%% if you are submitting an initial manuscript then you should have submission as an option here
%% if you are submitting a revised manuscript then you should have revision as an option here
%% otherwise options taken by the article class will be accepted
\documentclass[finalversion]{FPSAC2021}
\articlenumber{12}
%% but DO NOT pass any options (or change anything else anywhere) which alters page size / layout / font size etc

%% note that the class file already loads {amsmath, amsthm, amssymb}

\newtheorem{thm}{Theorem}[section]
\newtheorem{lem}{Lemma}[section]
\newtheorem{example}[thm]{Example}
\newtheorem{cor}{Corollary}[section]
\newtheorem{remark}[thm]{Remark}
%\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{conj}[thm]{Conjecture}


\usepackage{lipsum}


\usepackage{stmaryrd}
\usepackage{multirow}
\usepackage{tikz}
\usetikzlibrary{patterns}
\usetikzlibrary{matrix,arrows}
\usetikzlibrary{positioning}
\usetikzlibrary{fit}


\def\numset#1{{\mathbb #1}}
\def\blue{\textcolor{blue}}
\def\red{\textcolor{red}}
%\theoremstyle{plain}

\providecommand{\qbinom}[2]{\binom{#1}{#2}_{\!\!q}}


\usetikzlibrary{shapes}

%predefined: red, green, blue, cyan , magenta, yellow, black, gray, darkgray, lightgray, brown, lime, olive, orange, pink, purple, teal, violet and white

\tikzstyle{pathdefault}=[draw, line width=1, solid, color=black]
\tikzstyle{nodedefault}=[circle, inner sep=1.5, fill=black]
\tikzstyle{empty}=[]
\tikzstyle{nodeellipsis}=[circle, inner sep=0.5, fill=black]
\tikzstyle{pathcolor1}=[draw, line width=1.3, densely dashed, color=red]
\tikzstyle{pathcolor2}=[draw, line width=1.6, densely dotted, color=blue]
\tikzstyle{pathcolorlight}=[draw, line width=1, dotted, color=lightgray]

\tikzstyle{arbpathcolor0}=[line width=1, dashdotted, color=black]
\tikzstyle{arbpathcolor1}=[line width=1, densely dashed, color=red]
\tikzstyle{arbpathdefault}=[line width=1, densely dotted, color=blue]





\newcounter{id}
\newcommand{\drawlinedotswithstyle}[4]{
 \def\x{{#3}}
 \def\y{{#4}}
 \tikzstyle{thispathstyle}=[#1]
 \tikzstyle{thisnodestyle}=[#2]
 \setcounter{id}{-1} %start id at -1
 \foreach \j in {#3}{\stepcounter{id}} %id is one less than num of pts
 \foreach \i in {1,...,\the\value{id}}{  %loop through later indices
  \path[thispathstyle] (\x[\i],\y[\i]) --(\x[\i-1],\y[\i-1]); %draw edge
 }
 \foreach \i in {1,...,\the\value{id}}{  %loop through later indices
  \node[thisnodestyle] at (\x[\i],\y[\i]) {}; %draw node
 }
 \node[thisnodestyle] at (\x[0],\y[0]) {}; %draw first node outside of loop
}

\newcommand{\drawlinedotsloop}[4]{
 \foreach \i in {#1}{\stepcounter}
}

\definecolor{mhcblue}{HTML}{0077CC} %official Mount Holyoke color
\definecolor{davidsonred}{HTML}{AC1A2F} %official Davidson College color

\definecolor{green}{RGB}{0, 180, 0}
\definecolor{yellow}{RGB}{180, 180, 0}


\newcommand{\dbrac}[1]{{\llbracket#1\rrbracket}} 	% double brackets
\newcommand{\sep}{{\colon}}                         % Separator
\newcommand{\boks}[2]{({#1, #2})}   % A box with the old
% notation

%\newcommand{\dbrac}[1]{{\llbracket#1\rrbracket}} 	% double brackets
%\newcommand{\sep}{{\colon}}                         % Separator
%\newcommand{\boks}[2]{({#1, #2})}   % A box with the old
% notation
%\newcommand{\id}{\mathrm{id}}

\newcommand{\id}{\mathrm{id}}

\DeclareMathOperator{\des}{des}

\newcommand{\pattern}[4]{										% mesh pattern
	\raisebox{0.6ex}{
		\begin{tikzpicture}[scale=0.35, baseline=(current bounding box.center), #1]
		\foreach \x/\y in {#4}		\fill[pattern=north east lines] (\x,\y) rectangle +(1,1);
%		\foreach \x/\y in {#4}		\fill[pattern=north east lines] (\x,\y) rectangle +(1,1);
		\draw (0.01,0.01) grid (#2+0.99,#2+0.99);
		\foreach \x/\y in {#3}		\filldraw (\x,\y) circle (6pt);
		\end{tikzpicture}}
}




%% set
\def\N{\mathbb{N}}
\def\Z{\mathbb{Z}}
\def\S{\mathfrak{S}}
\def\D{\mathfrak{D}}
\def\L{\mathfrak{L}}
\def\A{\mathfrak{A}}
\def\M{\mathcal{M}}
\def\O{\mathcal{O}}
\def\R{\mathbb{R}}
\def\B{\mathcal{B}}
\def\I{\operatorname{\bf I}}

%% stats
\def\dist{\operatorname{dist}}
\def\ides{\operatorname{ides}}
\def\exc{\operatorname{exc}}
\def\wex{\textsf{wex}}
\def\dd{\textsf{dd}}
\def\des{\textsf{des}}
\def\asc{\textsf{asc}}
\def\ndes{\textsf{ndes}}
\def\inv{\mathrm{inv}}
\def\fix{\mathrm{fix}}
\def\fmax{\textsf{fmax}}
\def\cros{\textsf{cros}}
\def\nest{\textsf{nest}}
\def\icr{\textsf{icr}}
\def\ine{\textsf{ine}}
\def\st{\textrm{st}}
\def\MAD{\textsf{MAD}}
\def\ca-b{31\textsf{-}2}
\def\b-ca{2\textsf{-}31}
\def\stat{\textrm{stat}}
\def\val{\textrm{val}}
\def\pos{\textrm{pos}}
\def\Dyck{\text{Dyck}}
\def\rank{\mathsf{rank}}
\def\Av{\mathrm{Av}}
%%%%%%%
\def\ai{\mathrm{ai}}
\def\da{\mathrm{da}}
\def\dd{\mathrm{dd}}
\def\icr{\mathrm{icr}}
\def\ine{\mathrm{ine}}
\def\wex{\mathrm{wex}}
\def\MAD{\mathrm{MAD}}
\def\valley{\mathrm{val}}
\def\nest{\mathrm{nest}}
\def\cros{\mathrm{cros}}
\def\drop{\mathrm{drop}}
\def\cvalley{\mathrm{cval}}
\def\cval{\mathrm{cval}}
\def\cpeak{\mathrm{cpk}}
\def\peak{\mathrm{pk}}
\def\lpeak{\mathrm{lpk}}
\def\lvalley{\mathrm{lval}}
\def\ldd{\mathrm{ldd}}
\def\lda{\mathrm{lda}}
\def\cda{\mathrm{cda}}
\def\cdd{\mathrm{cdd}}
\def\cdes{\mathrm{cdes}}
\def\fdes{\mathrm{fdes}}
\def\fexc{\mathrm{fexc}}
\def\wexa{\mathrm{wex_A}}
\def\exca{\mathrm{exc_A}}
\def\wexc{\mathrm{wex_C}}
\def\neg{\mathrm{neg}}
\def\cyc{\mathrm{cyc}}
\def\Orb{\mathrm{Orb}}
\def\fmax{\mathrm{fmax}}
\def\amax{\mathrm{amax}}
\def\area{\mathrm{area}}
\def\cab{31\text{-}2}
\def\bca{2\text{-}31}
\def\acb{13\text{-}2}
\def\bac{2\text{-}13}
\def\SDE{\mathfrak D}
\def\SN{\mathfrak N}
\def\SB{{\mathfrak B}}
\def\SQ{{\mathfrak Q}}
\def\SD{{\mathfrak O}}
\def\SG{{\mathfrak S}}               %% Symmetric group
\def\SP{{\mathfrak P}} 
\def\r{\mathrm{r}}
\def\c{\mathrm{c}}
\def\Nr{\mathrm{Nr}}
             %% Seg permutations
\def\des{\operatorname{des}}         %% Number of descents
\def\seg{\operatorname{seg}}         %% Number of bars
\def\SCDes{\operatorname{SDes}}      %% Seg Composition of descents
\def\Des{\operatorname{Des}}         %% descent set
\def\Bar{\operatorname{Seg}}         %% Bar set of I
\def\segn{|\!|\!\!=}                 %% segmented compo of ..
\def\std{\operatorname{std}}         %% standardization of a word


\def\speak{\mathrm{spk}}
\def\svalley{\mathrm{sval}}
\def\sda{\mathrm{sda}}
\def\sdd{\mathrm{sdd}}

\def\scpeak{\mathrm{scpk}}
\def\scvalley{\mathrm{scval}}
\def\scda{\mathrm{scda}}
\def\scdd{\mathrm{scdd}}
\def\erec{\mathrm{erec}}
\def\rec{\mathrm{rec}}
\def\arec{\mathrm{arec}}
\def\earec{\mathrm{earec}}
\def\rar{\mathrm{rar}}
\def\nrar{\mathrm{nrar}}
\def\AREC{\mathrm{AREC}}







%% define your title in the usual way
\title[Equidistributions of mesh patterns of length two]{Equidistributions of mesh patterns of length two}

%% define your authors in the usual way
%% use \addressmark{1}, \addressmark{2} etc for the institutions, and use \thanks{} for contact details
\author{Bin Han\thanks{\href{mailto:han@math.biu.ac.il, han.combin@hotmail.com}{han@math.biu.ac.il, han.combin@hotmail.com}. 
 Supported by the Israel Science Foundation, grant no. 1970/18.
 }\addressmark{1} 
%Jianxi Mao \thanks{\href{mailto:maojianxi@hotmail.com}{maojianxi@hotmail.com}. 
 %Supported by the China Scholarship Council, part of the present work was done during his visit at Institut Camille Jordan, Universit\'e Claude Bernard Lyon 1 in 2018-2019.
%}\addressmark{2}
\and Jiang Zeng\thanks{\href{mailto:zeng@math.univ-lyon1.fr}{zeng@math.univ-lyon1.fr}.}\addressmark{2}}

%% then use \addressmark to match authors to institutions here
\address{
\addressmark{1}Department of Mathematics, Bar-Ilan University, Ramat-Gan 52900, Israel. \\
%\addressmark{2}School of Mathematic Sciences, Dalian University of Technology, Dalian 116024, P. R. China\\
\addressmark{2}Univ Lyon, Universit\'e Claude Bernard Lyon 1, CNRS UMR 5208, Institut Camille Jordan, 43 blvd. du 11 novembre 1918, F-69622 Villeurbanne cedex, France}





%\author[Optional shorter names]{Longer names me\thanks{\href{mailto:hello@world.c}{hello@world.c}. Longer names me was partially supported by Grant 2017.11.14.$\partial$\;supp.}\addressmark{1}, \and you\addressmark{2}}

%% then use \addressmark to match authors to institutions here
%\address{\addressmark{1}Department of This, University of That, There \\ \addressmark{2}School of Something, University of Something Else, Somewhere}

%% put the date of submission here
\received{\today}

%% leave this blank until submitting a revised version
%\revised{}

%% put your English abstract here, or comment this out if you don't have one yet
%% please don't use custom commands in your abstract / resume, as these will be displayed online
%% likewise for citations -- please don't use \cite, and instead write out your citation as something like (author year)
\abstract{A systematic study of \emph{avoidance} of mesh patterns of length 2 was conducted by Hilmarsson \emph{et al.} in 2015. In a recent paper
Kitaev and Zhang  examined the  distribution of 
the aforementioned 
patterns.    The aim of this paper is to  prove more equidistributions  of mesh pattern  and            confirm Kitaev and Zhang's four conjectures  by constructing two involutions on permutations.
}

%% put your French abstract here, or comment this out if you don't have one
%\resume{\lipsum[2]}

%% put your keywords here, or comment this out if you don't have them yet
\keywords{permutation, mesh pattern, distribution, involution, continued fraction, antirecord}

%% you can include your bibliography however you want, but using an external .bib file is STRONGLY RECOMMENDED and will make the editor's life much easier
%% regardless of how you do it, please use numerical citations, ie. [xx, yy] in the text

%% this sample uses biblatex, which (among other things) takes care of URLs in a more flexible way than bibtex
%% but you can use bibtex if you want
%\usepackage[backend=bibtex, maxnames=5]{biblatex}
\addbibresource{12Han.bib}
%% note the \printbibliography command at the end of the file which goes with these biblatex commands

\begin{document}

\maketitle
%% note that you DO NOT have to put your abstract here -- it is generated by \maketitle and the \abstract and \resume commands above

\section{Introduction}
Patterns in permutations and words have implicitly appeared in the mathematics literature for over a century, but interest in them has blown up  in the past four  decades (see~\cite{BrCl, HMZ20aam, HMZ20, Hilmarsson2015Wilf, Kit, KZ19, Stein} and references therein),  and the research of this area continues to increase gradually. 


A permutation $\sigma=\sigma(1)\cdots \sigma(n)$ of length $n$ is an arrangement
  of $1 \cdots n$. If $\pi$  and $\sigma$ are two permutations represented in this way, then $\pi$ 
is said to contain $\sigma$ as a pattern if some subsequence of the digits of
 $\pi$  has the same relative order as all of the digits of $\sigma$. For example, the permutation $31542$ contains two occurrences of the pattern $231$ as the two subwords $352$ and $342$ all have the same  ordering  as $231$.
  Let $S_n$ be the set of all permutations of length $n$. 
We call $\sigma(i)$ the value of $\sigma$ at position $i$ ($1\leq i\leq n$) 
and draw  a graphical presentation of $\sigma$ by putting
 a dot at $(i, \sigma(i)$  for $i=1, \ldots, n$ in the plan. 
For example, the permutation $231\in S_3$ is presented as follows, 
\[
\pattern{scale=1}{3}{1/2,2/3,3/1}{}
\]
where the horizontal lines represent the values and the vertical lines denote the positions in the permutation.
%A (classical permutation) pattern is a permutation $\tau\in S_n$.  We could draw the pattern 
In \cite{BrCl} Br\"and\'en and Claesson 
introduced and studied \emph{mesh patterns} as a common extension of several types of generalized permutation patterns.



%Let $S_n$ be the set of all permutations of length $n$. A (classical permutation) pattern is a permutation $\tau\in S_n$.  We could draw the pattern $231\in S_3$ as follows, where the horizontal lines represent the values and the vertical lines denote the positions in the pattern.
%\[
%\pattern{scale=1}{3}{1/2,2/3,3/1}{}
%\]
%To study the explicit expansions for certain permutation statistics as, possibly infinite, linear combinations of (classical) permutation patterns, Br\"and\'en and Claesson \cite{BrCl}  first introduced the notion of a \emph{mesh pattern}, which generalize several classes of patterns.



