\documentclass[12pt,reqno,dvipsnames]{article} %\usepackage[usenames]{color} \usepackage[rgb]{xcolor} \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{amscd} \usepackage{graphicx} \usepackage{tikz} \usepackage{enumitem} \usepackage{mathdots} \usepackage{breqn} \usepackage{psfig} \usepackage{latexsym} \usepackage{epsf} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{breakurl} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.1in} \setlength{\textheight}{8.4in} \newcommand{\seqnum}[1]{\href{https://oeis.org/#1}{\underline{#1}}} \newcommand{\fb}{{\mathfrak b}} \newcommand{\fr}{{\mathfrak r}} \newcommand{\fgg}{{\mathfrak g}} \newcommand{\Max}{{\rm Max}} \newcommand{\Omax}{{\rm Omax}} \newcommand{\omax}{{\rm omax}} \DeclareMathOperator{\Res}{Res} \DeclareMathOperator{\SDC}{SDC} \DeclareMathOperator{\DC}{DC} \DeclareMathOperator{\SP}{SP} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \begin{center} \vskip 1cm{\LARGE\bf Symmetric Dellac Configurations} \vskip 1cm \large Ange Bigeni\\ Department of Mathematics \\ HSE University\\ Usacheva str. 6 \\ Moscow 119048 \\ Russia\\ \href{mailto:ange.bigeni@gmail.com}{\tt ange.bigeni@gmail.com} \\ \ \\ Evgeny Feigin\\ Department of Mathematics \\ HSE University\\ Usacheva str. 6 \\ Moscow 119048 \\ Russia\\ and \\ Skolkovo Institute of Science and Technology\\ Nobelya Ulitsa 3 \\ Moscow 121205 \\ Russia\\ \href{mailto:evgfeig@gmail.com}{\tt evgfeig@gmail.com} \end{center} \vskip .2 in \begin{abstract} We define symmetric Dellac configurations as the Dellac configurations that are symmetrical with respect to their centers. The even-length symmetric Dellac configurations coincide with the Fang--Fourier symplectic Dellac configurations. Symmetric Dellac configurations generate the Poincar\'e polynomials of (odd or even) symplectic or orthogonal versions of degenerate flag varieties. We give several combinatorial interpretations of the Randrianarivony--Zeng polynomial extension of median Euler numbers in terms of objects that we call extended Dellac configurations. We show that the extended Dellac configurations generate symmetric Dellac configurations. As a consequence, the cardinalities of odd and even symmetric Dellac configurations are respectively given by two sequences $(1, 1, 3, 21, 267,\dots)$ and $(1,2,10,98,1594,\dots)$, defined as specializations of polynomial extensions of median Euler numbers. \end{abstract} \newcommand{\C}{\mathbb{C}} \newcommand{\PP}{\mathrm{P}} \newcommand{\B}{\mathbb{B}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\N}{\mathbb{N}} \newcommand{\R}{\mathbb{R}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\fS}{\mathfrak{S}} \newcommand{\Id}{\mathrm{Id}} \newcommand{\cone}{\mathrm{cone}} \newcommand{\conv}{\mathrm{conv}} \newcommand{\height}{\mathrm{height}} \newcommand{\im}{\mathrm{im}} \newcommand{\HH}{\mathcal{H}} \newcommand{\fpa}{\mathrm{FPA}} \newcommand{\pn}{\mathrm{part}} \newcommand{\RP}{\mathrm{RelPrime}} \newcommand{\rp}{\mathrm{relprime}} \newcommand{\rpac}{\mathrm{rpac}} \newcommand{\relprime}{\mathrm{relprime}} \newcommand{\ptn}{\mathrm{part}} \newcommand{\Rpac}{\mathrm{Rpac}} \newcommand{\Relprime}{\mathrm{Relprime}} \newcommand{\Ptn}{\mathrm{Part}} \def\v{{\boldsymbol v}} \def\t{{\boldsymbol t}} \newcommand{\A}{\mathcal{V}} \newcommand{\m}{\mathfrak{m}} \section{Notation and vocabulary}\label{sec:notation} For a pair of integers $n < m$, we denote the set $\{n,n+1,\dots,m\}$ by $[n,m]$. We denote the set $[1,n]$ by $[n]$. In this paper, we call \textit{tableau} (plural \textit{tableaux}) a rectangular grid made up of a finite number of columns and rows. We call \textit{box} the intersection of a column and a row. If a tableau $T$ has $n$ columns and $m$ rows, then for all $j \in [n]$ and for all $i \in [m]$, the $j$-th column of $T$ (from left to right) is denoted by $C_j^T$, and the $i$-th row of $T$ (from bottom to top) is denoted by $L_i^T$. The box $C_j^T \cap L_i^T$ is denoted by $(j:i)$. In the tableaux that we study, a box may be empty or contain a point $\bullet$. If the box $(j:i)$ contains a point referred to by a letter such as $p$, we also write $p = (j:i)$. If it is known that a row $L_i^T$ contains a single point, then this point is denoted by $p_i^T$. \section{Introduction}\label{sec:intro} Let $N$ be a positive integer. Recall that a Dellac configuration~\cite{De} $D \in\DC_N$ is a tableau made up of $N$ columns and $2N$ rows, which contains $2N$ points one per