $n \ge 4$.
Every element of $\mathcal{N}_{1,n}$ is obtained by turning some peaks of
$\widetilde{0}_n$ into valleys. However, in this procedure for the generation of
the elements of $\mathcal{N}_{1,n}$ we must turn at least one of each pair of
consecutive peaks into valleys.
Thus, if $A_1$ (resp. $A_2$) is the set that consists of all elements of
$\mathcal{N}_{1,n}$ that pass from the point $(2,0)$ (resp. $(2,2)$), then
$\{A_1, A_2\}$ is a partition of $\mathcal{N}_{1,n}$.
Clearly, $A_1 = \mathcal{N}_{1,n-1}$ and since the elements of $A_2$ must pass
from the point $(4,0)$, we have $A_2 = \mathcal{N}_{1,n-2}$, which gives the
required result. \hfill{$\Box$}
\bigskip
We note that using the previous proposition, we obtain by a simple summation
that $|\mathcal{N} \cap \mathcal{D}_n| = F_{n+3} - 3$, (\seqnum{A006327} of
\cite{Sloane}).
\section{Dyck paths and permutations}\label{section6}
We recall that a simple reduction of a permutation
$\pi = \pi(1)\pi(2)\cdots\pi(n)$ is a permutation obtained from $\pi$ by
interchanging some $\pi(i)$ with $\pi(i+1)$, provided that $\pi(i) > \pi(i+1)$.
The weak Bruhat order $\ltimes$ is defined on the symmetric group $S_n$ as
follows:
\centerline{$\sigma \ltimes \pi$ iff $\sigma$ can be obtained from $\pi$ by a
sequence of simple reductions.}
The poset $(S_n, \ltimes)$ is a well known distributive lattice, graded of rank ${n \choose 2}$ and
has many interesting properties \cite{EdelmanGreene,Stanley3}. In the following, we examine the
connection between the lattices $(S_n, \ltimes)$ and $(D_n, \preceq)$.
For this, we first define the set $\mathcal{L}_n$ of all finite sequences $(A_i)_{i \in [n]}$ of
pairwise disjoint subsets of $[n]$ such that $[\nu] \subseteq \bigcup\limits_{i=1}^{\nu} A_i$
for every $\nu \in [n]$ and $\bigcup\limits_{i=1}^n A_i = [n]$.
For example,
$\mathcal{L}_3 = \{ \left(\{1\},\{2\},\{3\}\right), \left(\{1\},\{2,3\},\emptyset \right),
\left(\{1,2\},\emptyset,\{3\}\right)$,
$\enspace \qquad \qquad \qquad \qquad \!\!\!\quad \left(\{1,2\},\{3\},\emptyset\right), \left(\{1,3\}, \{2\}, \emptyset\right)$,
$\left(\{1,2,3\}, \emptyset, \emptyset\right)\}$.
We will show that the sets $S_n$ and $\mathcal{L}_n$ can be identified.
Indeed, for $\sigma \in S_n$ and $i \in [n]$ we define $A_i^{\sigma}$ to be
the set of all elements $j \in [n]$ for which there exist exactly $i-1$
elements of $\sigma$, which are less than $j$ and lie on the left of $j$ in
$\sigma$.
We can easily check that the sequence $(A_i^{\sigma})_{i \in [n]}$ belongs to
$\mathcal{L}_n$.
Conversely, if $(A_i)_{i \in [n]} \in \mathcal{L}_n$ then there exists unique
$\sigma \in S_n$ such that $A_i^{\sigma} = A_i$ for every $i \in [n]$.
For the construction of $\sigma$ we define recursively a finite sequence
$(\sigma_j)_{j \in [n]}$, such that $\sigma_j \in S_j$ and for $j > 1$ with
$j \in A_i$ where $i \in [n]$, $\sigma_j$ is generated from $\sigma_{j-1}$ by
inserting $j$ before the $i^{\rm th}$ element of $\sigma_{j-1}$ if $i < j$, or by
placing $j$ at the end of $\sigma_{j-1}$ if $i = j$. Then, for $\sigma =
\sigma_n$ we have $A_i^{\sigma} = A_i$ for each $i \in [n]$.
\medskip
From the above discussion we have the following result.
\begin{prop}\label{prop6.1}
The mapping $\sigma \to (A_i^{\sigma})_{i \in [n]}$ is a bijection between the
sets $S_n$ and $\mathcal{L}_n$.
\end{prop}
\noindent \textbf{Remark}. Using the above bijection we can characterize the set of
all permutations of $S_n(312)$ (i.e. the ones avoiding the pattern 312) as follows:
$\sigma \in S_n(132)$ iff for every $j, k \in [n]$ with $j \in A_i^{\sigma}$, $k \in
A_l^{\sigma}$ and $i < l$ we have that $j < k$.