A
\emph{mesh pattern} of length $k$ is a pair $(\tau,R)$, where $\tau$ is a permutation of length $k$ and $R$ is a subset of $\dbrac{0,k} \times \dbrac{0,k}$
with  $\dbrac{0,k}=\{0,1, \ldots, k\}$.
Let $\boks{i}{j}$ denote the box whose corners have coordinates $(i,j), (i,j+1),
(i+1,j+1)$, and $(i+1,j)$. 
%Let the horizontal lines represent the values,  and the vertical lines denote the positions in the pattern. 
Mesh patterns can be depicted by shading the boxes in $R$. A mesh pattern with $\tau=231$ and $R = \{\boks{1}{2},\boks{2}{1}\}$ is drawn as follows.
\[
\pattern{scale=1}{3}{1/2,2/3,3/1}{1/2, 2/1}
\]
For example, the permutation $346512$ depicted in the following picture contains the mesh pattern $(231, \{\boks{1}{2},\boks{2}{1}\})$ since the subsequence $462$ forms the classical pattern $231$ and there are no points in the shaded
areas.
\[
346512 =
 \begin{tikzpicture}[scale=0.7, baseline=(current bounding box.center)]
    \draw (0.01,0.01) grid (6+0.99,6+0.99);
    \fill[pattern=north east lines] (2,4) rectangle +(1,2);
  \fill[pattern=north east lines] (3,2) rectangle +(3,2);
  \filldraw (1,3) circle (6pt);
  \filldraw (2,4) circle (6pt);
  \filldraw (3,6) circle (6pt);
  \filldraw (4,5) circle (6pt);
  \filldraw (5,1) circle (6pt);
  \filldraw (6,2) circle (6pt);
  % draw three circles around the points
  \draw (2,4) circle (12pt);
  \draw (3,6) circle (12pt);
  \draw (6,2) circle (12pt);
 \end{tikzpicture}
\]
%The permutation has an occurrence of the mesh pattern






Mesh patterns and their generalizations were studied in many papers; e.g.\ see \cite{AKV,BK21, Borie,HMZ21, HZ21, Hilmarsson2015Wilf,KL,KRT,KZ19,T2}. 
In the first systematic study of the mesh patterns avoidance, Hilmarsson et al.~\cite{Hilmarsson2015Wilf}  solved 25 out of 65 non-equivalent \emph{avoidance} cases of patterns of length 2.
%were solved. In other words, they gave the number of permutations avoiding the respective mesh patterns in the 25 cases. 
In a recent paper~\cite{KZ19}, Kitaev and Zhang further studied the distributions of mesh patterns considered in \cite{Hilmarsson2015Wilf} by giving 27 distribution results,  
%including 14 distributions for which avoidance was not known, 
see \cite[Table~1]{KZ19}.
%That is, they gave the distributions of mesh patterns in permutations in the 27 out of $65$ cases.
Moreover, for the unsolved case, they gave an equidistribution result and  conjectured 6 more equidistributions (see Table~\ref{tab-2}). We prove 3 conjectured equidistributions and 2 more equidistribuions (see Table~\ref{tab-3}) by constructing two involutions. This extended abstract is a summary of the recent paper \cite{HZ21}.
% 
%We note that from the distribution point of view, we  cannot consider just the 65 patterns presented in \cite{Hilmarsson2015Wilf}, since there are more patterns to consider. For example, the pattern Nr.\ 39 =  $\pattern{scale=0.6}{2}{1/1,2/2}{0/1,1/0}$ 
% was considered there, while its Wilf-equivalent pattern (by the \emph{Shading Lemma} in \cite{Hilmarsson2015Wilf}) 
% $\pattern{scale=0.6}{2}{1/1,2/2}{0/1,1/0,1/1}$ 
% was not considered. However, these two patterns have different distributions. Two patterns, $p_1$ and $p_2$, are said to be \emph{Wilf-equivalent} if for any $n\geq 0$, the number of permutations of length $n$ avoiding $p_1$ is equal to that avoiding $p_2$. 
%  

%
% Table~\ref{tab-1} overviews the  enumerative results (27 patterns) of Kitaev
% and Zhang. 
% In particular,  Nr.\ 3 = $\pattern{scale = 0.6}{2}{1/1,2/2}{0/0,0/1,1/2}$ is a conjectured distribution, Nr.\ 1 = $\pattern{scale = 0.6}{2}{1/1,2/2}{}$ is the well known distribution of non-inversions (same as the distribution of inversions) \cite[p. 21]{Stanley} given over permutations of length $n$ by
% \begin{equation}
% \label{inv-distr}
% (1+q)(1+q+q^2)\cdots(1+q+q^2+\cdots +q^{n-1}).
% \end{equation}
% 
% 
%  \begin{table}[t]
% 	
% 		\renewcommand{\arraystretch}{1.3}
% \begin{center} 
% 		\begin{tabular}{|c|c|c||c|c|c|}
% 			\hline
% 			{Nr.\ } & {Repr.\ $p$}  & {Distribution}  &  {Nr.\ } & {Repr.\ $p$}  & {Distribution}  
% 			\\[5pt]
% 			\hline		\hline
% 			1 & $\pattern{scale = 0.6}{2}{1/1,2/2}{}$ & \multirow{1}{2.5cm}{%\tiny{Non-inversions given by \eqref{inv-distr}; 
%			\cite[p. 21]{Stanley}}
% 		    %}
%		    
%			%{Non-inversions}
% 			&
% 			20 & $\pattern{scale = 0.6}{2}{1/1,2/2}{0/0,0/1,0/2,1/1,1/2,2/0,2/1}$ &  \cite[Theorem~2.8]{KZ19}
% 			\\[5pt]
% 			\hline
% 			3 & $\pattern{scale = 0.6}{2}{1/1,2/2}{0/0,0/1,1/2}$ & 
%		\cite[Conjecture~6.1]{KZ19}/Corollary~1.6
% 			& 
% 			21 & $\pattern{scale = 0.6}{2}{1/1,2/2}{0/1,1/2,0/0,2/0,2/2}$ &  \cite[Theorem~2.9]{KZ19}
% 			\\[5pt]
% 			\hline
% 			5 & $\pattern{scale = 0.6}{2}{1/1,2/2}{0/0,0/1,0/2}$ &  \cite[Theorem~2.1]{KZ19}
% 			&
% 			22 & $\pattern{scale = 0.6}{2}{1/1,2/2}{0/1,1/2,0/0,2/0,2/2,1/1}$ &  \cite[Theorem~2.10]{KZ19}
% 			\\[5pt] 
% 			\hline
% 		    8 & $\pattern{scale = 0.6}{2}{1/1,2/2}{0/0,0/1,1/0,1/1}$ &  \multirow{3}{2.5cm}{\cite[Theorem~4.1]{KZ19}} %\tiny{Unsigned Stirling numbers of the first kind, \cite[A132393]{OEIS}}
% 		    & 
% 			27 & $\pattern{scale=0.6}{2}{1/1,2/2}{0/1,2/0,2/2,1/0,1/1,0/2}$    &  \cite[Theorem~3.3]{KZ19}
% 		    \\[5pt] 
% 		    \cline{4-6}
% 			9 & $\pattern{scale = 0.6}{2}{1/1,2/2}{0/1,1/1,1/2,2/1}$ &
% 			% 			 Theorem~\ref{dis-patterns-8-9}
%%\tablefootnote{Unsigned Stirling numbers of the first kind, A132393.}
% 			&
%			28 & $\pattern{scale=0.6}{2}{1/1,2/2}{0/1,1/2,0/0,2/2,1/0,2/1}$    &   \cite[Theorem~3.4]{KZ19}
% 			\\[5pt]
% 			\hline
% 			10 & $\pattern{scale = 0.6}{2}{1/1,2/2}{0/0,0/1,0/2,2/0,2/1,2/2}$ &  \cite[Theorem~2.2]{KZ19}
% 			&
% 			30 & $\pattern{scale=0.6}{2}{1/1,2/2}{0/1,1/2,2/0,1/0,1/1,2/1,0/2}$ &  \cite[Theorem~3.5]{KZ19}
% 			\\[5pt]
% 			\hline
% 			11 & $\pattern{scale = 0.6}{2}{1/1,2/2}{0/0,0/1,0/2,1/0,1/1,1/2,2/0,2/1,2/2 }$ &  \cite[Theorem~2.3]{KZ19}
% 			&
% 			33 & $\pattern{scale=0.6}{2}{1/1,2/2}{0/1,1/2,2/0,1/0,0/2,2/1}$   &  \cite[Theorem~3.6]{KZ19} 			\\[5pt]
% 			\hline
% 			12 & $\pattern{scale = 0.6}{2}{1/1,2/2}{0/0,0/1,0/2,1/0,2/0}$ &  \cite[Theorem~2.4]{KZ19} 			&
% 			34 & $\pattern{scale=0.6}{2}{1/1,2/2}{0/1,1/2,0/0,2/2,1/0,1/1,2/1}$ &  \cite[Theorem~3.7]{KZ19}		
%			\\[5pt]
% 			\hline
% 			13 & $\pattern{scale = 0.6}{2}{1/1,2/2}{0/0,0/1,0/2,1/0,1/2,2/0,2/1,2/2}$ &   \cite[Theorem~2.5]{KZ19}
% 			&
% 			36 & $\pattern{scale=0.6}{2}{1/1,2/2}{0/1,1/2,0/0,1/0,1/1,2/1}$   &   \cite[Theorem~4.3]{KZ19}
%			
% 			\\[5pt]
% 			\hline
% 			14 & $\pattern{scale = 0.6}{2}{1/1,2/2}{0/1,1/1,1/2,1/0,1/2,2/1}$ & 
%			 \multirow{3}{2.5cm}{ \cite[Theorem~4.2]{KZ19} } %\footnotesize{small descents, \cite[A123513]{OEIS}}}
% 			&
% 			45 & $\pattern{scale=0.6}{2}{1/1,2/2}{0/1,1/2,1/0,1/1,2/1,0/2}$     &   \cite[Theorem~4.4]{KZ19}
% 			\\[5pt]
% 			\cline{4-6}
% 			15 & $\pattern{scale = 0.6}{2}{1/1,2/2}{0/1,0/2,1/0,1/1,1/2}$ &  
% 			&
% 			55 & $\pattern{scale = 0.6}{2}{1/1,2/2}{0/1,1/2,0/0,2/0,1/1,2/1}$ &  \cite[Theorem~3.8]{KZ19}
% 			\\[5pt]
% 			\hline
% 			16 & $\pattern{scale = 0.6}{2}{1/1,2/2}{0/1,2/0,1/0,0/2}$ &  \cite[Theorem~3.1]{KZ19}
% 			&
% 			56 & $\pattern{scale = 0.6}{2}{1/1,2/2}{0/1,1/2,0/0,2/2,1/1,2/1}$ &  \cite[Theorem~3.9]{KZ19}
% 			\\[5pt]
%			\hline
% 			17 & $\pattern{scale = 0.6}{2}{1/1,2/2}{0/1,1/2,0/0,2/0,1/0,0/2,2/1}$ &  \cite[Theorem~3.2]{KZ19}
% 			&
% 			63 & $\pattern{scale = 0.6}{2}{1/1,2/2}{0/1,1/2,0/0,2/1,2/0}$ &  \cite[Theorem~3.10]{KZ19}
% 			\\[5pt]
% 			\cline{1-3}
% 			18 & $\pattern{scale = 0.6}{2}{1/1,2/2}{0/0,0/1,0/2,1/2,2/0,2/2}$ &   \cite[Theorem~2.6]{KZ19}
% 			&
% 			64 & $\pattern{scale = 0.6}{2}{1/1,2/2}{0/1,1/2,2/0,0/2,1/1}$ &  \cite[Theorem~3.11]{KZ19}
% 			\\[5pt]
% 			\cline{1-3}
% 			19 & $\pattern{scale = 0.6}{2}{1/1,2/2}{0/1,0/2,1/1,1/2,2/0,2/2}$ &  \cite[Theorem~2.7]{KZ19}
% 			&
% 			65 & $\pattern{scale = 0.6}{2}{1/1,2/2}{0/1,1/0,0/0,1/1,2/2}$ &  \cite[Theorem~3.13]{KZ19}
% 			\\[5pt]
% 			\hline
% 		\end{tabular}
%\end{center} 	
%\medskip
% 	\caption{Known or conjectured distributions of mesh patterns of length 2. Pattern's number comes from \cite{Hilmarsson2015Wilf}. Absence of a horizontal line indicates  equidistributions.}\label{tab-1}
%% 	{Unsigned Stirling numbers of the first kind, A132393.}
%\end{table}

\begin{table}[htbp]\small{
		\renewcommand{\arraystretch}{1.3}
	\begin{center}
	\begin{tabular}{|c|c|c|c||c|c|c|}
		\hline
		&  	{Nr.\ } & {Repr.\ $p$}  & {Ref.}  &  {Nr.\ } & {Repr.\ $p$}  & {Ref.}  
		\\[6pt]
		\hline \hline
		{proved} 
		& 48 & $\pattern{scale = 0.6}{2}{1/1,2/2}{0/1,1/2,0/0,2/1,2/2}$&\multirow{2}{*}{\cite[Theorem~5.1]{KZ19}}&&&
		 \\[6pt]	\cline{2-3} 
     {equidistributions}   & 49 &	$\pattern{scale = 0.6}{2}{1/1,2/2}{0/1,1/2,0/0,1/1,2/0}$   &	 &&&	\\[6pt]	\hline
   	 & 	23 &	$\pattern{scale = 0.6}{2}{1/1,2/2}{0/0,0/2,1/0,1/1,1/2}$  & \multirow{2}{*}{Theorem~\ref{main2}}  &53 &	$\pattern{scale = 0.6}{2}{1/1,2/2}{0/1,1/2,0/0,2/1}$  & \multirow{2}{*}{Corollary~\ref{cor:main1}}
		\\[6pt]
	\multirow{2}{*}{conjectured} & 	24 &	$\pattern{scale = 0.6}{2}{1/1,2/2}{0/0,0/1,1/0,1/1,1/2}$ &&	54 &  $\pattern{scale = 0.6}{2}{1/1,2/2}{0/1,0/0,1/1,2/2}$   &			\\[6pt] \cline{2-7} 
	\multirow{3}{*}{equidistributions} &	 \multirow{2}{*}{48} & \multirow{2}{*}{$\pattern{scale = 0.6}{2}{1/1,2/2}{0/1,1/2,0/0,2/1,2/2}$}	  & 	
	&	57 & $\pattern{scale = 0.6}{2}{1/1,2/2}{0/1,1/2,1/1,2/0}$	&	 \multirow{2}{*}{N/A}	\\[6pt]		
	 	& \multirow{2}{*}{49} &   \multirow{2}{*}{$\pattern{scale = 0.6}{2}{1/1,2/2}{0/1,1/2,0/0,1/1,2/0}$}   &	 \multirow{2}{*}{Corollary~\ref{cor:main1}  and}
		&	58 &  $\pattern{scale = 0.6}{2}{1/1,2/2}{0/1,1/0,1/1,2/2}$ &	 	\\[6pt]	\cline{5-7}
 in \cite{KZ19}	&\multirow{2}{*}{50} &  \multirow{2}{*}{$\pattern{scale = 0.6}{2}{1/1,2/2}{0/1,1/2,0/0,1/1,2/2}$} &\multirow{2}{*}{\cite[Theorem~5.1]{KZ19}} & 61 & $\pattern{scale = 0.6}{2}{1/1,2/2}{0/1,1/2,0/0,2/0}$	&  \multirow{2}{*}{N/A}		\\[6pt]
		&  && &	62 & $\pattern{scale = 0.6}{2}{1/1,2/2}{0/1,1/0,0/0,2/2}$ &		\\[6pt]	\hline
	\end{tabular}
	\end{center}
}
\medskip
	\caption{Equidistributions for which enumeration is unknown. Pattern's numbers are adopted from  \cite{Hilmarsson2015Wilf,KZ19}}
 	\label{tab-2}
\end{table}

 
 

%
%$$(\pattern{scale=0.5}{2}{1/2,2/1}{1/0,2/0,2/2}, \pattern{scale=0.5}{2}{1/2,2/1}{1/0, 1/1,2/0,2/2},\pattern{scale=0.5}{2}{1/2,2/1}{1/0, 1/1, 1/2,2/0,2/2})(\pi)=(\pattern{scale=0.5}{2}{1/2,2/1}{2/1,1/0,2/0}, \pattern{scale=0.5}{2}{1/2,2/1}{2/1,1/0,2/0,1/2}, \pattern{scale=0.5}{2}{1/2,2/1}{2/1,1/0,2/0,1/1,1/2})
%\Psi(\pi)).$$


 
 
 
 
% Let $S_n$ be the set of all permutations of length $n$, which we call $n$-permutations. For example, $S_3=\{123, 132, 213, 231, 312, 321\}$. 
 For a pattern $p$ and a permutation $\pi$, we let $p(\pi)$ denote the number of occurrences of $p$ in $\pi$. 
% Also, let $S_n(p)$ denote the set of all permutations of length $n$ avoiding $p$ and $S(p)=\cup_{n\geq 0}S_n(p)$. 
Kitaev and Zhang~\cite[Conjecture~6.1]{KZ19} conjectured a Stieltjes continued fraction formula for 
the distribution of pattern Nr.\ $3$ = $\pattern{scale = 0.6}{2}{1/1,2/2}{0/0,0/1,1/2}$ (see \cite[$A200545$]{OEIS}), which 
is equivalent to the following identity.