row, two per column. In addition, all the points are located between the lines $y = x$ and $y = N+x$; in other words, if a box $(~j~:~i~)$ of $T$ contains a point, then $j \leq i \leq N+j$. The cardinality of $\DC_N$ is $h_N$ where $(h_N)_{N \geq 0} = (1,1,2,7,38,295,\dots)$ is the sequence \seqnum{A000366} of normalized median Genocchi numbers \cite{BD,B2,B3,B4,F1,F2,HZ1,HZ2,K}. For $D \in\DC_N$, let $\text{inv}(D)$ be the number of pairs $(p_1,p_2)$ of points of $D$ such that $p_1$ is located to the left of $p_2$ and above $p_2$ (such a pair is named an \textit{inversion} of $D$). For example, we depict in Figure~\ref{fig:DC123} the elements of $(DC_N)_{N \in [3]}$ (from top to bottom), whose inversions are represented by segments connecting the points involved. \begin{figure}[!htbp] \begin{center} \begin{tikzpicture}[scale=0.4] \begin{scope}[yshift=1cm] \begin{scope}[xshift=11.5cm] \draw [color=red] (-0.25,-0.25) rectangle (1.25,2.25); \draw (0,0) grid[step=1] (1,2); \draw (0,0) -- (1,1); \draw (0,1) -- (1,2); \fill (0.5,0.5) circle (0.2); \fill (0.5,1.5) circle (0.2); \end{scope} \end{scope} \begin{scope}[yshift=-4cm] \begin{scope}[xshift=12.5cm] \draw [color=red] (-0.25,-0.25) rectangle (2.25,4.25); \draw (0,0) grid[step=1] (2,4); \draw (0,0) -- (2,2); \draw (0,2) -- (2,4); \fill (0.5,0.5) circle (0.2); \fill (0.5,1.5) circle (0.2); \fill (1.5,2.5) circle (0.2); \fill (1.5,3.5) circle (0.2); \end{scope} \begin{scope}[xshift=9.5cm] \draw [color=red] (-0.25,-0.25) rectangle (2.25,4.25); \draw (0,0) grid[step=1] (2,4); \draw (0,0) -- (2,2); \draw (0,2) -- (2,4); \fill (0.5,0.5) circle (0.2); \fill (1.5,1.5) circle (0.2); \fill (0.5,2.5) circle (0.2); \fill (1.5,3.5) circle (0.2); \draw [line width=0.25mm] (0.5,2.5) -- (1.5,1.5); \end{scope} \end{scope} \begin{scope}[yshift=-11cm] \begin{scope}[xshift=21cm] \draw [color=red] (-0.25,-0.25) rectangle (3.25,6.25); \draw (0,0) grid[step=1] (3,6); \draw (0,0) -- (3,3); \draw (0,3) -- (3,6); \fill (0.5,0.5) circle (0.2); \fill (0.5,1.5) circle (0.2); \fill (1.5,2.5) circle (0.2); \fill (1.5,3.5) circle (0.2); \fill (2.5,4.5) circle (0.2); \fill (2.5,5.5) circle (0.2); \end{scope} \begin{scope}[xshift=17.5cm] \draw (0,0) grid[step=1] (3,6); \draw (0,0) -- (3,3); \draw (0,3) -- (3,6); \fill (0.5,0.5) circle (0.2); \fill (0.5,1.5) circle (0.2); \fill (1.5,2.5) circle (0.2); \fill (2.5,3.5) circle (0.2); \fill (1.5,4.5) circle (0.2); \fill (2.5,5.5) circle (0.2); \draw [line width=0.25mm] (1.5,4.5) -- (2.5,3.5); \end{scope} \begin{scope}[xshift=14cm] \draw (0,0) grid[step=1] (3,6); \draw (0,0) -- (3,3); \draw (0,3) -- (3,6); \fill (0.5,0.5) circle (0.2); \fill (1.5,1.5) circle (0.2); \fill (0.5,2.5) circle (0.2); \fill (1.5,3.5) circle (0.2); \fill (2.5,4.5) circle (0.2); \fill (2.5,5.5) circle (0.2); \draw [line width=0.25mm] (0.5,2.5) -- (1.5,1.5); \end{scope} \begin{scope}[xshift=7cm] \draw (0,0) grid[step=1] (3,6); \draw (0,0) -- (3,3); \draw (0,3) -- (3,6); \fill (0.5,0.5) circle (0.2); \fill (0.5,1.5) circle (0.2); \fill (2.5,2.5) circle (0.2); \fill (1.5,3.5) circle (0.2); \fill (1.5,4.5) circle (0.2); \fill (2.5,5.5) circle (0.2); \draw [line width=0.25mm] (1.5,3.5) -- (2.5,2.5); \draw [line width=0.25mm] (1.5,4.5) -- (2.5,2.5); \end{scope} \begin{scope}[xshift=10.5cm] \draw [color=red] (-0.25,-0.25) rectangle (3.25,6.25); \draw (0,0) grid[step=1] (3,6); \draw (0,0) -- (3,3); \draw (0,3) -- (3,6); \fill (0.5,0.5) circle (0.2); \fill (1.5,1.5) circle (0.2); \fill (0.5,2.5) circle (0.2); \fill (2.5,3.5) circle (0.2); \fill (1.5,4.5) circle (0.2); \fill (2.5,5.5) circle (0.2); \draw [line width=0.25mm] (0.5,2.5) -- (1.5,1.5); \draw [line width=0.25mm] (1.5,4.5) -- (2.5,3.5); \end{scope} \begin{scope}[xshift=3.5cm] \draw (0,0) grid[step=1] (3,6); \draw (0,0) -- (3,3); \draw (0,3) -- (3,6); \fill (0.5,0.5) circle (0.2); \fill (1.5,1.5) circle (0.2); \fill (1.5,2.5) circle (0.2); \fill (0.5,3.5) circle (0.2); \fill (2.5,4.5) circle (0.2); \fill (2.5,5.5) circle (0.2); \draw [line width=0.25mm] (0.5,3.5) -- (1.5,1.5); \draw [line width=0.25mm] (0.5,3.5) -- (1.5,2.5); \end{scope} \begin{scope}[xshift=0cm] \draw [color=red] (-0.25,-0.25) rectangle (3.25,6.25); \draw (0,0) grid[step=1] (3,6); \draw (0,0) -- (3,3); \draw (0,3) -- (3,6); \fill (0.5,0.5) circle (0.2); \fill (1.5,1.5) circle (0.2); \fill (2.5,2.5) circle (0.2); \fill (0.5,3.5) circle (0.2); \fill (1.5,4.5) circle (0.2); \fill (2.5,5.5) circle (0.2); \draw [line width=0.25mm] (0.5,3.5) -- (1.5,1.5); \draw [line width=0.25mm] (0.5,3.5) -- (2.5,2.5); \draw [line width=0.25mm] (1.5,4.5) -- (2.5,2.5); \end{scope} \end{scope} \end{tikzpicture} \end{center} \caption{The elements of $\DC_1,\DC_2,\DC_3$ from top to bottom; the elements of $\SDC_1,\SDC_2,\SDC_3$ are framed in red.