Indeed, assume that $\sigma \in S_n(312)$ and $j, k \in [n]$ with $j \in
A_i^{\sigma}$, $k \in A_l^{\sigma}$, $i < l$ and $j > k$. Since $i < l$, we
have that $j$ lies on the left of $k$ and there exists some element $m$
which is less that $k$ and lies between them in $\sigma$. Then, the triplet
$jmk$ is an appearance of the pattern $312$ in $\sigma$, which is a contradiction.
Conversely, assume that the sequence $(A_i^{\sigma})_{i \in [n]}$ satisfies the above
condition though $\sigma$ contains the pattern $312$. Let $jmk$ be the first
appearance of the pattern $312$ in $\sigma$, with $j \in A_i^{\sigma}$ and $k
\in A_l^{\sigma}$. It follows that each element lying on the left of $j$ in
$\sigma$ that is less than $j$ is also less than $k$. Thus, $i < l$ although $j > k$,
which is a contradiction.
\smallskip
From the above remark it follows that if $\sigma \in S_n(312)$, then
every non-empty $A_i^{\sigma}$ consists of consecutive integers.
\bigskip
Next, we consider the family of sets $(\Gamma_u)_{u \in \mathcal{D}_n}$, where
\[ \Gamma_u = \{ \sigma \in S_n : d(u) = (|A_i^{\sigma}|)_{i \in [n]} \}.\]
\begin{prop}\label{prop6.2}
The family $(\Gamma_u)_{u \in \mathcal{D}_n}$ is a partition of $S_n$ such that
if $\sigma \in \Gamma_u$, $\pi \in \Gamma_w$ and $\pi$ covers $\sigma$, then $w$
covers $u$.
\end{prop}
\noindent \textit{Proof}.
We first show that $\Gamma_u \ne \emptyset$, for every $u \in \mathcal{D}_n$.
Indeed, if $d(u) = (d_i)_{i \in [n]}$, set $A_1 = [d_1]$ and for $i > 1$,
$A_i = \emptyset$ iff $d_i = 0$ and
$A_i = [\sum\limits_{j=1}^{i-1} d_j + 1, \sum\limits_{j=1}^i d_j]$ iff $d_i \ne
0$. It follows that the sequence $(A_i)_{i \in [n]}$ belongs to $\mathcal{L}_n$
and $|A_i| = d_i$ for every $i \in [n]$, so that by Proposition \ref{prop6.1}
there exists $\sigma \in S_n$ such that $d(u) = (|A_i^{\sigma}|)_{i \in [n]}$ and
$\sigma \in \Gamma_u$.
Next, if $\sigma \in S_n$, since the sequence $(A_i^{\sigma})_{i \in [n]}$
belongs to $\mathcal{L}_n$, it follows that $(|A_i^{\sigma}|)_{i \in [n]}$ is a
dominating sequence, so that there exists unique $u \in \mathcal{D}_n$ such that
$d(u) = (|A_i^{\sigma}|)_{i \in [n]}$ and hence $\sigma \in \Gamma_u$.
This shows that $\bigcup\limits_{u \in \mathcal{D}_n} \Gamma_u = S_n$, and
since the sets $\Gamma_{u}$ are pairwise disjoint, the family
$(\Gamma_u)_{u \in \mathcal{D}_n}$ is a partition of $S_n$.