\begin{conj}[{\cite[Conjecture~6.1]{KZ19}}]\label{conj1} 
 We have 
 \begin{subequations}
\begin{align}\label{cf:sz}
\sum_{n\geq 0}t^n
 \sum_{\pi\in S_n}y^{\pattern{scale=0.5}{2}{1/1,2/2}{0/0,0/1,1/2}(\pi)}\;=\;
 \cfrac{1}
 {1-\cfrac{\alpha_1 t}
 {1-\cfrac{\alpha_2 t}
% {1-\cfrac{\alpha_3 t}
% {1-\cfrac{\alpha_4 t}
 {1-\cdots}}}
\end{align}
\text{with coefficients} 
\begin{align}
\alpha_{2k-1}=k,\quad 
\alpha_{2k}=y+k-1.
%\alpha_{2k-1}&=k\\
%\alpha_{2k}&=y+k-1
\end{align}
\end{subequations}
\end{conj}
%   \cfrac{1}{1 - \cfrac{t}{1 - \cfrac{yt}{1 - \cfrac{2t}{1- \cfrac{(y+1)t}{1 - \cfrac{3t}{1 - \cfrac{(y+2)t}{1-\cdots}}}}}}}.
%\end{align}
%$$(1,0,2,1,3,2,4,3,5,4,6,5,7,6,\ldots)\ \mbox\emph{DELTA } (0,1,0,1,0,1,0,1,0,1,\ldots)$$ where \emph{DELTA} is the operator defined in $A084938$ in \cite{OEIS} in terms of continued fraction: the triangle $[r_0,r_1,\ldots]\  \mbox{\emph{DELTA}}\  [s_0,s_1,\ldots]$ has generating function $$\frac{1}{1-\frac{r_0x+s_0xy}{1-\frac{r_1x+s_1xy}{1-\frac{r_2x+s_2xy}{1-\ldots}}}}.$$}












Presenting their  conjecture in this way, we notice that 
the S-continued fraction \eqref{cf:sz} 
 appears in a recent paper of Sokal and Zeng~\cite{SZ20}.
Let~us reformulate the relevant permutation statistics in \cite{SZ20} in terms of mesh patterns.
Given a permutation $\pi \in S_n$,
an index $i \in [n]$ (or a value $\pi(i) \in [n]$) is called 
%% (or an)
\begin{itemize}
\item  an \emph{excedance} if $\pi(i)>i$;
   
 \item a \emph{record} (rec) (or \emph{left-to-right maximum})
         if $\pi(j) < \pi(i)$ for all $j < i$
      [note in particular that the index 1 is always a record
       and that the value $n$ is always a record];
in other words, a record of $\pi$ is one occurrence of pattern $\pattern{scale=0.8}{1}{1/1}{0/1}$ of  $\pi$;
       
       
   \item an \emph{antirecord} (arec) (or \emph{right-to-left minimum})
         if $\pi(j) > \pi(i)$ for all $j > i$
      [note in particular that the index $n$ is always an antirecord
       and that the value 1 is always an antirecord];
in other words, an antirecord of $\pi$ is one occurrence of pattern $\pattern{scale=0.8}{1}{1/1}{1/0}$ of  $\pi$;      
       
   \item an \emph{exclusive record} (erec) if it is a record and not also
         an antirecord; in other words, an exclusive record of $\pi$ is one occurrence of pattern $\pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0}$ of  $\pi$, see \eqref{eq2:lemma1};       
         
   \item an \emph{exclusive antirecord} (earec) if it is an antirecord and not also
     a record; in other words, an exclusive antirecord of $\pi$ is one occurrence of pattern $\pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/2,2/0}$ of  $\pi$, see \eqref{eq1:lemma1};     
     
 \item a \emph{record-antirecord} (rar) (or \emph{pivot})
     if it is both a record and an antirecord;
   in other words, a record-antirecord of $\pi$ is one occurrence of pattern  \pattern{scale=0.8}{1}{1/1}{0/1,1/0}  of  $\pi$.     
     
     
%   \item a \emph{neither-record-antirecord} (nrar) if it is neither a record
%      nor an antirecord.
%     in other words, a neither-record-antirecord of $\pi$ is one occurrence of pattern  \pattern{scale=0.8}{1}{1/1}{0/0,1/1}  of  $\pi$.    
      
      
\end{itemize}
An \emph{inversion} of  a permutation $\pi\in S_n$ is a pair $(i, j)\in[n]\times[n]$ such that $i<j$ and $\pi(i)>\pi(j)$, in other words, an inversion of $\pi$ is one occurrence of pattern $\pattern{scale=0.5}{2}{1/2,2/1}{}$ of $\pi$.
%Every index $i$ thus belongs to exactly one of the latter four types;
%we refer to this classification as the \emph{record classification}.
We denote the number of excedances, records, antirecords, exclusive records, exclusive antirecords, record-antirecords and inversions in $\pi$
by $\exc(\pi)$, $\rec(\pi)$, $\arec(\pi)$, $\erec(\pi)$, $\earec(\pi)$, $\rar(\pi)$ and $\inv(\pi)$  respectively. 





 \begin{table}[thbp]
{
		\renewcommand{\arraystretch}{1.3}
	\begin{center}
	\begin{tabular}{|c|c|c|c||c|c|c|}
		\hline
		&  	{Nr.\ } & {Repr.\ $p$}  & {Ref.}  &  {Nr.\ } & {Repr.\ $p$}  & {Ref.}  
		\\[6pt]
		\hline \hline
		{proved} 
		& $1^*$ & $\pattern{scale=0.5}{2}{1/2,2/1}{1/0,2/0,2/2}$&\multirow{2}{*}{Theorem~\ref{main2}}& $3^*$&  $\pattern{scale=0.5}{2}{1/2,2/1}{1/0, 1/1,2/0,2/2}$ & $\multirow{2}{*}{Theorem~\ref{main2}}$
		 \\[6pt]	
     {equidistributions}   &  $2^*$ &	$\pattern{scale=0.5}{2}{1/2,2/1}{2/1,1/0,2/0}$  & 	 & $4^*$ &$\pattern{scale=0.5}{2}{1/2,2/1}{2/1,1/0,2/0,1/2}$  &	\\[6pt] 	\hline
	\end{tabular}
	\end{center}
}
\medskip
	\caption{More proved equidistributions. Pattern's numbers are not considered in \cite{Hilmarsson2015Wilf,KZ19}}
 	\label{tab-3}
\end{table} 








Dumont and Kreweras~\cite{DK88} gave the joint distribution of $(\pattern{scale=0.8}{1}{1/1}{0/1},\pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/2,2/0})$, Zeng \cite{Zeng89} gave the joint distribution of $(\pattern{scale=0.8}{1}{1/1}{0/1},\pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/2,2/0}, \pattern{scale = 0.5}{2}{1/2,2/1}{})$. 
Recently Sokal and Zeng~\cite{SZ20} proved  much more general results.
 For example,
define 
the generating function of the generalized Eulerian polynomials
\begin{align}\label{poly:dkz}
F(x,y,z, v, q; t)=\sum_{n=0}^\infty t^n\sum_{\sigma\in S_n}
x^{\arec(\sigma)}y^{\erec(\sigma)}z^{\rar(\sigma)}v^{\exc(\sigma)}q^{\inv(\sigma)}.
\end{align}
From \cite[Theorems~2.7 and 2.8]{SZ20} we derive   the following result.
\begin{thm}\label{thm:1.2}
We have 
\begin{subequations}
\begin{align}\label{cf-SZ}
F(x,y,z, v, q; t)=
\frac{F(x,y,1, v,q; t)}{1+x(1-z)tF(x,y, 1, v,q; t)},
\end{align}
where
\begin{align}\label{cf-pattern}
F(x,y,1, v,q; t)=\cfrac{1}{1-\cfrac{\alpha_1 t}{1-\cfrac{\alpha_2 t}
{1-\cdots}}}
\end{align}
\text{with coefficients} 
\begin{align}
\alpha_{2k-1}&=q^{k-1}(x+q+q^2+\cdots +q^{k-1})\\
\alpha_{2k}&=q^{k}v(y+q+q^2+\cdots +q^{k-1}).
\end{align}
\end{subequations}
\end{thm}
%\begin{proof} This follows from \cite[Theorems~2.7 and 2.8]{SZ20}
%by specializing the parameters.
%We just indicate the appropriate specialization  and refer the reader to \cite{SZ20}
%for further details.
%In the specialization (2.57) of \cite[Theorem~2.7]{SZ20},  if we   choose
%$w_0=xz$ (instead of $w_0=x$ in \cite{SZ20}) and 
%$$
%y=qv,\quad u=1,\quad v_1=v_2=qv, \quad p_{+}=p_{-}=q,\quad  q_{+}=q_{-}=q^2, 
%$$
%then equation (2.52) of \cite{SZ20} reduces to
%\begin{align}
%F(x,y,z, v, q; t)&=\frac{1}{1-\gamma_0 t-
%\cfrac{\beta_1 t}{1-\gamma_1 t-\cfrac{\beta_2 t}{1-\cdots}}}
%\end{align}
%with 
%\begin{align}
%\gamma_0=xz,\quad
%\gamma_n=\alpha_{2n}+\alpha_{2n+1},\quad 
%%q^n(x+q+\cdots +q^n)+q^{n}v(y+\cdots +q^{n-1})\\
%\beta_n=\alpha_{2n-1}\alpha_{2n}.
%%q^{2n-1}v(x+q+\cdots +q^n)(y+\cdots +q^{n-1}).
%\end{align}
%Therefore, the J-fraction formula can be written as (by contracting the 
%S-fraction starting from the second line),
%\begin{align}
%F(x,y,z, v, q; t)&=\frac{1}{1+x(1-z)t-
%\cfrac{\alpha_1 t}{1-\cfrac{\alpha_2 t}{1-\cdots}}},
%%&=\cfrac{1}{x(1-z)t+\cfrac{1}{F(x,y,1, v,q; t)}}.
%\end{align}
%which is equivalent to \eqref{cf-SZ}.
%\end{proof}
\begin{remark}
$\empty$
\begin{itemize}
\item We can also prove  \eqref{cf-SZ} by following the same
steps in  the special case  as in \cite{KZ19}  and then derive  \eqref{cf-pattern} directly from \cite[Theorem 2.8]{SZ20}. 
\item The case $x=y=v=q=1$ of Theorem~\ref{thm:1.2} is Theorem~1.1 in \cite{KZ19}.
\item Since 
$(\arec, \inv)\pi=(\rec, \inv)\pi^{-1}$ we derive from
 \cite{Zeng89} that  
\begin{equation}
F(x,1,1, 1,q; t)=\sum_{n=0}^\infty x(x+q)\ldots (x+q+\cdots q^{n-1})t^n.
\end{equation}
\end{itemize}
\end{remark}


%It is easy to see the number of $\erec(\pi)$ counts the number $\pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0} (\pi).$ and
%\begin{enumerate}

%\item
For $\pi=\pi(1)\ldots \pi(n)\in S_n$ we 
define the following three associated  permutations:
\begin{align}
\pi^{-1} &:=\pi^{-1}(1)\pi^{-1}(2)\cdots\pi^{-1}(n)\label{def:invpi}\\
\pi^\r &:=\pi(n)\cdots\pi(2)\pi(1)\label{def:revpi}\\
\pi^c &:=(n+1-\pi(1))(n+1-\pi(2))\cdots(n+1-\pi(n))\label{def:comlpi}
%\pi^{rc} &:=(n+1-\pi(n))\cdots(n+1-\pi(2))(n+1-\pi(1)).
\end{align}
Obviously 
we have %$\pi^{\r\circ \c}=\pi^{\c\circ \r}$ and 
$$
\pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0} (\pi)=
\pattern{scale=0.5}{2}{1/1,2/2}{0/0,1/2,2/2} (\pi^c)=
\pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/2,2/0} (\pi^{r\circ c})=
\pattern{scale=0.5}{2}{1/1,2/2}{0/0,1/0,2/2} (\pi^{r})
$$
and
\begin{align*}
&\pattern{scale=0.5}{2}{1/1,2/2}{0/0,0/1,1/2} (\pi)=
\pattern{scale=0.5}{2}{1/2,2/1}{0/1,0/2,1/0} (\pi^c)=
\pattern{scale=0.5}{2}{1/1,2/2}{1/0,2/1,2/2} (\pi^{r\circ c})=
\pattern{scale=0.5}{2}{1/2,2/1}{1/2,2/0,2/1} (\pi^{r})\\
=&\pattern{scale=0.5}{2}{1/1,2/2}{0/0,1/0,2/1}(\pi^{-1})
=\pattern{scale=0.5}{2}{1/2,2/1}{0/1,2/0,1/0} (\tau^r)=
\pattern{scale=0.5}{2}{1/1,2/2}{0/1,1/2,2/2} (\tau^{r\circ c})=
\pattern{scale=0.5}{2}{1/2,2/1}{2/1,0/2,1/2} (\tau^{c})
\end{align*}
with  $\tau=\pi^{-1}$.


\begin{lem}\label{lemma1:main1}
For $\pi\in S_n$, we have
\begin{align}
\earec(\pi)&=\pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/2,2/0} (\pi)
=\pattern{scale=0.5}{2}{1/2,2/1}{0/1,1/1,2/0} (\pi)
=\pattern{scale=0.5}{2}{1/2,2/1}{0/1,0/2,2/0} (\pi)
=\pattern{scale=0.5}{2}{1/2,2/1}{1/1,1/2,2/0} (\pi)
,\label{eq1:lemma1}\\
\erec(\pi)&=\pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0} (\pi)
=\pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/1,2/1} (\pi)
=\pattern{scale=0.5}{2}{1/2,2/1}{0/2,2/0,2/1} (\pi)
=\pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,1/1} (\pi).\label{eq2:lemma1}
\end{align}
\end{lem}
\begin{proof}
We just prove \eqref{eq1:lemma1} as the proof of \eqref{eq2:lemma1} is similar. In the entries placement representation of a permutation
$\pi\in S_n$ 
the entry $y=(i, \pi(i))$ is an exclusive antirecord iff there is
another entry $x=(j,\pi(j))$ at left of $y$, i.e., 
 $j<i$ and higher than $x$, i.e., $\pi(j)>\pi(i)$.
 Hence there are four unique choices for such a entry $x$: 
 the \emph{highest, lowest,  farthest} and \emph{nearest}. This corresponds
 to the four mesh patterns in \eqref{eq1:lemma1}, respectively. 
 \end{proof}
 \begin{remark}
As $\earec(\pi)=\erec(\pi^{\r\circ c})$ for $\pi\in S_n$, we can also derive 
\eqref{eq2:lemma1} from \eqref{eq1:lemma1}. 
 \end{remark}



%$\pattern{scale=0.5}{2}{1/2,2/1}{0/1,2/0,1/0}$  
%The first known operations are the symmetries reverse, complement and inverse.
%For a given mesh pattern $(\tau,R)$ of length $n$, we define
%\begin{equation*}
%    (\tau,R)^r = (\tau^r, R^r), \quad
%    (\tau,R)^c = (\tau^c,R^c), \quad
%    (\tau,R)^i = (\tau^i,R^i),
%\end{equation*}
%where $\tau^r$ is the usual reverse of the permutation $\tau$, $\tau^c$ the
%usual complement, $\tau^i$ the usual inverse, and
%\begin{align*}
%&R^r = \{ \boks{n-x}{y} \sep \boks{x}{y} \in R \}, \\
%&R^c = \{ \boks{x}{n-y} \sep \boks{x}{y} \in R \}, \\
%&R^i = \{ \boks{y}{x}   \sep \boks{x}{y} \in R \}.
%\end{align*}
%Hence, reverse is a reflection around the vertical center line, complement is a
%reflection around the horizontal center line and inverse is the reflection
%around the southwest to northeast diagonal. 
%%Figure~\ref{fig:meshsyms} is an
%%example of the use of these symmetries on the pattern $p
%%=(312,\{\boks{0}{1},\boks{1}{3},\boks{2}{2}\})$.
%
%It is well-known that a permutation $\pi$ avoids a mesh pattern $p$ if and only
%if the permutation $\pi^r$ avoids $p^r$. That is, the reverse operation
%preserves Wilf-equivalence. The same applies for complement and inverse or any
%composition of these three operations.
%\end{enumerate}