} \label{fig:DC123} \end{figure} Given an $N$-dimensional vector space $E$, a basis $(e_1,e_2,\dots,e_N)$ of $E$ and an integer $k \in [N]$, let $\text{pr}_k: E \rightarrow E$ be the projection along $e_k$ to the linear span of the rest basis vectors. Feigin~\cite{F1} defined the degenerate flag variety $\mathcal{F}_N^a$ as the set of collections $(V_1,V_2, \dots, V_{N-1})$ of subspaces of $E$ such that $\dim(V_k) = k$ and $\text{pr}_{k+1}(V_k) \subset V_{k+1}$ for all $k \in [N-1]$. Feigin~\cite{F1,F3,FFL} proved that the Poincar\'e polynomial $F_N$ of the degenerate flag variety $\mathcal{F}_N^a$ has the following combinatorial interpretation for all $N \geq 1$: \begin{equation*} F_N(q) = \sum_{D \in\DC_N} q^{\text{inv}(D)}. \end{equation*} For example, we can compute the Poincar\'e polynomials of $(\mathcal{F}_N^a)_{N\in[3]}$ from Figure~\ref{fig:DC123}: \begin{align*} F_1(q) &= 1,\\ F_2(q) & = q+1,\\ F_3(q) & = q^3+3q^2+2q+1. \end{align*} \subsection{Symmetric Dellac configurations} For a positive integer $n$, consider a $2n$-dimensional vector space $E$, a basis $(e_1,e_2,\dots,e_{2n})$ of $E$, and a non-degenerate skew-symmetric form defined on $E^2$ by $(e_i, e_{2n+1-i}) = 1$ for all $i \in [n]$ and $(e_i, e_j) = 0$ if $i+j\neq 2n$. Feigin et al.~\cite{FFL} defined the symplectic degenerate flag variety $\text{SpF}_{2n}^a$, as the subvariety of $\mathcal{F}_{2n}^a$ consisting of collections $(V_1, \dots , V_{2n-1})$ such that $V_k = V_{2n-k}^{\perp}$ for all $k \in [2n-1]$. Fang and Fourier~\cite{FF} considered the action of an algebraic torus on $\mathcal{F}_{2n}^a$~\cite{CFR,CLL}. They parametrized the torus fixed points of $\text{SpF}_{2n}^a \subset \mathcal{F}_{2n}^a$ (the points fixed by the action of the torus) using even symmetric Dellac configurations $D \in \SDC_{2n}$, which they introduced in their paper as \textit{symplectic Dellac configurations}. In a separate paper~\cite{BF}, we extended the definition of the symplectic degenerate flag varieties to $\text{SpF}_{N}^a$ for all $N \geq 1$. We proved that the Poincar\'e polynomials $S_N$ of these varieties have the following combinatorial interpretation: \begin{equation*} S_N(q) = \sum_{D \in \SDC_{N}} q^{\widetilde{\text{inv}}(D)}. \end{equation*} Here $\widetilde{\text{inv}}(D)$ is the number of inversions of $D$, modulo the rotation $r_{\pi}$. More precisely, we identify two inversions of $D$ if they are of the form $(p_1,p_2)$ and $(r_{\pi}(p_2),r_{\pi}(p_1))$. For example, below are the Poincar\'e polynomials of $(\text{SpF}_N^a)_{N \in [4]}$ (the cases $N \in [3]$ are given by Figure~\ref{fig:DC123}): \begin{align*} S_1(q) &= 1,\\ S_2(q) & = q+1,\\ S_3(q) & = q^2+q+1,\\ S_4(q) &= q^4 + 3q^3 + 3q^2 + 2q + 1. \end{align*} We also defined natural varieties $(\text{SOF}_N^a)_{N\geq1}$~\cite{BF}, named \textit{orthogonal degenerate flag varieties}, whose Poincar\'e polynomials $O_N$ have the following combinatorial interpretation: \begin{equation*} O_N(q) = \sum_{D \in \SDC_{N}} q^{\overline{\text{inv}}(D)}, \end{equation*} where $\overline{\text{inv}}(D)$ equals $\widetilde{\text{inv}}(D)$ minus the number of inversions of $D$ of the kind $(p,r_{\pi}(p))$. For example, the Poincar\'e polynomials of $(\text{SOF}_N^a)_{N\in[4]}$ (the cases $N\in[3]$ are given by Figure~\ref{fig:DC123}) are \begin{align*} O_1(q) &= 1,\\ O_2(q) & = 2,\\ O_3(q) & = 2q+1,\\ O_4(q) & = 4q^2 + 4q + 2. \end{align*} The primary goal of this paper is to determine the cardinalities of the sets $(\SDC_N)_{N\geq 1}$. According to the parity of $N$, they appear in two sequences: \begin{itemize} \item $(l_n)_{n \geq 0} = (1, 1, 3, 21,\dots)$~\seqnum{A098278}; and \item $(r_n)_{n \geq 0} = (1,2,10,98,\dots)$~\seqnum{A098279} \end{itemize} defined by Randrianarivony and Zeng~\cite{RZ} as specializations of polynomials interpolating median Euler numbers. We recall the necessary definitions below. \subsection{Extended median Euler numbers} \label{sec:DnPnlnrn} Consider the family of polynomials $(D_n(x))_{n \geq 0}$ defined by \begin{align*} D_0(x) &=1,\\ D_{n+1}(x) &= (x+1)(x+2)D_n(x+2) - x(x+1)D_n(x) \quad \forall n \geq 0. \end{align*} They were introduced by