Finally, since $\pi$ covers $\sigma$, there exists unique $k \in [n-1]$ such
that
\centerline{$\sigma(j) = \pi(j)$ for every $j \in [n] \setminus \{k,k+1\}$ and
$\sigma(k) = \pi(k+1) < \pi(k) = \sigma(k+1)$.}
Then, if $\pi(k) \in A_{\nu}^{\pi}$ we have that
\centerline{$\pi(k) \in A_{\nu+1}^{\sigma}$, $A_i^{\pi} = A_i^{\sigma}$ for every $i \in [n]
\setminus \{\nu,\nu+1\}$, $A_{\nu}^{\pi} = A_{\nu}^{\sigma} \cup \{\pi(k)\}$,
$A_{\nu+1}^{\pi} = A_{\nu+1}^{\sigma} \setminus \{\pi(k)\}$.}
Thus, since $d(u) = (|A_i^{\sigma}|)_{i \in [n]}$,
$d(w) = (|A_i^{\pi}|)_{i \in [n]}$, $|A_i^{\pi}| = |A_i^{\sigma}|$ for every
$i \in [n] \!\setminus\! \{\nu, \nu+1\}$, $|A_{\nu}^{\pi}| = |A_{\nu}^{\sigma}| + 1$,
and $|A_{\nu+1}^{\pi}| = |A_{\nu+1}^{\sigma}| -1$, we deduce that
$w$ covers $u$. \hfill{$\Box$}
\bigskip
\noindent \textbf{Remarks} \\
{\bf 1.} By Proposition \ref{prop6.2} it follows
easily that if $\sigma \in \Gamma_u$ and $\pi \in \Gamma_w$ with $\sigma \ltimes
\pi$, then $u \preceq w$. \\
{\bf 2.} Since $\Gamma_{0_n} = \{\hat{0}_n\}$ and $\Gamma_{1_n} = \{\hat{1}_n\}$
where $\hat{0}_n$ and $\hat{1}_n$ are the least and greatest element of
$S_n$ respectively, by Proposition \ref{prop6.2} it follows that for
$u \in \mathcal{D}_n$ we have
\centerline{$\rho(\sigma) = \rho(u)$ for every $\sigma \in \Gamma_u$,}
\noindent where $\rho$ denotes the rank function. \\
{\bf 3.} Following the first part of proof of Proposition \ref{prop6.2} we realize that the
sequence $(A_i^{\sigma})_{i \in [n]}$ constructed for every $u \in \mathcal{D}_n$ satisfies the
conditions of the previous remark and hence $\sigma \in S_n(312)$. Since the
number of permutations of $S_n(312)$ is equal to $C_n$, we deduce that
each $\Gamma_u$ contains exactly one element of $S_n(312)$. Thus, we
can define a bijection $f$ from $S_n(312)$ to $D_n$, such that $f(\sigma) = u$ iff $\sigma \in \Gamma_u$.
This bijection has been also presented in different ways in \cite{BK}, \cite{BBFP} and \cite{Fulmek}.
\medskip
In the following, we will evaluate the cardinal number of $\Gamma_u$ for a Dyck
path $u \in \mathcal{D}_n$ with $n \ge 2$. For this we need to consider the Dyck path $u' \in \mathcal{D}_{n-1}$
obtained from the path $u = a^{\nu} \bar{a} \tau \in \mathcal{D}_n$ if we delete its first peak,
i.e., the path $u' = a^{\nu-1} \tau$.
\begin{lemma}\label{lemma6.3}
For every $u \in \mathcal{D}_n$ with $d(u) = (d_i)_{i \in [n]}$ and
$d(u') = (d_i')_{i \in [n-1]}$, we have
\[ d'_1 = d_1 + d_2 -1, d'_i = d_{i+1} \textrm{ for every } i \in [2,n-1] \]
\noindent and
\begin{equation}\label{eq3} |\Gamma_u| = {d_1 + d_2 -1 \choose d_2}
|\Gamma_{u'}|.
\end{equation}
\end{lemma}
\noindent \textit{Proof}. Clearly, the proof of the first part of this lemma is evident.
So we restrict ourselves to the proof of relation \eqref{eq3}.
For $\sigma' \in \Gamma_{u'}$ and a subset $B \subseteq A_1^{\sigma'}$ with
$|B| =d_2$, we consider the finite sequence $(A_i)_{i \in [n]}$ defined as
follows :
$A_1 = \{1\} \cup \{x : x-1 \in A_1^{\sigma'} \setminus B\}$,
$A_2 = \{x : x -1 \in B\}$ and $A_i = \{x : x -1 \in A_{i-1}^{\sigma'}\}$
for $i \ge 3$.
Then, since $(A_i^{\sigma})_{i \in [n-1]} \in \mathcal{L}_{n-1}$, by
the previous construction it follows that $(A_i)_{i \in [n]} \in \mathcal{L}_n$,
so that by Proposition \ref{prop6.1} there exists unique $\sigma \in S_n$ such
that
$A_1^{\sigma} = \{1\} \cup \{x : x -1 \in A_1^{\sigma'} \setminus B \}$,
$A_2^{\sigma} = \{x : x - 1 \in B\}$ and
$A_i^{\sigma} = \{x : x -1 \in A_{i-1}^{\sigma'}\}$.
So $|A_1^{\sigma}| = 1 + (|A_1^{\sigma'}| \setminus |B|) = d_1$,
$|A_2^{\sigma}| = |B| = d_2$ and
$|A_i^{\sigma}| = |A_{i-1}^{\sigma'}| = d'_{i-1} = d_i$ for every $i \ge 3$.
Thus, $d(u) = (|A_i^{\sigma}|)_{i \in [n]}$ and $\sigma \in \Gamma_u$.
Moreover, we will show that every $\sigma \in \Gamma_u$ is generated by
a unique pair $(\sigma', B)$ as above.