%\begin{theorem}\label{Conj57}
%For $\pi\in\S_n$, we have
%$$\pattern{scale = 0.6}{2}{1/1,2/2}{0/1,1/2,1/1,2/0} (\pi)=\pattern{scale = 0.6}{2}{1/1,2/2}{0/1,1/0,1/1,2/2}(\pi^c).$$
%\end{theorem}



\begin{thm}\label{main1} There exists an involution $\Phi$ on $S_n$ such that for $\pi\in S_n$,
$$(\pattern{scale=0.5}{2}{1/2,2/1}{0/1,2/0,1/0},  \pattern{scale=0.5}{2}{1/2,2/1}{0/1,0/2,2/0,1/0,1/2},\pattern{scale=0.5}{2}{1/2,2/1}{0/1,1/0,2/0,1/2})\pi=
(\pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0},  \pattern{scale=0.5}{2}{1/2,2/1}{0/1,0/2,1/0,2/0,1/1}, \pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0,1/1})\Phi(\pi).
$$
\end{thm}




\begin{cor}\label{cor:main1}
The triple pattern $(\Nr.3, \Nr.48, \Nr.53)$ is equidistributed with 
the triple pattern $(\erec, \Nr.50, \Nr.54)$ on $S_n$.
\end{cor}
\begin{proof} For any $\pi\in S_n$ we have 
\begin{align}\label{eq:triple}
\begin{bmatrix}
\Nr.3\\
\\
\Nr.48\\
\\
\Nr.53
\end{bmatrix}\pi
=
\begin{bmatrix}
\pattern{scale=0.5}{2}{1/1,2/2}{0/0,0/1,1/2}\\
\\
\pattern{scale=0.5}{2}{1/1,2/2}{ 0/0,0/1,1/2,2/1,2/2}\\
\\
\pattern{scale=0.5}{2}{1/1,2/2}{ 0/0,0/1,1/2,2/1}
\end{bmatrix}\pi
=
\begin{bmatrix}
\pattern{scale=0.5}{2}{1/1,2/2}{0/0,1/0,2/1} \\
\\
\pattern{scale=0.5}{2}{1/1,2/2}{0/0,1/0,1/2,2/1,2/2}\\
\\
\pattern{scale=0.5}{2}{1/1,2/2}{0/0,1/0,2/1,1/2}
\end{bmatrix}\pi^{-1}
=
\begin{bmatrix}
\pattern{scale=0.5}{2}{1/2,2/1}{0/1,2/0,1/0} \\
\\
\pattern{scale=0.5}{2}{1/2,2/1}{ 0/1,0/2,1/0,1/2,2/0}  \\
\\
\pattern{scale=0.5}{2}{1/2,2/1}{ 0/1,1/0,1/2,2/0}
\end{bmatrix}(\pi^{-1})^\r
\end{align}
and 
\begin{align}\label{eq:double}
(\Nr.50, \Nr.54)\pi=
(\pattern{scale=0.5}{2}{1/1,2/2}{ 0/0,0/1,1/1,2/2,1/2},
\pattern{scale=0.5}{2}{1/1,2/2}{ 0/0,0/1,1/1,2/2})\pi
= 
 (\pattern{scale=0.5}{2}{1/1,2/2}{0/0,1/0,1/1,2/1,2/2}, 
 \pattern{scale=0.5}{2}{1/1,2/2}{0/0,1/0,1/1,2/2}) \pi^{-1}
 =
(\pattern{scale=0.5}{2}{1/2,2/1}{ 0/1,0/2,1/0,1/1,2/0},
\pattern{scale=0.5}{2}{1/2,2/1}{ 0/2,1/0,1/1,2/0})(\pi^{-1})^\r.
\end{align}
By Theorem~\ref{main1} the result follows from \eqref{eq2:lemma1}, \eqref{eq:triple} and \eqref{eq:double}.
\end{proof}









\begin{cor}\label{cor:conj1}
Conjecture~\ref{conj1} holds true.
\end{cor}
\begin{proof}
% Mesh pattern Nr.~3 $\pattern{scale=0.5}{2}{1/1,2/2}{0/0,0/1,1/2}$
%is equidistributed with $\pattern{scale=0.5}{2}{1/2,2/1}{0/1,2/0,1/0}$
%as
%\begin{align*}
%\pattern{scale=0.5}{2}{1/1,2/2}{0/0,0/1,1/2} (\pi)
%=\pattern{scale=0.5}{2}{1/1,2/2}{0/0,1/0,2/1}(\pi^{-1})
%=\pattern{scale=0.5}{2}{1/2,2/1}{0/1,2/0,1/0} ((\pi^{-1})^{\r})\quad (\forall \pi\in S_n)
%%\pattern{scale=0.5}{2}{1/1,2/2}{0/1,1/2,2/2} ((\pi^{r\circ c})^{-1})=
%%\pattern{scale=0.5}{2}{1/2,2/1}{2/1,0/2,1/2} ((\pi^r)^{-1}).
%\end{align*}
%By Theorem~\ref{main1}  mesh pattern Nr. $3$ is equidistributed with 
%the pattern $\pattern{scale=0.5}{2}{1/2,2/1}{0/1,2/0,1/0}$.  Thus 
By Corollary~\ref{cor:main1} this follows from \eqref{cf-pattern} with $x=v=q=1$. 
\end{proof}
%
%\begin{corollary}
%Mesh patterns Nr.$53$ and Nr. $54$ are equidistributed on $S_n$.
%\end{corollary}
%\begin{proof}
%For mesh patterns Nr.$53$ and Nr. $54$, we have
%\begin{align*}
%&\pattern{scale=0.5}{2}{1/1,2/2}{ 0/0,0/1,1/2,2/1} (\pi)= \pattern{scale=0.5}{2}{1/1,2/2}{0/0,1/0,2/1,1/2} (\pi^i)=
%\pattern{scale=0.5}{2}{1/2,2/1}{ 0/1,1/0,1/2,2/0} (\pi^{r\circ i}),\\
%&\pattern{scale=0.5}{2}{1/1,2/2}{ 0/0,0/1,1/1,2/2} (\pi)= \pattern{scale=0.5}{2}{1/1,2/2}{0/0,1/0,1/1,2/2} (\pi^i)=
%\pattern{scale=0.5}{2}{1/2,2/1}{ 0/2,1/0,1/1,2/0} (\pi^{r\circ i}),
%\end{align*}
%\end{proof}

As the equidistribution of $\Nr. 48$ and $\Nr. 49$ is known~\cite[Theorem~5.1]{KZ19}, Corollary~\ref{cor:main1} confirms  two conjectured equidistributions in Table~\ref{tab-2}.
%\begin{corollary}
%Mesh patterns Nr.$48$ and Nr. $50$ are equidistributed on $S_n$.
%\end{corollary}
%\begin{proof}
%and for mesh patterns Nr.$48$ and Nr. $50$, we have
%\begin{align*}
%&\pattern{scale=0.5}{2}{1/1,2/2}{ 0/0,0/1,1/2,2/1,2/2} (\pi)= \pattern{scale=0.5}{2}{1/1,2/2}{0/0,1/0,1/2,2/1,2/2} (\pi^i)=
%\pattern{scale=0.5}{2}{1/2,2/1}{ 0/1,0/2,1/0,1/2,2/0} (\pi^{r\circ i}),\\
%&\pattern{scale=0.5}{2}{1/1,2/2}{ 0/0,0/1,1/1,2/2,1/2} (\pi)= \pattern{scale=0.5}{2}{1/1,2/2}{0/0,1/0,1/1,2/1,2/2} (\pi^i)=
%\pattern{scale=0.5}{2}{1/2,2/1}{ 0/1,0/2,1/0,1/1,2/0} (\pi^{r\circ i}).
%\end{align*}
%Theorem~\ref{main1} leads two conjectured equidistributions in Table~\ref{tab-2}.
%\end{proof}


\begin{thm}\label{main2}
 There exist an involution $\Psi$ on $S_n$ such that
 for $\pi\in S_n$,
$$(\pattern{scale=0.5}{2}{1/2,2/1}{1/0,2/0,2/2}, \pattern{scale=0.5}{2}{1/2,2/1}{1/0, 1/1,2/0,2/2},\pattern{scale=0.5}{2}{1/2,2/1}{1/0, 1/1, 1/2,2/0,2/2})(\pi)=(\pattern{scale=0.5}{2}{1/2,2/1}{2/1,1/0,2/0}, \pattern{scale=0.5}{2}{1/2,2/1}{2/1,1/0,2/0,1/2}, \pattern{scale=0.5}{2}{1/2,2/1}{2/1,1/0,2/0,1/1,1/2})
\Psi(\pi)).
$$
\end{thm}




%\begin{remark}
For the patterns $\Nr.23$ and $\Nr.24$, we have
\begin{align*}
(\Nr.23, \Nr.24)\pi=
(\pattern{scale=0.5}{2}{1/1,2/2}{1/0, 1/1, 1/2,0/0,0/2},
\pattern{scale=0.5}{2}{1/1,2/2}{0/1,1/0,0/0,1/1,1/2})
\pi=(\pattern{scale=0.5}{2}{1/2,2/1}{1/0, 1/1, 1/2,2/0,2/2},
\pattern{scale=0.5}{2}{1/2,2/1}{2/1,1/0,2/0,1/1,1/2})
\pi^r.
\end{align*}
By Theorem~\ref{main2}, we confirm another conjecture in Table~\ref{tab-2}, i.e., the patterns $\Nr.23$ and $\Nr.24$ are equidistributed.
%\end{remark}













%\begin{conj}For $\pi\in S_n$, there exist a bijection $\phi$, Nr.$49$ and Nr. $50$,
%$$\pattern{scale=0.5}{2}{1/2,2/1}{0/1,1/1,1/0,2/0,2/2}(\pi)=\pattern{scale=0.5}{2}{1/2,2/1}{0/1,0/2,1/0,2/0,1/1}
%(\phi(\pi)).$$
%\end{conj}
%
%\begin{remark}
%\begin{align*}
%&\pattern{scale=0.5}{2}{1/1,2/2}{ 0/0,0/1,1/1,1/2,2/0} (\pi)= \pattern{scale=0.5}{2}{1/1,2/2}{0/0,0/2,1/0,1/1,2/1} (\pi^i)=
%\pattern{scale=0.5}{2}{1/2,2/1}{ 0/1,1/0,1/1,2/0,2/2} (\pi^{r\circ i}),\\
%&\pattern{scale=0.5}{2}{1/1,2/2}{ 0/0,0/1,1/1,2/2,1/2} (\pi)= \pattern{scale=0.5}{2}{1/1,2/2}{0/0,1/0,1/1,2/1,2/2} (\pi^i)=
%\pattern{scale=0.5}{2}{1/2,2/1}{ 0/1,0/2,1/0,1/1,2/0} (\pi^{r\circ i}).
%\end{align*}
%\end{remark}


We shall prove Theorem~\ref{main1} and Theorem~\ref{main2}
in Section~2 and Section~3, respectively. %and make a connection  between
  %pattern Nr. 14 and the statistic \emph{succession}  in permutations in Section~4.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%Section 2
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Proof outlines of Theorem~\ref{main1}}






% In Figure~\ref{Fig:1}, $y$ is an exclusive antirecord, there exist a $\red{x}$ in the left of $y$ in first pattern. In the second pattern $x$ is the maximum element in the left of $y$.  In the third pattern $x$ is the minimum element in the left of $y$. 
%\begin{figure}[!htb]
%	\begin{center}
%		\begin{tikzpicture}[scale=0.8, baseline=(current bounding box.center)]
%		\foreach \x/\y in {2/0}		    
%		\fill[blue!20] (\x,\y) rectangle +(1,1);
%		\draw (0.01,0.01) grid (2+0.99,2+0.99);
%		\filldraw (1,2) circle (3pt) node[below left] {$\red{x}$};
%		\filldraw (2,1) circle (3pt) node[above right] {$y$};
%\end{tikzpicture}
%		$\xrightarrow{=}$
%	       \begin{tikzpicture}[scale=0.8, baseline=(current bounding box.center)]
%		\foreach \x/\y in {0/2,1/2,2/0}		    
%		\fill[blue!20] (\x,\y) rectangle +(1,1);
%		\draw (0.01,0.01) grid (2+0.99,2+0.99);
%		\filldraw (1,2) circle (3pt) node[below left] {$x$};
%		\filldraw (2,1) circle (3pt) node[above right] {$y$};
%\end{tikzpicture}
%				$\xrightarrow{=}$
%		\begin{tikzpicture}[scale=0.8, baseline=(current bounding box.center)]
%		\foreach \x/\y in {0/1,1/1,2/0}		    
%		\fill[blue!20] (\x,\y) rectangle +(1,1);
%		\draw (0.01,0.01) grid (2+0.99,2+0.99);
%		\filldraw (1,2) circle (3pt) node[below left] {$y$};
%		\filldraw (2,1) circle (3pt) node[above right] {$x$};
%\end{tikzpicture}
%%				$\xrightarrow{?}$
%%		\begin{tikzpicture}[scale=0.8, baseline=(current bounding box.center)]
%%		\foreach \x/\y in {0/1,1/0,2/0}		    
%%		\fill[blue!20] (\x,\y) rectangle +(1,1);
%%		\draw (0.01,0.01) grid (2+0.99,2+0.99);
%%		\filldraw (1,2) circle (3pt) node[below left] {$y$};
%%		\filldraw (2,1) circle (3pt) node[above right] {$x$};
%%\end{tikzpicture}
%
%		
%		\end{center}
%	\caption{The graph description of $\earec$}\label{Fig:1}
%\end{figure}








%$\pmb{Algorithm\,\, \phi:}$

For $\pi\in S_n$ let $\AREC(\pi)=(i_1, i_2, \ldots, i_l)$ be the sequence of antirecord positions of $\pi$ from left to right. So 
$\pi(i_1)=1$, $i_1<\cdots <i_l$ and $i_l=n$.
For each antirecord position $i_k$ % $a:=\pi(i_k)$,  let $b=\pi(i_{k-1})$ and 
 define two mappings
 \begin{subequations}\label{eq:defanti1}
\begin{align}
&\varphi_1^{(i_k)}: \pi\mapsto \pi'\\ 
&\varphi_2^{(i_k)}: \pi\mapsto \pi''
\end{align}
\end{subequations}
as follows: 
\begin{itemize}
\item let  $w=w_1\ldots w_r$  be  the subword of $\pi$ 
consisting of letters  greater than $\pi(i_k)$ on the left of $\pi(i_k)$ (resp. $\pi(i_{k-1})$);
\item let 
$w'=w_1'\ldots w_r'$   be the word obtained by substituting 
the $j$th largest letter  with  the $j$th smallest letter in $w$
for $j=1,\ldots, r$;
\item  let $\pi'$ (resp. $\pi''$) be the word obtained by 
replacing $w_j$  with  $w_j'$ for $j=1,\ldots, r$ 
 in $\pi$. 
\end{itemize}

%\begin{itemize}
%\item let  $w=w_1\ldots w_r$ (resp. $x=x_1\ldots x_q$) be  the subword of $\pi$ 
%consisting of letters  greater than $\pi(i_k)$ on the left of $\pi(i_k)$ (resp. $\pi(i_{k-1})$);
%\item let 
%$w'=w_1'\ldots w_r'$ (resp. $x'=x'_1\ldots x'_q$)  be the word obtained by substituting 
%the $j$th largest letter  with  the $j$th smallest letter in $w$ (resp. $x$)
%for $j=1,\ldots, r$ (resp. $q$);
%\item  let $\pi'$ (resp. $\pi''$) be the word obtained by 
%replacing $w_j$ (resp. $x_j$) with  $w_j'$ (resp. $x_j'$) for $j=1,\ldots, r$ (resp. $q$)
% in $\pi$. 
%\end{itemize}
\begin{remark}
By convention, we define $\varphi_2^{(i_1)}$ to be the identity mapping.
Clearly  the two operations keep the sequence of 
 antirecords for both values and positions, that is,
 \begin{subequations}
 \begin{align}
 \AREC(\pi)&=\AREC(\pi')=\AREC(\pi'')\\
\pi'(i_k)&=\pi''(i_k)=\pi(i_k)\quad \text{for} \quad k=1, \ldots l.
\end{align}
\end{subequations}
\end{remark}



 

Let $P=\{p_1<\cdots <p_r\}$ and $Q=\{q_1<\cdots <q_r\}$ be two ordered sets and 
 $\pi=p_{1}\ldots p_{r}$ and $\tau=q_{1}\ldots q_{r}$ are permutations of $P$ and $Q$, respectively.  We say that $\pi$ and $\tau$ are  \emph{order isomorphic} and write $\pi\sim \tau$ 
   if for any two  indices $r$ and $s$ we have the equivalence 
 $p_r<p_s$ if and only if $q_r<q_s$. In other words, $\tau$ is the
  permutation obtained from $\pi$ by substituting $p_i$ by $q_i$ for $i=1, \ldots, r$.