Randrianarivony and Zeng~\cite{RZ} as interpolations of the median Euler numbers $(L_n)_{n \geq 0} = (1, 1, 4, 46, \dots)$~\seqnum{A000657} and $(R_n)_{n \geq 0} = (1, 3, 24, 402,\dots)$~\seqnum{A002832}, through the equalities $L_n = D_n(1/2)$ and $R_n = D_n(-1/2)$. It is easy to check that for $n \geq 1$, the polynomial $D_n(x)$ is of the form $2^n(x+1)P_{n}(x+1)$ for some polynomial $P_{n}(x)$ of degree $n-1$ with positive integer coefficients. The polynomials $P_{n}(x)$ satisfy the following recursive formulas: \begin{align*} P_1(x) &=1,\\ P_{n+1}(x) &= \frac{(x+2)(x+1)}{2} P_n(x+2) - \frac{x(x-1)}{2} P_n(x) \quad \forall n \geq 1. \end{align*} The first few such polynomials are \begin{align*} P_1(x) & = 1,\\ P_2(x) & = 2x+1,\\ P_3(x) & = 6x^2 + 10x + 5, \\ P_4(x) & = 24x^3 + 84x^2 + 110x + 49,\\ P_5(x) & = 120x^4 + 720x^3 + 1758x^2 + 1954x + 797. \end{align*} For all $n \geq 1$, let $c_{n,n-1} x^{n-1} + c_{n,n-2} x^{n-2} + \dots+ c_{n,1}x + c_{n,0} = P_n(x)$. It is easy to check the following recursive formulas: $c_{1,0} = 1$ and \begin{align} \label{eq:an0} c_{n,0} &= \sum_{i = 0}^{n-2} 2^i c_{n-1,i},\\ \label{eq:ank} c_{n,k} &= (k+1)c_{n-1,k-1} + \sum_{i = k}^{n-2} 2^{i-k} \left( \binom{i+1}{k} + 2\binom{i+1}{k-1} \right) c_{n-1,i},\\ \label{eq:anmoins1} c_{n,n-1} &= n c_{n-1,n-2} \end{align} for all $n \geq 2$ and $k \in [n-2]$. In particular, we have $c_{n,n-1} = n!$ for all $n \geq 1$. The sequences $(l_n)_{n \geq 0} = (1, 1, 3, 21,\dots)$ and $(r_n)_{n \geq 0} = (1,2,10,98,\dots)$ are defined by $l_n = D_n(0)/2^n$ and $r_n = D_n(1)/2^n$ for all $n \geq 0$, which, in terms of $P_n$, is equivalent to $l_0 = r_0 = 1$ and \begin{align*} l_n &= P_n(1),\\ r_n &= 2P_n(2) = 2P_{n+1}(0) \end{align*} for all $n \geq 1$. From Randrianarivony and Zeng~\cite[Th\'eor\`eme 24]{RZ}, it follows that $$1+ \sum_{n \geq 1} xP_n(x)t^n = \dfrac{1}{1- \dfrac{xt}{1- \dfrac{(x+1)t}{1- \dfrac{2(x+2)t}{1- \dfrac{2(x+3)t}{1-\dfrac{3(x+4)t}{1-\dfrac{3(x+5)t}{\ddots}}}}}}}.$$ Afterwards, recall~\cite{Du,DR,DZ,ZZ} that a \textit{surjective pistol} of size $n$ is a surjective map $f:~[2n]~\twoheadrightarrow~\{2,4,6,\dots,2n\}$ such that $f(j) \geq j$ for all $j \in [2n]$. Let $\SP_n$ be the set of the surjective pistols of size $n$. For all $f \in \SP_n$, we denote by $\max(f)$ the number of maximal points of $f$ (integers $j \in [2n-2]$ such that $f(j) = 2n$), and by $\text{fd}(f)$ the number of doubled fixed points of $f$ (integers $j \in [2n-2]$ such that there exists $j'< j$ with $f(j') = f(j) = j$). For example $\SP_2$ has three elements: $$\begin{pmatrix} 1&2&3&4\\ 2&2&4&4 \end{pmatrix},\begin{pmatrix} 1&2&3&4\\ 4&2&4&4 \end{pmatrix}\text{ and }\begin{pmatrix} 1&2&3&4\\ 2&4&4&4 \end{pmatrix},$$ whose numbers of maximal points are respectively $0$,$1$ and $1$, and whose numbers of doubled fixed points are respectively $1$,$0$ and $0$. Still due to Randrianarivony and Zeng~\cite[Proposition~25]{RZ}, it is known that \begin{equation} \label{eq:PnintermsofSPn} P_n(x) = \sum_{f \in \SP_n} 2^{n-1-\max(f)-\text{fd}(f)} x^{\max(f)} \end{equation} for all $n \geq 1$. For example, considering the three elements of $\SP_2$ shown above, we indeed have the equality $P_2(x) = 2x+1 = 2^{1-0-1} x^0 + 2^{1-1-0} x^1 + 2^{1-1-0} x^1$. \subsection{Even symmetric Dellac configurations} One of the goals of this paper is to give a new proof of the statement that the cardinality of $\SDC_{2n}$ (for $n \geq 1)$ is the number $r_n$ from the sequence $(r_n)_{n \geq 0} = (1,2,10,98,1594,\dots)$ (another proof can be found in a previous paper~\cite{B1}). The first step in this direction is to reduce $\SDC_{2n}$ to smaller elements that we name even extended Dellac configurations, whose set is denoted by $\mathcal{T}_n^e$. The main result of this section is Theorem~\ref{theo:PnintermsofTne}, which interprets $P_n(x)$ combinatorially in terms of $\mathcal{T}_n^e$, and which implies the equality $|\SDC_{2n}| = 2P_n(2) = r_n$. \subsubsection{Generation of $\SDC_{2n}$} An \textit{even extended Dellac configuration} $T \in \mathcal{T}_n^e$ is a tableau made up of $n$ columns and $2n$ rows, which contains $2n$ points (one per row, two per column) on or above the line $y = x$ (in other words, this is the definition of $\DC_n$, except that we remove the condition that the points must be under the line $y = n+x$). By considering the two points of the last column (from left to right) of a tableau $T \in \mathcal{T}_n^e$, it is easy to obtain the formula $|\mathcal{T}_n^e| = \frac{(n+1)n}{2} |\mathcal{T}_{n-1}^e| = (n+1)!n!