Indeed, given $\sigma \in \Gamma_u$ we consider the sequence $(A'_i)_{i \in [n-1]}$
of sets in $[n-1]$ defined by
\[ A_1' = \{ x : x + 1 \in A_1^{\sigma} \cup A_2^{\sigma} \}, \enspace A_i' = \{ x : x+1 \in A_{i+1}^{\sigma} \}, \]
\noindent as well as the set
\[ B = \{ x : x + 1 \in A_2^{\sigma} \} \subseteq A_1'. \]
Then, $(A_i') \in \mathcal{L}_{n-1}$ and so by Proposition \ref{prop6.1} there exists a unique $\sigma' \in S_{n-1}$ such
that $A_i' = A_i^{\sigma'}$ for every $i \in [n-1]$. It follows that $\sigma' \in \Gamma_{u'}$ and hence $(\sigma',B)$ is
the required pair.
Thus, since $|A_1^{\sigma'}| = d_1 + d_2 - 1$ and $|B| = d_2$, we deduce that
each permutation \mbox{$\sigma' \in \Gamma_{u'}$} generates exactly
${d_1 + d_2 - 1 \choose d_2}$ permutations $\sigma \in \Gamma_u$ according to
the above procedure.
This shows that
\[ |\Gamma_u| = {d_1 + d_2 -1 \choose d_2} |\Gamma_{u'}|. \]
$\enspace$\\[-48pt]
\hfill{$\Box$}
\medskip
\begin{prop}
If $u \in \mathcal{D}_n$ with $d(u) = (d_i)_{i \in [n]}$, we have that
\[ |\Gamma_u| = \prod\limits_{j=1}^{n-1} \frac{\sum\limits_{i=1}^j d_i
- j + 1}{d_j !}. \]
\end{prop}
\noindent \textit{Proof}.
We consider the finite sequence $(u_j)_{j \in [n]}$ of Dyck paths, where
$u_1 = u$ and for $j > 1$ the Dyck path $u_j$ is obtained from $u_{j-1}$ by
deleting its first peak.
Clearly, $u_j \in \mathcal{D}_{n-j+1}$ for every $j \in [n]$; if
$d(u_j) = (d_i^{j})_{i \in [n-j+1]}$, by Lemma \ref{lemma6.3} we have that
\[ d_1^j = d_1^{j-1} + d_2^{j-1} - 1, d_i^j = d_{i+1}^{j-1} \]
\noindent and
\begin{equation}\label{eq4}
|\Gamma_{u_{j-1}}| = {d_1^{j-1} + d_2^{j-1} - 1 \choose d_2^{j-1}} | \Gamma_{u_j}|
\end{equation}
\noindent for every $j \in [2,n]$.
It is easy to check that $d_2^{j-1} = d_j$ and
$d_1^{j-1} + d_2^{j-1} - 1 = \sum\limits_{i=1}^j d_i - j +1$, for every
$j \in [2,n]$.
Furthermore, using the previous equalities and applying formula \eqref{eq4} for
every $j \in [2,n]$, we obtain that
\begin{align*}|\Gamma_u|
& = \prod\limits_{j=2}^n {\sum\limits_{i=1}^{j} d_i - j + 1 \choose d_j} |\Gamma_{u_n}| \\
& = \prod\limits_{j=1}^{n-1} \frac{\sum\limits_{i=1}^{j} d_i - j + 1}{d_j!}.
\end{align*}
$\enspace$\\[-48pt]
\hfill{$\Box$}
\bigskip
We note that since the family $(\Gamma_u)_{u \in \mathcal{D}_n}$ is
a partition of $S_n$, by the previous proposition we obtain an
identity for the factorial number, i.e.,
\[ \sum\limits_{d} \prod\limits_{j=1}^{n-1} \frac{\sum\limits_{i=1}^{j} d_i - j + 1}{d_j!} = n!\]
\noindent where the sum is taken over all dominating sequences $d = (d_i)_{i \in [n]}$.
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\bigskip
\hrule
\bigskip
\noindent 2000 {\it Mathematics Subject Classification}:
Primary 06A07; Secondary 06D99.
\noindent \emph{Keywords:} Dyck paths, lattices, dominating sequences.
\bigskip
\hrule
\bigskip
\noindent (Concerned with sequences \seqnum{A000045}, \seqnum{A000108},
\seqnum{A006327},
and \seqnum{A086581}.)
\bigskip
\hrule
\bigskip
\vspace*{+.1in}
\noindent
Received September 15 2005;
revised version received April 18 2006.
Published in {\it Journal of Integer Sequences},
May 19 2006.
\bigskip
\hrule
\bigskip
\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
\vskip .1in
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