Let $w=w_1\ldots w_n$ be a permutation of $a_1<a_2<\cdots<a_n$. We define
the \emph{complement} of $w$ by $w^c$\footnote[1]{When $a_i=i$, $w^c$ reduces to $\pi^c$, see \eqref{def:comlpi}.}, which is the word obtained by substituting $a_i$ by $a_{n+1-i}$ in $w$ for $i=1, \ldots, n$. 
If $x$ is a subset of  the letters in $w$, we write $[w]_x$ as the subword of $w$ consisting of the letters $a\in x$. 
 %%%%%
\begin{lem}\label{property:compl}
\begin{enumerate}
\item If $w=w_1w_2$ and $w^c=w_1'w_2'$, then $(w_1')^c\sim w_1$.\footnote[2]{The word $w_1'$ is the complement of $w_1$ in the alphabet of $w$, while $(w_1')^c$ is the complement of $w_1'$ in the alphabet of $w_1'$.}
\item  Let  $w=w_1w_2w_3$ and  $v=v_1v_2v_3$ with $|w_1|=|v_1|$. If $w_1w_2\sim v_1v_2$ with
$(w_1w_2)^c=w_1'w_2'$ and $(v_1v_2)^c=v_1'v_2'$,  then $w_1\sim v_1$, $w_2\sim v_2$, $w_1'\sim v_1'$ and $w_2'\sim v_2'$. 
%(Here 
%%$w_1^c$ (resp. $v_1$) is the word obtained by replacing each letter in $w_1$ (resp. %$v_1$) by its complement in the alphabet of $w$ (resp. $v$).) 
Moreover,
we have    
$(w'_1)^c\sim (v'_1)^c$ and $(w'_2)^c\sim (v'_2)^c$.
%where $[w]$ means the complement word of $w$ with respect to its alphabet.
\item  If $w\sim v$ and $[w]_{x} =[v]_{x}$ for some set $x$ of some common 
letters in $w$ and $v$, then 
\begin{itemize}
\item $w^c\sim v^c$ and $[w^c]_{x}=[v^c]_{x}$.
\item $[w]_{y}\sim [v]_{z}$, where $y$ (resp. $z$) is the complementary of $x$ in the alphabet of $w$ (resp. $v$).
\end{itemize}
\end{enumerate}
\end{lem}
%\begin{proof}
%The verification is easy and left to the reader.
%\end{proof}


For example, 
if $w=3\,5\,9\,1\,4\,7\,2\,8\,6$, then $w^c=7\,5\,1\,9\,6\,3\,8\,2\,4$. Let $w=w_1w_2$ with 
$w_1=3\,5\,9\,1\,4\,7$ and $w_2=2\,8\,6$, then $w_1'=7\,5\,1\,9\,6\,3$ and $(w_1')^c=3\,6\,9\,1\,5\,7$.
Clearly $(w_1')^c\sim w_1$ and $[(w_1')^c]_x=[w_1]_x$ with
$x=\{1,\,3, \,9\}$.  We see that $w_1^c=7\,4\,1\,9\,5\,3$ and $[w_1']_x=[w_1^c]_x=1\,9\,3$.


\begin{lem}\label{lem:opera1}
For any antirecord position $i$ of $\pi\in S_n$ 
the mappings $\varphi_1^{(i)}$  and  $\varphi_2^{(i)}$ are involutions and 
commute, namely,
\begin{equation}\label{eq1:lem2.5}
\varphi_1^{(i)}\circ \varphi_1^{(i)}(\pi)=
\varphi_2^{(i)}\circ \varphi_2^{(i)}(\pi)=\pi
\end{equation}
and 
\begin{equation}\label{eq2:lem2.5}
\varphi_2^{(i)}\circ \varphi_1^{(i)}(\pi)=\varphi_1^{(i)}\circ \varphi_2^{(i)}(\pi).
\end{equation}
\end{lem}
\begin{proof}
From the definitions of $\varphi_1^{(i)}$ and $\varphi_2^{(i)}$ in Eq.~\eqref{eq:defanti1}, it is easy to check Eq.~\eqref{eq1:lem2.5} holds and 
$$
\varphi_2^{(i)}\circ \varphi_1^{(i)}(\pi) \sim \varphi_1^{(i)}\circ \varphi_2^{(i)}(\pi).
$$
Since the set of letters greater than $\pi(i)$ on the left of $\pi(i)$ are invariant under the operation $\varphi_1^{(i)}$ and  $\varphi_2^{(i)}$ on $\pi$, we obtain Eq.\eqref{eq2:lem2.5} immediately. 
\end{proof}





 Let  $\pi\in S_n$ with  sequence of antirecord positions 
$\AREC(\pi)=(i_1,i_2,\ldots, i_l).$ 
 We define the operation  $\Phi$ on $\pi$ by 
 \begin{align}\label{eq:defPhi}
 \Phi(\pi)=\varphi^{(i_1)}\circ\varphi^{(i_2)}\circ\cdots\circ
 \varphi^{(i_l)}(\pi)
 \end{align}
 with $\varphi^{(i_k)}=\varphi_2^{(i_k)}\circ \varphi_1^{(i_k)}$
for $k=1, \ldots, l$.




%% In Figure~\ref{Fig:1}, $y$ is an exclusive antirecord, there exist a $\red{x}$ in the left of $y$ in first pattern. In the second pattern $x$ is the maximum element in the left of $y$.  In the third pattern $x$ is the minimum element in the left of $y$. 
%\begin{figure}[t]
%	\begin{center}
%		\begin{tikzpicture}[scale=1.2, baseline=(current bounding box.center)]
%%		\foreach \x/\y in {3/0}		    
%%		\fill[blue!20] (\x,\y) rectangle +(1,1);
%		\draw (0.01,0.01) grid (3+0.99,3+0.99);
%		\filldraw (3,3) circle (3pt) node[below left] {$i_k$};		
%		\filldraw (2,2) circle (3pt) node[below left] {$i_{k-1}$};
%		\filldraw (1,1) circle (3pt) node[below left] {$i_{k-2}$};
%		\draw (0.5,2.5) node{$u_2$};
%		\draw (0.5,3.5) node{$u_1$};	
%		\draw (1.5,2.5) node{$v_2$};
%		\draw (1.5,3.5) node{$v_1$};
%                 \draw (2.5,3.5) node{$w$};	
%\end{tikzpicture}
%		\end{center}
%		\medskip
%	\caption{The decomposition of three consecutive  anti-records of $\pi$}\label{Fig:1}
%\end{figure}

\begin{lem}\label{lemma2.6:main1} 
For $\pi\in S_n$ with $\AREC(\pi)=\{i_1, \ldots, i_l\}$. The mappings  
$g:=\varphi^{(i_{k-1})}$ and $f:=\varphi^{(i_k)}$ commute, i.e., 
$$
g\circ f(\pi)=f\circ g(\pi).
%\varphi^{(i_{k-1})}\circ \varphi^{(i_k)}(\pi)=\varphi^{(i_k)}\circ \varphi^{(i_{k-1})}(\pi).
$$
\end{lem}
%%%%%%%%%
%\begin{proof}
% We write the permutation  $\pi=\pi(1)\ldots \pi(n)$ as 
%$\pi=u\pi(i_{k-2})v\pi(i_{k-1})w\pi(i_{k})x$ for $3\leq k\leq l$, and 
%\begin{itemize}
%\item[(i)] $u_1$ (resp. $v_1$,  $w$) as  the subword consisting of letters greater than $\pi(i_k)$ in $u$ (resp. $v$, $w$);
%\item[(ii)] $u_2$ (resp. $v_2$) as  the subword consisting of letters between 
% $\pi(i_{k-1})$  and $\pi(i_k)$  in $u$ (resp. $v$); 
%\end{itemize}
% see    Figure~\ref{Fig:1}.
%For convenience, we introduce the following notations:
%\begin{subequations}
%\begin{align}
%f(\pi):&=u'\pi(i_{k-2})v'\pi(i_{k-1})w'\pi(i_{k})x\label{eq:f}\\
%g\circ f (\pi):&=\tilde u\pi(i_{k-2})\tilde v\pi(i_{k-1})w'\pi(i_{k})x,\label{eq:gf}\\
%g(\pi):&=u''\pi(i_{k-2})v''\pi(i_{k-1})w\pi(i_{k})x,\label{eq:g}\\
%f\circ g (\pi):&=\hat u\pi(i_{k-2})\hat v\pi(i_{k-1})w'\pi(i_{k})x.\label{eq:fg}
%\end{align}
%\end{subequations}
%We will use the similar notations $u_i', \hat{u}_i, \tilde{u_i}, u_i'', v_i', \hat{v}_i, \tilde{v_i}, v_i''$ as in (i) and (ii) for $i=1, 2$.
%
%By the definition of  $f$ and $g$ and Lemma~\ref{property:compl}, we have the following facts:
%\begin{enumerate}
%\item Under the operation $f$ (cf.\ \eqref{eq:f}), as $u_1v_1\sim u_1'v_1'$, $u_2'v_2'=u_2v_2$, we have $u'v'\sim uv$;
%\item Applying $g$ to  $f(\pi)$ and $\pi$ (cf.\ \eqref{eq:gf} and \eqref{eq:g}), respectively,   we see that $\tilde u\tilde v\sim u''v''$  by (1) and Lemma~\ref{property:compl}, hence
%$\tilde u_2\tilde v_2=u''_2v''_2$;
%\item Applying $f$ to $g(\pi)$ (cf.\ \eqref{eq:g} and \eqref{eq:fg}),  we have  $u_2''v_2''=\hat u_2\hat v_2$, 
% combining  with  (2) yields 
%  $\hat u_2\hat v_2=\tilde u_2\tilde v_2$, that is $\hat u_2=\tilde u_2$ and 
%  $\hat v_2=\tilde v_2$;
% \item Applying $f$ to $g(\pi)$ (cf.\ \eqref{eq:g} and \eqref{eq:fg}),  using Lemma~\ref{property:compl}
% we have   $\hat u\hat v\sim u''v''$, from $u''v''\sim \tilde u\tilde v$ (cf.\ (2)) we derive 
% $\hat u\hat v\sim \tilde u\tilde v$.  Combining with Lemma~\ref{property:compl} and    (3) yields
% $\hat u_1\hat v_1\sim \tilde u_1\tilde v_1$.  
% \item 
% Let $\{w\}$ denote the set of letters in $w$.  
% \begin{itemize}
%\item                        Applying $f$ to $\pi$ we have 
% $\{u_1'v_1'w'\}=\{ u_1  v_1 w\}$,
% \item applying $g$ to $f(\pi)$ we have 
% $\{u_1'v_1'w'\}=\{\tilde u_1\tilde v_1w'\}$, 
%  \item applying $g$ to $\pi$ we have 
% $\{u_1''v_1''w\}=\{u_1 v_1 w\}$,
% \item applying $f$ to $g(\pi)$ we have 
% $\{u_1''v_1''w\}=\{\hat u_1\hat  v_1 w'\}$.
%\end{itemize}
% Thus  $\{\tilde u_1\tilde v_1\}=\{\hat u_1\hat  v_1\}$. It follows from (4) that 
% $\hat u_1= \tilde u_1$ (resp. $\hat v_1=\tilde v_1$).
% \end{enumerate}
% Summarizing the above facts we have proved $f\circ g=g\circ f$.
% \end{proof}



 

\begin{lem}\label{lemma2.7:main1} The mapping  $\varphi^{(i_k)}$ is an involution   such that for $\pi\in S_n$ and $r\neq k$,
\begin{subequations}
\begin{align}\label{eq:phia}
(\pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0}, \pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0,1/1}, 
\pattern{scale=0.5}{2}{1/2,2/1}{0/1,0/2,1/0,2/0,1/1})_k\;\pi
=(\pattern{scale=0.5}{2}{1/2,2/1}{0/1,2/0,1/0}, \pattern{scale=0.5}{2}{1/2,2/1}{0/1,1/0,2/0,1/2}, \pattern{scale=0.5}{2}{1/2,2/1}{0/1,0/2,2/0,1/0,1/2})_k\;\varphi^{(i_k)}(\pi),\\
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
(\pattern{scale=0.5}{2}{1/2,2/1}{0/1,2/0,1/0}, \pattern{scale=0.5}{2}{1/2,2/1}{0/1,1/0,2/0,1/2}, \pattern{scale=0.5}{2}{1/2,2/1}{0/1,0/2,2/0,1/0,1/2})_r\;\pi
=
(\pattern{scale=0.5}{2}{1/2,2/1}{0/1,2/0,1/0}, \pattern{scale=0.5}{2}{1/2,2/1}{0/1,1/0,2/0,1/2}, \pattern{scale=0.5}{2}{1/2,2/1}{0/1,0/2,2/0,1/0,1/2})_r\;\varphi^{(i_k)}(\pi),\label{eq:phib}\\
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
(\pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0}, \pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0,1/1}, 
\pattern{scale=0.5}{2}{1/2,2/1}{0/1,0/2,1/0,2/0,1/1})_r\;\pi
=
(\pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0}, \pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0,1/1}, 
\pattern{scale=0.5}{2}{1/2,2/1}{0/1,0/2,1/0,2/0,1/1})_r\;\varphi^{(i_k)}(\pi),
\label{eq:phic}
\end{align}
\end{subequations}
where $(\textrm{pattern})_k$ means the number of the patterns between $\pi(i_{k-1})$ and $\pi(i_k)$.
\end{lem}
%\begin{proof}
%If  the pair
%$(\pi(j),\pi(i_k))$ with $j<i_k$ contributes the pattern $\pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0}$(resp. $\pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0,1/1}, \pattern{scale=0.5}{2}{1/2,2/1}{0/1,0/2,1/0,2/0,1/1})$,
%then  $j>i_{k-1}$ because $\pi(i_{k-1})<\pi(i_{k})$ and 
%$\pi(i)>\pi(i_k)$ for $j\leq i<i_k$.
%Also, for  $i<j$, we have the equivalence 
% $$
%   \pi(i)<\pi(j)\Longleftrightarrow \pi(i_k)<\varphi_1^{(i_k)}(\pi(i))>\varphi_1^{(i_k)}(\pi(j)),
%   $$
%as $\varphi_2^{(i_k)}$ will affect only the letters at the left of 
%$\pi(i_{k-1})$. Thus  we have proved \eqref{eq:phia}.
%
%Next, recall that the operation $\varphi^{(i_k)}$ keeps the 
%sequence of antirecords for both positions and values. The two identities \eqref{eq:phib} and \eqref{eq:phic} are clear if $r>k$.
%Assume that $r<k$ and $[\pi]_r=\pi(1)\ldots \pi(i_{k-1})$. 
%%Let $w=w_1\ldots w_r$ be the subword of $[\pi]_r$ consisting of letters greater 
%%than $\pi(i_k)$. Then, 
%By Lemma~\ref{property:compl}, 
%after two operations
%$\varphi_1^{(i_k)}$ and $\varphi_2^{(i_k)}$  the permutation
%$\varphi^{(i_k)}([\pi]_r)$ is  isomorphic with $[\pi]_r$. This proves \eqref{eq:phib} and \eqref{eq:phic}.
% 
%
%%the pair $(\tau(j),\tau(i))$ contributes the pattern
%% $\pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0}$ (resp. $\pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0,1/1}, \pattern{scale=0.5}{2}{1/2,2/1}{0/1,0/2,1/0,2/0,1/1}$), 
%%%where$\phi(\pi)=\sigma$ and $\pi(i)$ is a antirecord in $\pi$ and $\sigma$.
%%Also, If the pair $(\pi(j),\pi(i))$ contributes the the pattern  $\pattern{scale=0.5}{2}{1/2,2/1}{0/1,2/0,1/0}$ (resp. $\pattern{scale=0.5}{2}{1/2,2/1}{0/1,1/0,2/0,1/2}, \pattern{scale=0.5}{2}{1/2,2/1}{0/1,0/2,2/0,1/0,1/2}$), 
%%then the pair $(\sigma(j),\sigma(i))$ contributes the pattern $\pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0}$ (resp. $\pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0,1/1}, \pattern{scale=0.5}{2}{1/2,2/1}{0/1,0/2,1/0,2/0,1/1}$). 
%%%where $\phi(\pi)=\sigma$ and $\pi(i)$ is a antirecord in $\pi$ and $\sigma$.
%%It is easy to see the algorithm $\Phi$ is a bijection by considering the reverse process of algorithm $\Phi$. 
%%Moreover, it is easy to check 
%%$$\Phi^2(\pi)=id\,\, \text{if}\,\, \pi\in S_n.$$
%%If the pair $(\pi(j),\pi(i))$ contributes the the pattern \pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0}, 
%%then the pair $(\sigma(j),\pi(i))$ contributes the pattern \pattern{scale=0.5}{2}{1/2,2/1}{0/1,2/0,1/0}, where
%%$\Phi(\pi)=\sigma$ and $\pi(i)$ is a antirecord in $\pi$ and $\sigma$.
%%Also, if the pair $(\pi(j),\pi(i))$ contributes the the pattern  \pattern{scale=0.5}{2}{1/2,2/1}{0/1,2/0,1/0}, 
%%then the pair $(\sigma(j),\pi(i))$ contributes the pattern \pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0}, where
%%$\Phi(\pi)=\sigma$ and $\pi(i)$ is an antirecord in $\pi$ and $\sigma$.
%\end{proof}