/2^n$. For example, in Figure~\ref{fig:T2e} we represent the 3 elements of $T_2^e$. Note that the points located on or above the line $y = 2n-x$ (represented by a dashed line) are represented by stars; in general, we name such points \textit{free points}, and we denote the number of free points of $T \in \mathcal{T}_n^e$ by $\text{fr}(T)$. In Figure~\ref{fig:T2e}, the numbers of free points are 2, 1 and 2 from left to right. \begin{figure}[!htbp] \begin{center} \begin{tikzpicture}[scale=0.4] \begin{scope}[shift={(0,0)}] \draw (0,0) grid[step=1] (2,4); \draw (0,0) -- (2,2); \draw [dashed] (0,4) -- (2,2); \fill (0.5,0.5) circle (0.2); \fill (0.5,1.5) circle (0.2); \draw (1.5,2.5) node[scale=0.8]{$\bigstar$}; \draw (1.5,3.5) node[scale=0.8]{$\bigstar$}; \end{scope} \begin{scope}[shift={(3,0)}] \draw (0,0) grid[step=1] (2,4); \draw (0,0) -- (2,2); \draw [dashed] (0,4) -- (2,2); \fill (0.5,0.5) circle (0.2); \fill (1.5,1.5) circle (0.2); \fill (0.5,2.5) circle (0.2); \draw (1.5,3.5) node[scale=0.8]{$\bigstar$}; \end{scope} \begin{scope}[shift={(6,0)}] \draw (0,0) grid[step=1] (2,4); \draw (0,0) -- (2,2); \draw [dashed] (0,4) -- (2,2); \fill (0.5,0.5) circle (0.2); \fill (1.5,1.5) circle (0.2); \draw (1.5,2.5) node[scale=0.8]{$\bigstar$}; \draw (0.5,3.5) node[scale=0.8]{$\bigstar$}; \end{scope} \end{tikzpicture} \end{center} \caption{The 3 elements of $\mathcal{T}_2^e$.} \label{fig:T2e} \end{figure} There exists a one-to-one correspondence between $\SDC_{2n}$ and the set of tableaux $T \in \mathcal{T}_n^e$ whose free points are labeled with the number 0 or 1. For such a tableau $T$, the principle is: \begin{itemize} \item to draw an empty tableau $\widetilde{T}$ made up of $2n$ columns and $4n$ rows; \item for a box $(j:i)$ of $T$ containing a point $p$, \begin{itemize} \item if $p$ is not free, or if $p$ is a free point whose label is $1$, then we draw a point in the box $(j:i)$ of $\widetilde{T}$; \item otherwise $p$ is free and its label is $0$, and we draw a point in the box $(2n+1-j:i)$ of $\widetilde{T}$; \end{itemize} \item to apply the central reflection with respect to the center $(n,2n)$ of $\widetilde{T}$ to the $2n$ points that have been drawn in $\widetilde{T}$, which indeed makes $\widetilde{T}$ an element of $\SDC_{2n}$. \end{itemize} For example, in Figure~\ref{fig:Ttilde} we depict how a labeled tableau $T_0 \in \mathcal{T}_2^e$ is mapped to its corresponding element $\widetilde{T_0} \in \SDC_4$. \begin{figure}[!htbp] \begin{center} \begin{tikzpicture}[scale=0.4] \begin{scope}[shift={(0,0)}] \fill [color=red,opacity = 0.5] (0,3) rectangle (1,4); \fill [color=blue,opacity = 0.5] (1,2) rectangle (2,3); \draw (0,0) grid[step=1] (2,4); \draw (0,0) -- (2,2); \draw [dashed] (0,4) -- (2,2); \fill (0.5,0.5) circle (0.2); \fill (1.5,1.5) circle (0.2); \draw (1.5,2.5) node[scale=1]{$0$}; \draw (0.5,3.5) node[scale=1]{$1$}; \end{scope} \draw (3,2) node[scale=1]{$\longrightarrow$}; \begin{scope}[shift={(4,0)}] \fill [color=red,opacity = 0.5] (0,3) rectangle (1,4); \fill [color=blue,opacity = 0.5] (1,2) rectangle (2,3); \fill [color=red,opacity = 0.2] (3,3) rectangle (4,4); \fill [color=blue,opacity = 0.2] (2,2) rectangle (3,3); \draw (0,0) grid[step=1] (2,4); \draw (0,0) -- (4,4); \draw (2,6) -- (4,8); \draw [dashed] (0,4) -- (2,2); \fill (0.5,0.5) circle (0.2); \fill (1.5,1.5) circle (0.2); \fill (0.5,3.5) circle (0.2); \draw [black,opacity =1] (0,0) grid[step=1] (2,8); \draw [black,opacity =1] (0,0) -- (2,2); \draw [black,opacity =1] (0,4) -- (2,6); \begin{scope}[shift={(4,8)},rotate=180] \draw [black,opacity =1] (0,0) grid[step=1] (2,8); \end{scope} \fill [black,opacity = 1] (2.5,2.5) circle (0.2); \end{scope} \draw (9,4) node[scale=1]{$\longrightarrow$}; \begin{scope}[shift={(10,0)}] \fill [color=red,opacity = 0.5] (0,3) rectangle (1,4); \fill [color=blue,opacity = 0.5] (1,2) rectangle (2,3); \fill [color=red,opacity = 0.2] (3,3) rectangle (4,4); \fill [color=blue,opacity = 0.2] (2,2) rectangle (3,3); \draw (0,0) grid[step=1] (2,4); \draw (0,0) -- (2,2); \draw [dashed] (0,4) -- (2,2); \fill (0.5,0.5) circle (0.2); \fill (1.5,1.5) circle (0.2); \fill (1.5,5.5) circle (0.2); \fill (0.5,3.5) circle (0.2); \draw [black,opacity =1] (0,0) grid[step=1] (2,8); \draw [black,opacity =1] (0,0) -- (2,2); \draw [black,opacity =1] (0,4) -- (2,6); \begin{scope}[shift={(4,8)},rotate=180] \fill [color=red,opacity = 