\begin{proof}[Proof of Theorem~\ref{main1}]
By \eqref{eq:defPhi}  the reverse of the mapping $\Phi$ is given by 
\begin{align}\label{eq:defPhireverse}
 \Phi^{-1}(\pi)=\varphi^{(i_l)}\circ\cdots\circ \varphi^{(i_2)}\circ
 \varphi^{(i_1)}(\pi).
 \end{align}
Theorem~\ref{main1} follows from Lemma~\ref{lem:opera1},  Lemma~\ref{lemma2.6:main1} and Lemma~\ref{lemma2.7:main1}.
\end{proof}


\begin{figure}[htb]
\begin{center}
\begin{tikzpicture}[scale=0.4] 	
\draw[step=1,lightgray,thin] (0,0) grid (9,9); 
	\tikzstyle{ridge}=[draw, line width=1, dotted, color=black] 
	\path[ridge] (1,2)--(2,5)--(3,7)--(4,1)--(5,8)--(6,9)--(7,4)--(8,6)--(9,3); 
	\tikzstyle{node0}=[circle, inner sep=2, fill=black] 
	\tikzstyle{node1}=[rectangle, inner sep=3, fill=mhcblue] 
	\tikzstyle{node2}=[diamond, inner sep=2, fill=davidsonred] 
	\node[node0] at (1,2) {}; 
	\node[node1] at (2,5) {}; 
	\node[node0] at (3,7) {}; 
	\node[node2] at (4,1) {}; 
	\node[node1] at (5,8) {}; 
	\node[node0] at (6,9) {}; 
	\node[node0] at (7,4) {}; 
	\node[node0] at (8,6) {}; 
	\node[node2] at (9,3) {}; 
	\draw[thick,blue,-] (0,3) to (9,3);	
	\draw[thick,blue,-] (9,3) to (9,9);		
	%\node[node0] at (9,1) {}; 
	%\node[node0] at (10,10) {}; 
	%\tikzstyle{hop1}=[draw, line width = 1.5, color=davidsonred,->]
	%\tikzstyle{hop2}=[draw, line width = 1.5, color=davidsonred,<-] 
	%\path[hop2] (0.8,8)--(1.5,8);
	%\path[hop1] (5.4,12)--(6.4, 12);
	%\path[hop2](5.2,11)--(6.5, 11);
	\tikzstyle{pi}=[above=-1] 
	%\node[pi] at (0,0) {0}; 
	\node[pi] at (1,0) {$2$}; 
	%\node[pi] at (2,0) {1^1}; 
	\node[pi] at (2,0) {$5$}; 
	\node[pi] at (3,0) {$7$}; 
         \node[pi, color=davidsonred] at (4,0) {$1$}; 
         \node[pi] at (5,0) {$8$};
	\node[pi] at (6,0) {$9$}; 
	\node[pi] at (7,0) {$4$}; 
	\node[pi] at (8,0) {$6$}; 	
         \node[pi, color=davidsonred] at (9,0) {$3$}; 

	%\node[pi] at (10,0) {$\infty$}; 
	\path[draw,line width=1,->] (11,5)--(15,5); 
	\node[pi] at (13,8){$\varphi_1^{(3)}$};
	\begin{scope}[shift={(16,0)}] 
	\draw[step=1,lightgray,thin] (0,0) grid (9,9); 
	\path[ridge] (1,2)--(2,8)--(3,6)--(4,1)--(5,5)--(6,4)--(7,9)--(8,7)--(9,3); 
	%\node[node0] at (0,0) {}; 
	\node[node0] at (1,2) {}; 
	\node[node1] at (2,8) {}; 
	\node[node0] at (3,6) {}; 
	\node[node2] at (4,1) {}; 
	\node[node1] at (5,5) {}; 
	\node[node0] at (6,4) {}; 
	\node[node0] at (7,9) {}; 
	\node[node0] at (8,7) {};
	\node[node2] at (9,3) {};
       % \node[pi] at (0,0) {0}; 
       	\draw[thick,blue,-] (0,3) to (9,3);	
	\draw[thick,blue,-] (9,3) to (9,9);	
		\draw[thick,blue,-] (0,1) to (4,1);	
	\draw[thick,blue,-] (4,1) to (4,9);				       
       
	\node[pi] at (1,0) {$2$}; 
	\node[pi] at (2,0) {$8$}; 
	\node[pi] at (3,0) {$6$}; 
      \node[pi, color=davidsonred] at (4,0) {$1$}; 
	\node[pi] at (5,0) {$5$}; 
	\node[pi] at (6,0) {$4$};
	\node[pi] at (7,0) {$9$}; 
	\node[pi] at (8,0) {$7$};
      \node[pi, color=davidsonred] at (9,0) {$3$}; 
	\end{scope}
	
		\path[draw,line width=1,->] (-5,-7)--(-1,-7); 
	\node[pi] at (-3,-4){$\varphi_2^{(3)}$};
	\begin{scope}[shift={(0,-12)}] 
	\draw[step=1,lightgray,thin] (0,0) grid (9,9); 
	\path[ridge] (1,2)--(2,6)--(3,8)--(4,1)--(5,5)--(6,4)--(7,9)--(8,7)--(9,3); 
	%\node[node0] at (0,0) {}; 
	\node[node0] at (1,2) {}; 
	\node[node1] at (2,6) {}; 
	\node[node0] at (3,8) {}; 
	\node[node2] at (4,1) {}; 
	\node[node1] at (5,5) {}; 
	\node[node0] at (6,4) {}; 
	\node[node0] at (7,9) {}; 
	\node[node0] at (8,7) {};
	\node[node2] at (9,3) {};
		\draw[thick,blue,-] (0,1) to (4,1);	
	\draw[thick,blue,-] (4,1) to (4,9);			
	
       % \node[pi] at (0,0) {0}; 
	\node[pi] at (1,0) {$2$}; 
	\node[pi] at (2,0) {$6$}; 
	\node[pi] at (3,0) {$8$}; 
      \node[pi, color=davidsonred] at (4,0) {$1$}; 
	\node[pi] at (5,0) {$5$}; 
	\node[pi] at (6,0) {$4$};
	\node[pi] at (7,0) {$9$}; 
	\node[pi] at (8,0) {$7$};
      \node[pi, color=davidsonred] at (9,0) {$3$}; 
	\end{scope}	
	
	
			\path[draw,line width=1,->] (11,-7)--(15,-7); 
	\node[pi] at (13,-4){$\varphi_1^{(1)}$};
	\begin{scope}[shift={(16,-12)}] 
	\draw[step=1,lightgray,thin] (0,0) grid (9,9); 
	\path[ridge] (1,8)--(2,6)--(3,2)--(4,1)--(5,5)--(6,4)--(7,9)--(8,7)--(9,3); 
	%\node[node0] at (0,0) {}; 
	\node[node0] at (1,8) {}; 
	\node[node1] at (2,6) {}; 
	\node[node0] at (3,2) {}; 
	\node[node2] at (4,1) {}; 
	\node[node1] at (5,5) {}; 
	\node[node0] at (6,4) {}; 
	\node[node0] at (7,9) {}; 
	\node[node0] at (8,7) {};
	\node[node2] at (9,3) {};
			\draw[thick,blue,-] (0,1) to (4,1);	
	\draw[thick,blue,-] (4,1) to (4,9);		
	
	
	
       % \node[pi] at (0,0) {0}; 
	\node[pi] at (1,0) {$8$}; 
	\node[pi] at (2,0) {$6$}; 
	\node[pi] at (3,0) {$2$}; 
      \node[pi, color=davidsonred] at (4,0) {$1$}; 
	\node[pi] at (5,0) {$5$}; 
	\node[pi] at (6,0) {$4$};
	\node[pi] at (7,0) {$9$}; 
	\node[pi] at (8,0) {$7$};
      \node[pi, color=davidsonred] at (9,0) {$3$}; 
	\end{scope}		
	
	
	
	
\end{tikzpicture}
\end{center}

\caption{The involution $\Phi$ on the permutation $257189463$}\label{fig:2}
\end{figure}




\begin{example}
We show the process of the involution $\Phi$ in Figure~\ref{fig:2}, 
For $\pi= 2\,5\,7\,1\,8\,9\,4\,6\,3$, we have $\AREC(\pi)=(4, 9)$.
 We proceed from right to left. 
 
\begin{enumerate} 
\item For position $9$ with value $3$, w have 
$w=5\, 7\, 8\, 9\, 4\, 6$ and $w'=8\,6\,5\,4\,9\,7$.
Thus  $\varphi_1^{(9)}: \pi\mapsto \pi'=2\,8\,6\,1\,5\,4\,9\,7\,3$. Next,  we have 
$w=8\,6$ and $w'=6\,8$. 
Thus $\varphi_2^{(9)}: \pi'\mapsto \pi''=2\,6\,8\,1\,5\,4\,9\,7\,3$.
%\item 
%For  position $4$ with value $1$, we have 
%$w=5\,9\,6\,7\,4$ and $w'=7\,4\,6\,5\,9$. So
%$\psi_1^{(6)}(\pi'')=2\,8\,1\,7\,4\,3\,6\,5\,9$.
%Next, we have $x=6\,5\,9$ and $x'=6\,9\,5$.
%Thus we have $\psi^{(6)}(\pi'')=2\,8\,1\,7\,4\,3\,6\,9\,5$.
\item 
For position $4$ with value $1$ we have 
$w=2\,6\,8$ and $w'=8\,6\,2$. Finally we obtain
$\Phi(\pi)=8\,6\,2\,1\,5\,4\,9\,7\,3$.
\end{enumerate}

Now, we check the mesh patterns.
\begin{itemize}
\item
First, $\varphi_1^{(9)}: \pi= 2\,5\,7\,1\,8\,9\,4\,6\,3 \mapsto \pi'=2\,8\,6\,1\,5\,4\,9\,7\,3$, 
 the pair $(8,3)$ of $\pi$ contributes the pattern $\pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0}$ without the patterns $\pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0,1/1}, \pattern{scale=0.5}{2}{1/2,2/1}{0/1,0/2,1/0,2/0,1/1}$), 
the pair $(5,3)$ of $\pi'$ contributes the pattern $\pattern{scale=0.5}{2}{1/2,2/1}{0/1,2/0,1/0}$ without the patterns $\pattern{scale=0.5}{2}{1/2,2/1}{0/1,1/0,2/0,1/2}, \pattern{scale=0.5}{2}{1/2,2/1}{0/1,0/2,2/0,1/0,1/2}$,
 the operations $\varphi_2^{(9)}, \varphi_1^{(1)}$ do not change the corresponding mesh pattens at position $9$ of $\pi'$.

\item
Second, $\varphi_2^{(9)}\circ \varphi_1^{(9)}:\pi=2\,5\,7\,1\,8\,9\,4\,6\,3 \mapsto \pi''=2\,6\,8\,1\,5\,4\,9\,7\,3$  it is easy to see that $257\sim 268$. The pair $(5, 1)$ of $\pi$ contributes the patterns $\pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0}, \pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0,1/1}$ without the pattern $\pattern{scale=0.5}{2}{1/2,2/1}{0/1,0/2,1/0,2/0,1/1}$,  
the pair $(6, 1)$ of $\pi''$ also contributes the pattern $\pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0}$,$\pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0,1/1}$ without the pattern $\pattern{scale=0.5}{2}{1/2,2/1}{0/1,0/2,1/0,2/0,1/1}$,  
$\varphi_1^{(1)}:\pi''=2\,6\,8\,1\,5\,4\,9\,7\,3 \mapsto \pi'''=8\,6\,2\,1\,5\,4\,9\,7\,3$, the pair $(6,1)$ of $\pi'''$ contributes the pattern $\pattern{scale=0.5}{2}{1/2,2/1}{0/1,2/0,1/0}$, $\pattern{scale=0.5}{2}{1/2,2/1}{0/1,1/0,2/0,1/2}$ without the pattern $\pattern{scale=0.5}{2}{1/2,2/1}{0/1,0/2,2/0,1/0,1/2}$.
 % the operations $\psi_2^{(6)}, \psi_1^{(9)}$ do not change the mesh pattens at $(4,3)$ of $\pi''$.
\end{itemize}



%We show the process of the involution $\phi$ in Figure~\ref{fig:2}, 
%For $\pi= 257189463$ with antirecord $3$ we have 
%$w=5\,7\,8\,9\,4\,6$ and $w'=8\,6\,5\,4\,9\,7$.
%%$\pi(2)=5$, we have $\pi^1(2)=8$, other elements are similar. 
%We have $\Phi_1(\pi)=2\,8\,6\,1\,5\,4\,9\,7\,3$.
%
%Consider the first shifted  left antirecord complement on the antirecord $3$,
%
%$$B=\{8, 6\}\,\,\text{and}\,\,B'=68,$$
%$\pi^1(2)=8$, we have $\pi^2(2)=6$, other elements are similar. We have $\pi^2=268154973$.
%
%%%%%%%%%%%%%%%%%%%%%%%
%Consider the second left antirecord complement on the antirecord $1$,
%
%
%$$A=\{2,6,8\}\,\,\text{and}\,\,A'=268,$$
%$\pi^2(2)=6$, we have $\pi^3(2)=6$. We have $\pi^3=862154973$.
%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%For $\pi=257189463$, the \pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0} and \pattern{scale=0.5}{2}{1/2,2/1}{0/1,2/0,1/0} appear in the following pairs
%$$\{(2,1), (5, 1), (7,1), (8,3),(9,3)\}\,\,\text{and}\,\,\{(2,1), (4,3)\}.$$
%For $\pi=862154973$, the pattern \pattern{scale=0.5}{2}{1/2,2/1}{0/1,2/0,1/0} and \pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0} appear in the following pairs
%$$\{(8,1), (6, 1), (2,1), (5,3),(4,3)\}\,\,\text{and}\,\,\{(8,1), (9,3)\}.$$
%Then we have
%$$\pattern{scale=0.5}{2}{1/2,2/1}{0/1,2/0,1/0}(862154973)=\pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0}(257189463)=5\,\, \text{and}\,\,\pattern{scale=0.5}{2}{1/2,2/1}{0/1,2/0,1/0}(257189463)=\pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0}(862154973)=2.$$
%%%%%%%%
%For $\pi=257189463$, the patterns $\pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0,1/1}$ and $\pattern{scale=0.5}{2}{1/2,2/1}{0/1,2/0,1/0,1/2}$ appear in the following pairs
%$$\{(2,1), (5, 1), (7,1)\}\,\,\text{and}\,\,\,\emptyset.$$
%For $\pi=862154973$, the patterns $\pattern{scale=0.5}{2}{1/2,2/1}{0/1,2/0,1/0,1/2}$ and $\pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0,1/1}$ appear in the following pairs
%$$\{(8,1), (6, 1), (2,1)\}\,\,\text{and}\,\,\,\emptyset.$$
%Then we have
%$$\pattern{scale=0.5}{2}{1/2,2/1}{0/1,2/0,1/0,1/2}(862154973)=\pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0,1/1}(257189463)=3\,\, \text{and}\,\,\pattern{scale=0.5}{2}{1/2,2/1}{0/1,2/0,1/0,1/2}(257189463)=\pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0,1/1}(862154973)=0.$$
%%%%%%%%%%%%%%%%%%%%%%%%
%For $\pi=257189463$, the patterns  $\pattern{scale=0.5}{2}{1/2,2/1}{0/1,0/2,2/0,1/0,1/1}$ and  $\pattern{scale=0.5}{2}{1/2,2/1}{0/1,0/2,1/0,2/0,1/2}$ and appear in the following pairs
%$$\{2,1\} \,\,\text{and}\,\,\,  \emptyset .$$
%For $\pi=862154973$, the patterns $\pattern{scale=0.5}{2}{1/2,2/1}{0/1,0/2,1/0,2/0,1/2}$ and $\pattern{scale=0.5}{2}{1/2,2/1}{0/1,0/2,2/0,1/0,1/1}$ appear in the following pairs
%$$\{(8,1)\}\,\,\text{and}\,\,\,\emptyset.$$
%Then we have
%$$\pattern{scale=0.5}{2}{1/2,2/1}{0/2,0/1,2/0,1/0,1/2}(862154973)=\pattern{scale=0.5}{2}{1/2,2/1}{0/2,0/1,1/0,2/0,1/1}(257189463)=1\,\, \text{and}\,\,\pattern{scale=0.5}{2}{1/2,2/1}{0/2,0/1,2/0,1/0,1/2}(257189463)=\pattern{scale=0.5}{2}{1/2,2/1}{0/1,0/2,1/0,2/0,1/1}(862154973)=0.$$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%(\pattern{scale=0.5}{2}{1/2,2/1}{0/1,2/0,1/0}, \pattern{scale=0.5}{2}{1/2,2/1}{0/1,1/0,2/0,1/2}, \pattern{scale=0.5}{2}{1/2,2/1}{0/1,0/2,2/0,1/0,1/2})(\pi)=%\pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0}
%(\pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0}, \pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0,1/1}, \pattern{scale=0.5}{2}{1/2,2/1}{0/1,0/2,1/0,2/0,1/1})(\phi(\pi))





\end{example}





\section{Proof outlines of Theorem~\ref{main2}}
First we introduce two mappings different from Section~2. For $\pi\in S_n$, 
%let $\AREC(\pi)=(i_1,i_2,\ldots, i_l)$ be the sequence of antirecord positions  of $\pi$.
recall that $\AREC(\pi)=(i_1,i_2,\ldots, i_l)$ be the sequence of antirecord positions  of $\pi$ from left to right. 
For any antirecord position $i_k$ we define two mappings
\begin{subequations}\label{eq:defanti2}
\begin{align}
&\psi_1^{(i_k)}: \pi\mapsto \pi'\\
&\psi_2^{(i_k)}: \pi\mapsto \pi''
\end{align}
\end{subequations}
as follows: 
\begin{itemize}
\item let  $w=w_1\ldots w_r$ is  the subword of $\pi$ 
consisting of letters  greater than $\pi(i_k)$ on the right side of $\pi(i_{k-1})$ (resp. $\pi(i_{k})$)   with $\pi(i_0)=0$;
\item let
$w'=w_1'\ldots w_r'$  be the word obtained by substituting 
the $j$th largest letter  with  the $j$th smallest letter in $w$ for $j=1,\ldots, r$;
\item let  $\pi'$ (resp. $\pi''$) is defined to be the word obtained by 
replacing $w_j$ with  $w_j'$ in $\pi$. 
\end{itemize}
 Note that  $\pi'(i_k)=\pi(i_k)$.



\begin{lem}\label{lem:opera2}
For any antirecord positions $i_{k-1}$ and $i_k$ of $\pi\in S_n$ 
the mappings $\psi_1^{(i_k)}$  and  $\psi_2^{(i_k)}$ are involutions and 
commutate, namely,
\begin{equation}\label{eq1:lem3.1}
\psi_1^{(i_k)}\circ \psi_1^{(i_k)}(\pi)=
\psi_2^{(i_k)}\circ \psi_2^{(i_k)}(\pi)=\pi
\end{equation}
and
\begin{equation}\label{eq2:lem3.1}
\psi_2^{(i_k)}\circ \psi_1^{(i_k)}(\pi)=\psi_1^{(i_k)}\circ \psi_2^{(i_k)}(\pi).
\end{equation}
Let  $\psi^{(i_k)}=\psi_2^{(i_k)}\circ \psi_1^{(i_k)}$.  Then $\psi^{(i_k)}(\pi)$ and $\pi$ have the same  sequence of antirecord 
positions.
\end{lem}
 
%\begin{proof}
%From the definitions of $\psi_1^{(i_k)}$ and $\psi_2^{(i_k)}$ in Eq.~\eqref{eq:defanti2}, it is easy to check Eq.~\eqref{eq1:lem3.1} holds and 
%$$
%\psi_2^{(i_k)}\circ \psi_1^{(i_k)}(\pi) \sim \psi_1^{(i_k)}\circ \psi_2^{(i_k)}(\pi).
%$$
%Since the set of letters greater than $\pi(i_k)$ on the right of $\pi(i_{k-1})$ are invariant under the operation $\psi_1^{(i_k)}$ and  $\psi_2^{(i_k)}$ on $\pi$, we obtain Eq.~\eqref{eq2:lem3.1} immediately. 
%\end{proof}









% with $\psi^{(i_k)}=\phi_2^{(i_k)}\circ \phi_1^{(i_k)}$
%for $k=1, \ldots, l$.
% 

\begin{lem}\label{lemma2:main2} 
For $\pi\in S_n$ with $\AREC(\pi)=\{i_1, \ldots, i_l\}$. For $k=2, \ldots, l$ the mappings  
$\psi^{(i_{k-1})}$ and $\psi^{(i_k)}$ commute, i.e., 
$$
\psi^{(i_k)}\circ \psi^{(i_{k-1})}(\pi)=\psi^{(i_{k-1})}\circ \psi^{(i_k)}(\pi).
%\varphi^{(i_{k-1})}\circ \varphi^{(i_k)}(\pi)=\varphi^{(i_k)}\circ \varphi^{(i_{k-1})}(\pi).
$$
\end{lem}
%\begin{proof}
%For  the permutation $\pi=\pi(1)\ldots \pi(n)$ we write
%$\pi=u\pi(i_{k-1})v\pi(i_{k})w$ and 
%%For convenience, we introduce the following notations:
%\begin{subequations}
%\begin{align}
%\psi^{(i_{k-1})}(\pi):&=\pi'=u'\pi(i_{k-1})v'\pi'(i_{k})w'\label{eq:barf}\\
%\psi^{(i_k)}\circ \psi^{(i_{k-1})} (\pi):&=\tilde \pi= u'\pi(i_{k-1})\tilde v\pi'(i_{k})\tilde w,\label{eq:bargf}\\
%\psi^{(i_k)}(\pi):&=\pi''=u\pi(i_{k-1})v''\pi(i_{k})w'',\label{eq:barg}\\
%\psi^{(i_{k-1})}\circ \psi^{(i_k)} (\pi):&=\hat \pi= u'\pi(i_{k-1})\hat v \hat \pi(i_{k})\hat w.\label{eq:barfg}
%\end{align}
%\end{subequations}
%By the definition of  $\psi^{(i_{k})}$, we have the following facts.
%\begin{enumerate}
%\item Applying $\psi^{(i_{k-1})}$ to $\pi$ (cf.\ \eqref{eq:barf}) we have   $v'\pi'(i_k)w'\sim v\pi(i_k)w$;
%\item Applying $\psi^{(i_k)}$ to $\psi^{(i_{k-1})}(\pi)$ and $\pi$ (cf.\eqref{eq:bargf} and \eqref{eq:barg}), respectively,  we take complement of $v'$ and $v$ once, while twice for
%$w'$ and $w$. By Lemma~\ref{property:compl} and (1) we get
%$\tilde v\pi'(i_k)\tilde w \sim v''\pi(i_k)w''$;
% \item Applying $\psi^{(i_{k-1})}$ to $\psi^{(i_{k})}(\pi)$ (cf.\ \eqref{eq:barg} and \eqref{eq:barfg}),  by (1) we have   $\hat v\hat\pi(i_k)\hat w\sim v''\pi(i_k)w''$,  combining with 
% by (2) yields 
% $\hat v\hat \pi(i_k)\hat w\sim \tilde v\pi'(i_k)\tilde w$.
%  \item 
% Let $\{w\}$ denote the set of letters in $w$.  
% \begin{itemize}
% \item Applying $\psi^{(i_{k})}$ to $\psi^{(i_{k-1})}(\pi)$ we have 
% $\{u'v'\pi'(i_k)w'\}=\{u'\tilde v\pi'(i_k)\tilde w\}$, 
% \item    applying $\psi^{(i_{k-1})}$ to $\pi$ we have 
% $\{u'v'\pi'(i_{k})w'\}=\{ uv\pi(i_{k})w\}$,
%  \item applying $\psi^{(i_{k})}$ to $\pi$ we have 
% $\{uv''\pi(i_{k})w''\}=\{ uv \pi(i_{k})w\}$,
% \item applying $\psi^{(i_{k-1})}$ to $\psi^{(i_{k})}(\pi)$ we have 
% $\{u'\hat v\hat \pi(i_{k})\hat w\}=\{u v''\pi(i_{k})w''\}$,
% \item it follows that $ \{u'\hat v\hat \pi(i_k)\hat w\}=\{u'\tilde v\pi'(i_k)\tilde w\}$ and  $\{\hat v \hat \pi(i_k)\hat w\}=\{\tilde v \pi'(i_k) \tilde w\}$.
%\end{itemize}
% 
%\item  It follows from (3) and (4)  that $\hat v \hat \pi(i_k)\hat w=\tilde v \pi'(i_k) \tilde w$, that is, 
% Thus 
% $\hat v= \tilde v$, $\hat \pi(i_k)=\pi'(i_k)$ and $\hat w=\tilde w$.
% \end{enumerate}
% \end{proof}




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{lem}\label{lem:opera2bis} The mapping $\psi^{(i)}$ is an involution  such that for  $\pi\in S_n$ and $r\neq k$
\begin{subequations}
\begin{align}\label{eq:psia}
(\pattern{scale=0.5}{2}{1/2,2/1}{1/0,2/0,2/2}, \pattern{scale=0.5}{2}{1/2,2/1}{1/0, 1/1,2/0,2/2},\pattern{scale=0.5}{2}{1/2,2/1}{1/0, 1/1, 1/2,2/0,2/2})_k\;\pi
=
(\pattern{scale=0.5}{2}{1/2,2/1}{2/1,1/0,2/0}, \pattern{scale=0.5}{2}{1/2,2/1}{2/1,1/0,2/0,1/2}, \pattern{scale=0.5}{2}{1/2,2/1}{2/1,1/0,2/0,1/1,1/2})_k\;\psi^{(i_k)}(\pi),\\
(\pattern{scale=0.5}{2}{1/2,2/1}{1/0,2/0,2/2}, \pattern{scale=0.5}{2}{1/2,2/1}{1/0, 1/1,2/0,2/2},\pattern{scale=0.5}{2}{1/2,2/1}{1/0, 1/1, 1/2,2/0,2/2})_r\;\pi
=
(\pattern{scale=0.5}{2}{1/2,2/1}{1/0,2/0,2/2}, \pattern{scale=0.5}{2}{1/2,2/1}{1/0, 1/1,2/0,2/2},\pattern{scale=0.5}{2}{1/2,2/1}{1/0, 1/1, 1/2,2/0,2/2})_r\;\psi^{(i_k)}(\pi),\label{eq:psib}\\
(\pattern{scale=0.5}{2}{1/2,2/1}{2/1,1/0,2/0}, \pattern{scale=0.5}{2}{1/2,2/1}{2/1,1/0,2/0,1/2}, \pattern{scale=0.5}{2}{1/2,2/1}{2/1,1/0,2/0,1/1,1/2})_r\;\pi
=
(\pattern{scale=0.5}{2}{1/2,2/1}{2/1,1/0,2/0}, \pattern{scale=0.5}{2}{1/2,2/1}{2/1,1/0,2/0,1/2}, \pattern{scale=0.5}{2}{1/2,2/1}{2/1,1/0,2/0,1/1,1/2})_r\;\psi^{(i_k)}(\pi).
\label{eq:psic}
\end{align}
\end{subequations}
where $(\textrm{pattern})_k$ means the number of the patterns between $\pi(i_{k-1})$ and $\pi(i_k)$.
\end{lem}


%\begin{proof}
%If  the pair
%$(\pi(j),\pi(i_k))$ with $j<i_k$ contributes the pattern $\pattern{scale=0.5}{2}{1/2,2/1}{2/2,2/0,1/0}$(resp. $\pattern{scale=0.5}{2}{1/2,2/1}{1/1,1/0,2/0,2/2}, \pattern{scale=0.5}{2}{1/2,2/1}{1/1,2/2,2/0,1/0,1/2})$,
%then  $j>i_{k-1}$ because $\pi(i_{k-1})<\pi(i_{k})$ and 
%$\pi(i)>\pi(i_k)$ for $j\leq i<i_k$.
%Also, for  $j<i_k<i$, we have the equivalence 
% $$
%  \pi(i_k)<\pi(i)<\pi(j)\Longleftrightarrow \pi(i_k)<\psi_1^{(i_k)}(\pi(j))<\psi_1^{(i_k)}(\pi(i)),
%   $$
%as $\psi_2^{(i_k)}$ will affect only the letters at the right of 
%$\pi(i_{k})$. Thus  we have proved \eqref{eq:psia}.
%
%Next, recall that the operation $\psi^{(i_k)}$ keeps the 
%sequence of antirecord positions. The two identities \eqref{eq:psib} and \eqref{eq:psic} are clear if $r<k$.
%Assume that $r>k$ and $[\pi]_{>r}=\pi(i_k+1)\ldots \pi(n)$. 
%%Let $w=w_1\ldots w_r$ be the subword of $[\pi]_{>r}$ consisting of letters greater 
%%than $\pi(i_{k-1})$ on the right side of $\pi(i_k)$. 
%By Lemma~\ref{property:compl} after two operations
%$\psi_1^{(i_k)}$ and $\psi_2^{(i_k)}$  the permutation
%$\psi^{(i_k)}([\pi]_{>r})$ is  isomorphic with $[\pi]_{>r}$. This proves \eqref{eq:psib} and \eqref{eq:psic}.
% 
%
%%the pair $(\tau(j),\tau(i))$ contributes the pattern
%% $\pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0}$ (resp. $\pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0,1/1}, \pattern{scale=0.5}{2}{1/2,2/1}{0/1,0/2,1/0,2/0,1/1}$), 
%%%where$\phi(\pi)=\sigma$ and $\pi(i)$ is a antirecord in $\pi$ and $\sigma$.
%%Also, If the pair $(\pi(j),\pi(i))$ contributes the the pattern  $\pattern{scale=0.5}{2}{1/2,2/1}{0/1,2/0,1/0}$ (resp. $\pattern{scale=0.5}{2}{1/2,2/1}{0/1,1/0,2/0,1/2}, \pattern{scale=0.5}{2}{1/2,2/1}{0/1,0/2,2/0,1/0,1/2}$), 
%%then the pair $(\sigma(j),\sigma(i))$ contributes the pattern $\pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0}$ (resp. $\pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0,1/1}, \pattern{scale=0.5}{2}{1/2,2/1}{0/1,0/2,1/0,2/0,1/1}$). 
%%%where $\phi(\pi)=\sigma$ and $\pi(i)$ is a antirecord in $\pi$ and $\sigma$.
%%It is easy to see the algorithm $\Phi$ is a bijection by considering the reverse process of algorithm $\Phi$. 
%%Moreover, it is easy to check 
%%$$\Phi^2(\pi)=id\,\, \text{if}\,\, \pi\in S_n.$$
%%If the pair $(\pi(j),\pi(i))$ contributes the the pattern \pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0}, 
%%then the pair $(\sigma(j),\pi(i))$ contributes the pattern \pattern{scale=0.5}{2}{1/2,2/1}{0/1,2/0,1/0}, where
%%$\Phi(\pi)=\sigma$ and $\pi(i)$ is a antirecord in $\pi$ and $\sigma$.
%%Also, if the pair $(\pi(j),\pi(i))$ contributes the the pattern  \pattern{scale=0.5}{2}{1/2,2/1}{0/1,2/0,1/0}, 
%%then the pair $(\sigma(j),\pi(i))$ contributes the pattern \pattern{scale=0.5}{2}{1/2,2/1}{0/2,1/0,2/0}, where
%%$\Phi(\pi)=\sigma$ and $\pi(i)$ is an antirecord in $\pi$ and $\sigma$.
%\end{proof}
\begin{proof}[Proof of Theorem~\ref{main2}]

For  $\pi\in S_n$ and 
%with  sequence of antirecord positions 
$\AREC(\pi)=(i_1,i_2,\ldots, i_l),$ 
 we define the operation  $\Psi$ on $\pi$ by 
 \begin{align}\label{eq:defPsi}
 \Psi(\pi)=\psi^{(i_l)}\circ\cdots\circ \psi^{(i_2)}\circ
 \psi^{(i_1)}(\pi).
 \end{align}

By \eqref{eq:defPsi} the mapping $\Psi$ is reversible with reverse
$$
\Psi^{-1}(\pi)=\psi^{(i_1)}\circ\psi^{(i_2)}\circ\ldots \circ\psi^{(i_l)}(\pi).
$$
%with $\AREC(\pi)=(i_1,i_2,\ldots, i_l)$.
% is  the sequence of antirecord positions  of $\pi$.
Theorem~\ref{main2} follows from Lemma~\ref{lem:opera2}, Lemma~\ref{lemma2:main2} and Lemma~\ref{lem:opera2bis}.
\end{proof}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
An example of the involution $\Psi$ could be found in \cite[Example~3.4]{HZ21}.





%\begin{figure}[htb]\scriptsize
%\begin{center}
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%\draw[step=1,lightgray,thin] (0,0) grid (9,9); 
%	\tikzstyle{ridge}=[draw, line width=1, dotted, color=black] 
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%	\tikzstyle{node0}=[circle, inner sep=2, fill=black] 
%	\tikzstyle{node1}=[rectangle, inner sep=3, fill=mhcblue] 
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%	%\tikzstyle{hop2}=[draw, line width = 1.5, color=davidsonred,<-] 
%	%\path[hop2] (0.8,8)--(1.5,8);
%	%\path[hop1] (5.4,12)--(6.4, 12);
%	%\path[hop2](5.2,11)--(6.5, 11);
%	\tikzstyle{pi}=[above=-1] 
%	%\node[pi] at (0,0) {0}; 
%	\node[pi] at (1,0) {$9$}; 
%	%\node[pi] at (2,0) {1^1}; 
%	\node[pi] at (2,0) {$3$}; 
%         \node[pi, color=davidsonred] at (3,0) {$1$}; 
%         \node[pi] at (4,0) {$5$};
%	\node[pi] at (5,0) {$8$}; 
% \node[pi, color=davidsonred] at (6,0) {$2$}; 	
%	\node[pi] at (7,0) {$6$}; 
%	\node[pi] at (8,0) {$7$}; 	
%         \node[pi, color=davidsonred] at (9,0) {$4$}; 
%
%	%\node[pi] at (10,0) {$\infty$}; 
%	\path[draw,line width=1,->] (11,5)--(15,5); 
%	\node[pi] at (13,8){$\psi_1^{(3)}$};
%	\begin{scope}[shift={(16,0)}] 
%	\draw[step=1,lightgray,thin] (0,0) grid (9,9); 
%	\path[ridge] (1,2)--(2,8)--(3,1)--(4,6)--(5,3)--(6,9)--(7,5)--(8,4)--(9,7); 
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%       % \node[pi] at (0,0) {0}; 
%       	\draw[thick,blue,-] (0,1) to (9,1);	
%	\draw[thick,blue,-] (3,1) to (3,9);	
%		%\draw[thick,blue,-] (0,1) to (4,1);	
%	%\draw[thick,blue,-] (4,1) to (4,9);				       
%       
%	\node[pi] at (1,0) {$2$}; 
%	\node[pi] at (2,0) {$8$}; 
% \node[pi, color=davidsonred] at (3,0) {$1$}; 
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%      \node[pi, color=davidsonred] at (9,0) {$7$}; 
%	\end{scope}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%	
%		\path[draw,line width=1,->] (-5,-7)--(-1,-7); 
%	\node[pi] at (-3,-4){$\psi_2^{(3)}$};
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%	
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%	
%	
%			\path[draw,line width=1,->] (11,-7)--(15,-7); 
%	\node[pi] at (13,-4){$\psi_1^{(6)}$};
%	\begin{scope}[shift={(16,-12)}] 
%	
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%	
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%      \node[pi, color=davidsonred] at (9,0) {$9$}; 
%	\end{scope}		
%	
%	%%%%%%%%%%%%%%%%%%%%%%%%%
%		\path[draw,line width=1,->] (-5,-19)--(-1,-19); 
%	\node[pi] at (-3,-16){$\psi_2^{(6)}$};
%	\begin{scope}[shift={(0,-24)}] 	
%		\draw[step=1,lightgray,thin] (0,0) grid (9,9); 
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%	
%       % \node[pi] at (0,0) {0}; 
%	\node[pi] at (1,0) {$2$}; 
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%      \node[pi, color=davidsonred] at (9,0) {$5$}; 
%	\end{scope}		
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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%	\node[pi] at (13,-16){$\psi_1^{(9)}$};
%	\begin{scope}[shift={(16,-24)}] 
%			\draw[step=1,lightgray,thin] (0,0) grid (9,9); 
%	\path[ridge] (1,2)--(2,8)--(3,1)--(4,7)--(5,4)--(6,3)--(7,9)--(8,6)--(9,5); 
%	%\node[node0] at (0,0) {}; 
%	\node[node1] at (1,2) {}; 
%	\node[node0] at (2,8) {}; 
%	\node[node2] at (3,1) {}; 
%	\node[node0] at (4,7) {}; 
%	\node[node1] at (5,4) {}; 
%	\node[node2] at (6,3) {}; 
%	\node[node0] at (7,9) {}; 
%	\node[node0] at (8,6) {};
%	\node[node2] at (9,5) {};
%	%\draw[thick,blue,-] (6,3) to (6,9);			
%	% \node[pi] at (0,0) {0}; 
%	\node[pi] at (1,0) {$2$}; 
%	\node[pi] at (2,0) {$8$}; 
%	\node[pi, color=davidsonred] at (3,0) {$1$}; 
%      \node[pi] at (4,0) {$7$}; 
%	\node[pi] at (5,0) {$4$}; 
%	\node[pi, color=davidsonred] at (6,0) {$3$};
%	\node[pi] at (7,0) {$9$}; 
%	\node[pi] at (8,0) {$6$};
%      \node[pi, color=davidsonred] at (9,0) {$5$}; 
%	\end{scope}	
%	
%	
%\end{tikzpicture}
%\end{center}
%
%\caption{The involution $\Psi$ on the permutation $931582674$}\label{fig:3}
%\end{figure}



%\begin{proof}[Proof of ]


%Let $\bar{\phi}$ denote the composition of the operations  \emph{shifted right antirecord complement} and 
%\emph{right antirecord complement}. Let $\bar{\phi}_{i_k}$ donote the $\bar{\phi}$ act on $\pi(i_k)$. Define $\Psi:=\bar{\phi}_{i_k}\bar{\phi}_{i_{k-1}}\ldots\bar{\phi}_{i_1}$. By Lemma~\ref{lem:opera2}, we obtain the equidistributions of Theorem~\ref{thm:main2}.




%It is easy to see the algorithm $\phi_2$ is a bijection by considering the reverse process of algorithm $\phi_2$. Moreover, it is easy to check 
%$$\phi_2^2(\pi)=id\,\, \text{if}\,\, \pi\in S_n.$$
%If the pair $(\pi(j),\pi(i))$ contributes the the pattern $\pattern{scale=0.5}{2}{1/2,2/1}{2/2,1/0,2/0}$, 
%then the pair $(\sigma(j),\sigma(i))$ contributes the pattern \pattern{scale=0.5}{2}{1/2,2/1}{2/1,2/0,1/0}, where
%$\pi(i)$, $\sigma(i)$ are antirecords in $\pi$ and $\sigma$ and $\phi_2(\pi)=\sigma$. 
% .Also, If the pair $(\pi(j),\pi(i))$ contributes the the pattern  $\pattern{scale=0.5}{2}{1/2,2/1}{2/1,2/0,1/0}$, 
%then the pair $(\sigma(j),\pi(i))$ contributes the pattern $\pattern{scale=0.5}{2}{1/2,2/1}{2/2,1/0,2/0}$, where
%$\pi(i)$, $\sigma(i)$ are antirecords in $\pi$ and $\sigma$ and $\phi_2(\pi)=\sigma$. 


%\begin{example} Figure~\ref{fig:3}.
%For $\pi= 9\,3\,1\,5\,8\,2\,6\,7\,4$, we have $\AREC(\pi)=(3, 6,9)$.
% We procced from left to right. 
% 
%\begin{enumerate} 
%\item For position $3$ with value $1$, we have $w=9\,3\,5\, 8\, 2\, 6\, 7\, 4$ and $w^c=2\,8\,6\,3\,9\,5\,4\,7$ Thus  $\psi_1^{(3)}: \pi\mapsto \pi'=2\,8\,1\,6\,3\,9\,5\,4\,7$. 
%Next, we have $w=6\,3\, 9\, 5\, 4\, 7$ and $w^c=5\,9\,3\,6\,7\,4$
%%The subword of $\pi$ 
%%c%onsisting of letters  greater than $\pi(3)$ on the right side of $\pi(3)$
%%is $6\,3\, 9\, 5\, 4\, 7$ 
%%since $(6\,3\, 9\, 5\, 4\, 7)^c=5\,9\,3\,6\,7\,4$. 
%Thus $\psi_2^{(3)}: \pi'\mapsto \pi''=2\,8\,1\,5\,9\,3\,6\,7\,4$.
%\item 
%For  position $6$ with value $3$, we have 
%$w=5\,9\,6\,7\,4$ and $w^c=7\,4\,6\,5\,9$. So
%$\psi_1^{(6)}(\pi'')=2\,8\,1\,7\,4\,3\,6\,5\,9$. Next, we have $w=6\,5\,9$ and $w^c=6\,9\,5$.
%Thus we have $\psi^{(6)}(\pi'')=2\,8\,1\,7\,4\,3\,6\,9\,5$.
%\item 
%For position $9$ with value $5$ we have 
%$w=6\,9$ and $w^c=9\,6$. Finally we obtain
%$\Psi(\pi)=2\,8\,1\,7\,4\,3\,9\,6\,5$.
%\end{enumerate}
%
%Now, we check the mesh patterns.
%\begin{itemize}
%\item
%First, $\psi_1^{(3)}: \pi=9\,3\,1\,5\,8\,2\,6\,7\,4 \mapsto \pi'=2\,8\,1\,6\,3\,9\,5\,4\,7$, the pair $(9, 1)$ contributes the pattern $\pattern{scale=0.5}{2}{1/2,2/1}{2/2,2/0,1/0}$ without the patterns $\pattern{scale=0.5}{2}{1/2,2/1}{1/1,1/0,2/0,2/2}$, $\pattern{scale=0.5}{2}{1/2,2/1}{2/2,1/2,2/0,1/0,1/1}$, 
%then the pair $(2,1)$ of $\pi'$ contributes the pattern $\pattern{scale=0.5}{2}{1/2,2/1}{2/1,2/0,1/0}$ without the patterns $\pattern{scale=0.5}{2}{1/2,2/1}{1/2,1/0,2/0,2/1}$, $\pattern{scale=0.5}{2}{1/2,2/1}{2/1,1/2,1/0,2/0,1/1}$, the operations $\psi_2^{(i)}(i=3,6), \psi_1^{(j)}(j=6,9)$ do not change the corresponding mesh pattens at $(2,1)$ of $\pi'$.
%\item
%Second, $\psi_2^{(3)}\circ \psi_1^{(3)}:\pi=9\,3\,1\,5\,8\,2\,6\,7\,4 \mapsto \pi''=2\,8\,1\,5\,9\,3\,6\,7\,4$,
%it is easy to see $5\,8\,2\,6\,7\,4\sim 5\,9\,3\,6\,7\,4$. The pair $(8, 2)$ of $\pi$ contributes the pattern $\pattern{scale=0.5}{2}{1/2,2/1}{2/2,2/0,1/0}$ (resp. $\pattern{scale=0.5}{2}{1/2,2/1}{1/1,1/0,2/0,2/2}$, $\pattern{scale=0.5}{2}{1/2,2/1}{2/2,1/2,2/0,1/0,1/1}$ ),  then the pair $(9, 3)$ of $\pi''$ also contributes the pattern $\pattern{scale=0.5}{2}{1/2,2/1}{2/2,2/0,1/0}$ (resp. $\pattern{scale=0.5}{2}{1/2,2/1}{1/1,1/0,2/0,2/2}$, $\pattern{scale=0.5}{2}{1/2,2/1}{2/2,1/2,2/0,1/0,1/1}$ ),  
% $\psi_1^{(6)}:\pi''=2\,8\,1\,5\,9\,3\,6\,7\,4 \mapsto \pi'''=2\,8\,1\,7\,4\,3\,6\,5\,9$, the pair $(4,3)$ contributes the pattern $\pattern{scale=0.5}{2}{1/2,2/1}{2/1,2/0,1/0}$(resp.  $\pattern{scale=0.5}{2}{1/2,2/1}{1/2,1/0,2/0,2/1}$, $\pattern{scale=0.5}{2}{1/2,2/1}{2/1,1/2,1/0,2/0,1/1}$), the operations $\psi_2^{(6)}, \psi_1^{(9)}$ do not change the corresponding mesh pattens at $(4,3)$ of $\pi'''$.
%\end{itemize}

%\begin{itemize}
%\item
%
%For $\pi=391582674$, the \pattern{scale=0.5}{2}{1/2,2/1}{1/0,2/1,2/0} and \pattern{scale=0.5}{2}{1/2,2/1}{1/0,2/0,2/2} appear in the following pairs
%$$\{(6, 4), (7,4)\}\,\,\text{and}\,\,\{(9,1), (8,2), (7,4)\}.$$
%For $\pi=821743965$, the patterns \pattern{scale=0.5}{2}{1/2,2/1}{1/0,2/0,2/2} and \pattern{scale=0.5}{2}{1/2,2/1}{1/0,2/1,2/0}  appear in the following pairs
%$$\{(6,5), (9,5)\}\,\,\, \text{and}\,\,\,\{(2,1),(4,3),(9,5)\}.$$
%Then we have
%$$\pattern{scale=0.5}{2}{1/2,2/1}{1/0,2/1,2/0}(391582674)=\pattern{scale=0.5}{2}{1/2,2/1}{1/0,2/0,2/2}(821743965)=2\,\, \text{and}\,\,\pattern{scale=0.5}{2}{1/2,2/1}{1/0,2/1,2/0}(821743965)=\pattern{scale=0.5}{2}{1/2,2/1}{1/0,2/0,2/2}(391582674)=3.$$
%
%\item
%For $\pi=391582674$, the \pattern{scale=0.5}{2}{1/2,2/1}{1/0,1/2,2/1,2/0} and \pattern{scale=0.5}{2}{1/2,2/1}{1/0,1/1,2/0,2/2} appear in the following pairs
%$$\{(7,4)\}\,\,\text{and}\,\,\{(9,1), (8,2), (7,4)\}.$$
%For $\pi=821743965$, the patterns \pattern{scale=0.5}{2}{1/2,2/1}{1/0,1/1,2/0,2/2} and \pattern{scale=0.5}{2}{1/2,2/1}{1/0,1/2,2/1,2/0}  appear in the following pairs
%$$\{(6,5)\}\,\,\, \text{and}\,\,\,\{(2,1),(4,3),(9,5)\}..$$
%Then we have
%$$\pattern{scale=0.5}{2}{1/2,2/1}{1/0,1/2,2/1,2/0}(391582674)=\pattern{scale=0.5}{2}{1/2,2/1}{1/0,1/1,2/0,2/2}(821743965)=1\,\, \text{and}\,\,\pattern{scale=0.5}{2}{1/2,2/1}{1/0,1/2,2/1,2/0}(821743965)=\pattern{scale=0.5}{2}{1/2,2/1}{1/0,1/1,2/0,2/2}(391582674)=3.$$
%
%\item 
%For $\pi=391582674$, the \pattern{scale=0.5}{2}{1/2,2/1}{1/0,1/1,1/2,2/1,2/0} and \pattern{scale=0.5}{2}{1/2,2/1}{1/0,1/1,1/2,2/0,2/2} appear in the following pairs
%$$\emptyset\,\,\text{and}\,\,\{(9,1), (8,2), (7,4)\}.$$
%For $\pi=821743965$, the patterns \pattern{scale=0.5}{2}{1/2,2/1}{1/0,1/1,1/2,2/0,2/2} and \pattern{scale=0.5}{2}{1/2,2/1}{1/0,1/1,1/2,2/1,2/0}  appear in the following pairs
%$$\emptyset\,\,\, \text{and}\,\,\,\{(2,1),(4,3),(9,5)\}..$$
%Then we have
%$$\pattern{scale=0.5}{2}{1/2,2/1}{1/0,1/1,1/2,2/1,2/0}(391582674)=\pattern{scale=0.5}{2}{1/2,2/1}{1/0,1/1,1/2,2/0,2/2}(821743965)=0\,\, \text{and}\,\,\pattern{scale=0.5}{2}{1/2,2/1}{1/0,1/1,1/2,2/1,2/0}(821743965)=\pattern{scale=0.5}{2}{1/2,2/1}{1/0,1/1,1/2,2/0,2/2}(391582674)=3.$$
%\end{itemize}


%\end{example}

%\acknowledgements{\lipsum[20]}

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\end{document}