0.5] (0,3) rectangle (1,4); \fill [color=blue,opacity = 0.5] (1,2) rectangle (2,3); \fill [color=red,opacity = 0.2] (3,3) rectangle (4,4); \fill [color=blue,opacity = 0.2] (2,2) rectangle (3,3); \draw [black,opacity =1] (0,0) grid[step=1] (2,8); \draw [black,opacity =1] (0,0) -- (2,2); \draw [black,opacity =1] (0,4) -- (2,6); \draw [black,opacity =1] [dashed] (0,4) -- (2,2); \fill [black,opacity =1] (0.5,0.5) circle (0.2); \fill [black,opacity =1] (1.5,1.5) circle (0.2); \fill [black,opacity =1] (1.5,5.5) circle (0.2); \fill [black,opacity =1] (0.5,3.5) circle (0.2); \end{scope} \fill [black,opacity = 1] (2.5,2.5) circle (0.2); \end{scope} \end{tikzpicture} \end{center} \caption{Construction of an even symmetric Dellac configuration.} \label{fig:Ttilde} \end{figure} \begin{figure}[!htbp] \begin{center} \begin{tikzpicture}[scale=0.3] \draw (0,0) grid[step=1] (2,4); \draw (0,0) -- (2,2); \draw [dashed] (0,4) -- (2,2); \fill (0.5,0.5) circle (0.2); \fill (0.5,1.5) circle (0.2); \draw (1.5,2.5) node[scale=0.6]{$\bigstar$}; \draw (1.5,3.5) node[scale=0.6]{$\bigstar$}; \draw [gray,opacity = 0.7] (0,0) grid[step=1] (2,8); \draw [gray,opacity = 0.7] (0,0) -- (2,2); \draw [gray,opacity = 0.7] (0,4) -- (2,6); \begin{scope}[shift={(4,8)},rotate=180] \draw [gray,opacity = 0.7] (0,0) grid[step=1] (2,8); \draw [gray,opacity = 0.7] (0,0) -- (2,2); \draw [gray,opacity = 0.7] (0,4) -- (2,6); \draw [gray,opacity = 0.7] [dashed] (0,4) -- (2,2); \fill [gray,opacity = 0.7] (0.5,0.5) circle (0.2); \fill [gray,opacity = 0.7] (0.5,1.5) circle (0.2); \draw [gray,opacity = 0.7] (1.5,2.5) node[scale=0.6]{$\bigstar$}; \draw [gray,opacity = 0.7] (1.5,3.5) node[scale=0.6]{$\bigstar$}; \end{scope} \begin{scope}[shift={(5,0)}] \draw (0,0) grid[step=1] (2,4); \draw (0,0) -- (2,2); \draw [dashed] (0,4) -- (2,2); \fill (0.5,0.5) circle (0.2); \fill (0.5,1.5) circle (0.2); \draw (1.5,2.5) node[scale=0.6]{$\bigstar$}; \draw (2.5,3.5) node[scale=0.6]{$\bigstar$}; \draw [gray,opacity = 0.7] (0,0) grid[step=1] (2,8); \draw [gray,opacity = 0.7] (0,0) -- (2,2); \draw [gray,opacity = 0.7] (0,4) -- (2,6); \begin{scope}[shift={(4,8)},rotate=180] \draw [gray,opacity = 0.7] (0,0) grid[step=1] (2,8); \draw [gray,opacity = 0.7] (0,0) -- (2,2); \draw [gray,opacity = 0.7] (0,4) -- (2,6); \draw [gray,opacity = 0.7] [dashed] (0,4) -- (2,2); \fill [gray,opacity = 0.7] (0.5,0.5) circle (0.2); \fill [gray,opacity = 0.7] (0.5,1.5) circle (0.2); \draw [gray,opacity = 0.7] (1.5,2.5) node[scale=0.6]{$\bigstar$}; \draw [gray,opacity = 0.7] (2.5,3.5) node[scale=0.6]{$\bigstar$}; \end{scope} \end{scope} \begin{scope}[shift={(0,-9)}] \draw (0,0) grid[step=1] (2,4); \draw (0,0) -- (2,2); \draw [dashed] (0,4) -- (2,2); \fill (0.5,0.5) circle (0.2); \fill (0.5,1.5) circle (0.2); \draw (2.5,2.5) node[scale=0.6]{$\bigstar$}; \draw (1.5,3.5) node[scale=0.6]{$\bigstar$}; \draw [gray,opacity = 0.7] (0,0) grid[step=1] (2,8); \draw [gray,opacity = 0.7] (0,0) -- (2,2); \draw [gray,opacity = 0.7] (0,4) -- (2,6); \begin{scope}[shift={(4,8)},rotate=180] \draw [gray,opacity = 0.7] (0,0) grid[step=1] (2,8); \draw [gray,opacity = 0.7] (0,0) -- (2,2); \draw [gray,opacity = 0.7] (0,4) -- (2,6); \draw [gray,opacity = 0.7] [dashed] (0,4) -- (2,2); \fill [gray,opacity = 0.7] (0.5,0.5) circle (0.2); \fill [gray,opacity = 0.7] (0.5,1.5) circle (0.2); \draw [gray,opacity = 0.7] (2.5,2.5) node[scale=0.6]{$\bigstar$}; \draw [gray,opacity = 0.7] (1.5,3.5) node[scale=0.6]{$\bigstar$}; \end{scope} \end{scope} \begin{scope}[shift={(5,-9)}] \draw (0,0) grid[step=1] (2,4); \draw (0,0) -- (2,2); \draw [dashed] (0,4) -- (2,2); \fill (0.5,0.5) circle (0.2); \fill (0.5,1.5) circle (0.2); \draw (2.5,2.5) node[scale=0.6]{$\bigstar$}; \draw (2.5,3.5) node[scale=0.6]{$\bigstar$}; \draw [gray,opacity = 0.7] (0,0) grid[step=1] (2,8); \draw [gray,opacity = 0.7] (0,0) -- (2,2); \draw [gray,opacity = 0.7] (0,4) -- (2,6); \begin{scope}[shift={(4,8)},rotate=180] \draw [gray,opacity = 0.7] (0,0) grid[step=1] (2,8); \draw [gray,opacity = 0.7] (0,0) -- (2,2); \draw [gray,opacity = 0.7] (0,4) -- (2,6); \draw [gray,opacity = 0.7] [dashed] (0,4) -- (2,2); \fill [gray,opacity = 0.7] (0.5,0.5) circle (0.2); \fill [gray,opacity = 0.7] (0.5,1.5) circle (0.2); \draw [gray,opacity = 0.7] (2.5,2.5) node[scale=0.6]{$\bigstar$}; \draw [gray,opacity = 0.7] (2.5,3.5) node[scale=0.6]{$\bigstar$}; \end{scope} \end{scope} \draw (-1,-10) rectangle (10,9); \begin{scope}[shift={(3.5,10)}] \draw (0,0) grid[step=1] (2,4); \draw (0,0) -- (2,2); \draw [dashed] (0,4) -- (2,2); \fill (0.5,0.5) circle (0.2); \fill (0.5,1.5) circle (0.2); \draw (1.5,2.5) node[scale=0.6]{$\bigstar$}; \draw (1.5,3.5) node[scale=0.6]{$\bigstar$}; \end{scope} \begin{scope}[shift={(11,0)}] \draw (0,0) grid[step=1] (2,4); \draw (0,0) -- (2,2); \draw [dashed] (0,4) -- (2,2); \fill (0.5,0.5) circle (0.2); \fill (1.5,1.5) circle (0.2); \fill (0.5,2.5) circle (0.2); \draw (1.5,3.5) node[scale=0.6]{$\bigstar$}; \draw [gray,opacity = 0.7] (0,0) grid[step=1] (2,8); \draw [gray,opacity = 0.7] (0,0) -- (2,2); \draw [gray,opacity = 0.7] (0,4) -- (2,6); \begin{scope}[shift={(4,8)},rotate=180] \draw [gray,opacity = 0.7] (0,0) grid[step=1] (2,8); \draw [gray,opacity = 0.7] (0,0) -- (2,2); \draw [gray,opacity = 0.7] (0,4) -- (2,6); \draw [gray,opacity = 0.7] [dashed] (0,4) -- (2,2); \fill [gray,opacity = 0.7] (0.5,0.5) circle (0.2); \fill [gray,opacity = 0.7] (1.5,1.5) circle (0.2); \fill [gray,opacity = 0.7] (0.5,2.5) circle (0.2); \draw [gray,opacity = 0.7] (1.5,3.5) node[scale=0.6]{$\bigstar$}; \end{scope} \begin{scope}[shift={(0,-9)}] \draw (0,0) grid[step=1] (2,4); \draw (0,0) -- (2,2); \draw [dashed] (0,4) -- (2,2); \fill (0.5,0.5) circle (0.2); \fill (1.5,1.5) circle (0.2); \fill (0.5,2.5) circle (0.2); \draw (2.5,3.5) node[scale=0.6]{$\bigstar$}; \draw [gray,opacity = 0.7] (0,0) grid[step=1] (2,8); \draw [gray,opacity = 0.7] (0,0) -- (2,2); \draw [gray,opacity = 0.7] (0,4) -- (2,6); \begin{scope}[shift={(4,8)},rotate=180] \draw [gray,opacity = 0.7] (0,0) grid[step=1] (2,8); \draw [gray,opacity = 0.7] (0,0) -- (2,2); \draw [gray,opacity = 0.7] (0,4) -- (2,6); \draw [gray,opacity = 0.7] [dashed] (0,4) -- (2,2); \fill [gray,opacity = 0.7] (0.5,0.5) circle (0.2); \fill [gray,opacity = 0.7] (1.5,1.5) circle (0.2); \fill [gray,opacity = 0.7] (0.5,2.5) circle (0.2); \draw [gray,opacity = 0.7] (2.5,3.5) node[scale=0.6]{$\bigstar$}; \end{scope} \end{scope} \draw (-1,-10) rectangle (5,9); \begin{scope}[shift={(1,10)}] \draw (0,0) grid[step=1] (2,4); \draw (0,0) -- (2,2); \draw [dashed] (0,4) -- (2,2); \fill (0.5,0.5) circle (0.2); \fill (1.5,1.5) circle (0.2); \fill (0.5,2.5) circle (0.2); \draw (1.5,3.5) node[scale=0.6]{$\bigstar$}; \end{scope} \end{scope} \begin{scope}[shift={(17,0)}] \draw (0,0) grid[step=1] (2,4); \draw (0,0) -- (2,2); \draw [dashed] (0,4) -- (2,2); \fill (0.5,0.5) circle (0.2); \fill (1.5,1.5) circle (0.2); \draw (1.5,2.5) node[scale=0.6]{$\bigstar$}; \draw (0.5,3.5) node[scale=0.6]{$\bigstar$}; \draw [gray,opacity = 0.7] (0,0) grid[step=1] (2,8); \draw [gray,opacity = 0.7] (0,0) -- (2,2); \draw [gray,opacity = 0.7] (0,4) -- (2,6); \begin{scope}[shift={(4,8)},rotate=180] \draw [gray,opacity = 0.7] (0,0) grid[step=1] (2,8); \draw [gray,opacity = 0.7] (0,0) -- (2,2); \draw [gray,opacity = 0.7] (0,4) -- (2,6); \draw [gray,opacity = 0.7] [dashed] (0,4) -- (2,2); \fill [gray,opacity = 0.7] (0.5,0.5) circle (0.2); \fill [gray,opacity = 0.7] (1.5,1.5) circle (0.2); \draw [gray,opacity = 0.7] (1.5,2.5) node[scale=0.6]{$\bigstar$}; \draw [gray,opacity = 0.7] (0.5,3.5) node[scale=0.6]{$\bigstar$}; \end{scope} \begin{scope}[shift={(0,-9)}] \draw (0,0) grid[step=1] (2,4); \draw (0,0) -- (2,2); \draw [dashed] (0,4) -- (2,2); \fill (0.5,0.5) circle (0.2); \fill (1.5,1.5) circle (0.2); \draw (2.5,2.5) node[scale=0.6]{$\bigstar$}; \draw (0.5,3.5) node[scale=0.6]{$\bigstar$}; \draw [gray,opacity = 0.7] (0,0) grid[step=1] (2,8); \draw [gray,opacity = 0.7] (0,0) -- (2,2); \draw [gray,opacity = 0.7] (0,4) -- (2,6); \begin{scope}[shift={(4,8)},rotate=180] \draw [gray,opacity = 0.7] (0,0) grid[step=1] (2,8); \draw [gray,opacity = 0.7] (0,0) -- (2,2); \draw [gray,opacity = 0.7] (0,4) -- (2,6); \draw [gray,opacity = 0.7] [dashed] (0,4) -- (2,2); \fill [gray,opacity = 0.7] (0.5,0.5) circle (0.2); \fill [gray,opacity = 0.7] (1.5,1.5) circle (0.2); \draw [gray,opacity = 0.7] (2.5,2.5) node[scale=0.6]{$\bigstar$}; \draw [gray,opacity = 0.7] (0.5,3.5) node[scale=0.6]{$\bigstar$}; \end{scope} \end{scope} \begin{scope}[shift={(5,0)}] \draw (0,0) grid[step=1] (2,4); \draw (0,0) -- (2,2); \draw [dashed] (0,4) -- (2,2); \fill (0.5,0.5) circle (0.2); \fill (1.5,1.5) circle (0.2); \draw (1.5,2.5) node[scale=0.6]{$\bigstar$}; \draw (3.5,3.5) node[scale=0.6]{$\bigstar$}; \draw [gray,opacity = 0.7] (0,0) grid[step=1] (2,8); \draw [gray,opacity = 0.7] (0,0) -- (2,2); \draw [gray,opacity = 0.7] (0,4) -- (2,6); \begin{scope}[shift={(4,8)},rotate=180] \draw [gray,opacity = 0.7] (0,0) grid[step=1] (2,8); \draw [gray,opacity = 0.7] (0,0) -- (2,2); \draw [gray,opacity = 0.7] (0,4) -- (2,6); \draw [gray,opacity = 0.7] [dashed] (0,4) -- (2,2); \fill [gray,opacity = 0.7] (0.5,0.5) circle (0.2); \fill [gray,opacity = 0.7] (1.5,1.5) circle (0.2); \draw [gray,opacity = 0.7] (1.5,2.5) node[scale=0.6]{$\bigstar$}; \draw [gray,opacity = 0.7] (3.5,3.5) node[scale=0.6]{$\bigstar$}; \end{scope} \end{scope} \begin{scope}[shift={(5,-9)}] \draw (0,0) grid[step=1] (2,4); \draw (0,0) -- (2,2); \draw [dashed] (0,4) -- (2,2); \fill (0.5,0.5) circle (0.2); \fill (1.5,1.5) circle (0.2); \draw (2.5,2.5) node[scale=0.6]{$\bigstar$}; \draw (3.5,3.5) node[scale=0.6]{$\bigstar$}; \draw [gray,opacity = 0.7] (0,0) grid[step=1] (2,8); \draw [gray,opacity = 0.7] (0,0) -- (2,2); \draw [gray,opacity = 0.7] (0,4) -- (2,6); \begin{scope}[shift={(4,8)},rotate=180] \draw [gray,opacity = 0.7] (0,0) grid[step=1] (2,8); \draw [gray,opacity = 0.7] (0,0) -- (2,2); \draw [gray,opacity = 0.7] (0,4) -- (2,6); \draw [gray,opacity = 0.7] [dashed] (0,4) -- (2,2); \fill [gray,opacity = 0.7] (0.5,0.5) circle (0.2); \fill [gray,opacity = 0.7] (1.5,1.5) circle (0.2); \draw [gray,opacity = 0.7] (2.5,2.5) node[scale=0.6]{$\bigstar$}; \draw [gray,opacity = 0.7] (3.5,3.5) node[scale=0.6]{$\bigstar$}; \end{scope} \end{scope} \draw (-1,-10) rectangle (10,9); \begin{scope}[shift={(3.5,10)}] \draw (0,0) grid[step=1] (2,4); \draw (0,0) -- (2,2); \draw [dashed] (0,4) -- (2,2); \fill (0.5,0.5) circle (0.2); \fill (1.5,1.5) circle (0.2); \draw (1.5,2.5) node[scale=0.6]{$\bigstar$}; \draw (0.5,3.5) node[scale=0.6]{$\bigstar$}; \end{scope} \end{scope} \end{tikzpicture} \end{center} \caption{Construction of $\SDC_4$ from $\mathcal{T}_2^e$.} \label{fig:SDC4generatedbyT2e} \end{figure} This bijection between the tableaux $T \in \mathcal{T}_n^e$ whose free points are labeled with 0 or 1 and $\SDC_{2n}$ implies the following formula: \begin{equation} \label{eq:TnegeneratesSpDC2n} |\SDC_{2n}| = \sum_{T \in \mathcal{T}_n^e} 2^{\text{fr}(T)}. \end{equation} For example, for $n = 2$, the $10 = 2^2+2^1+2^2$ elements of $\SDC_{4}$ are generated by the 3 elements of $\mathcal{T}_2^e$ as depicted in Figure~\ref{fig:SDC4generatedbyT2e}. The elements of $\SDC_4$ in the same frame under each tableau $T \in \mathcal{T}_2^e$ correspond to the $2^{\text{fr}(T)}$ different labellings of the free points of $T$. \subsubsection{A combinatorial interpretation of $P_n(x)$ in terms of $\mathcal{T}_n^e$} \label{sec:theorem1} \begin{definition} \label{defi:evenpath} Let $T \in \mathcal{T}_n^e$. For a given point $p_0 = (j:i)$ of $T$, we first define a path $(p_0,p_1,\dots)$ of points of $T$ as follows: for all $k \geq 0$, assume that the points $p_0,\dots,p_k$ are defined, and let $(j_k:i_k) = p_k$; \begin{enumerate}[label=(\alph*)] \item if $j_k = n$, then $p_{k+1}$ is defined as $p_k$; \item if $j_k < n$ and $p_k$ is the upper point of its column, then $p_{k+1}$ is defined as $p_{2n-j_k}^T$ (the point of the $(2n-j_k)$-th row $L_{2n-j_k}^T$ of $T$, see Section 1); \item if $j_k < n$ and $p_k$ is the lower point of its column, then $p_{k+1}$ is defined as $p_{j_k}^T$. \end{enumerate} We then define a finite subsequence $S_T(i) = (p_{k_0},\dots,p_{k_f})$ of $(p_0,p_1,\dots)$ by $k_0 = 0$ and, if we assume that $k_0,\dots,k_q$ are defined for some $q \geq 0$, \begin{itemize} \item if $j_{k_q} < \max \{j_k: k \geq 0\}$, then $k_{q+1} = \min \{k > k_q: j_k > j_{k_q}\}$; \item otherwise $f = q$. \end{itemize} We finally define $B(T)$ as $S_T(n)$ and $R(T)$ as $S_T(2n)$. Let $\text{b}(T)+1$ and $\text{r}(T)+1$ be their respective number of elements. \end{definition} For example, in Figure~\ref{fig:examplepaths} we represent a tableau $T_1 \in \mathcal{T}_7^e$ such that $B(T_1)$ is the tuple $((2:7),(3:12),(7:11))$ and $R(T_1)$ is the tuple $((4:14),(7:10))$. In general, when depicting a tableau $T \in \mathcal{T}_n^e$, the first $\text{b}(T)$ elements of $B(T)$ and the first $\text{r}(T)$ elements of $R(T)$ are painted in blue and red respectively. The meaning of the green color in Figure~\ref{fig:examplepaths} will be explained further (see Definition~\ref{defi:green}). \begin{figure}[!htbp] \begin{center} \begin{tikzpicture}[scale=0.5] \draw (-0.5,0.5) node[scale=1]{$1$}; \draw (-0.5,1.5) node[scale=1]{$2$}; \draw (-0.5,2.5) node[scale=1]{$3$}; \draw (-0.5,3.5) node[scale=1]{$4$}; \draw (-0.5,4.5) node[scale=1]{$5$}; \draw (-0.5,5.5) node[scale=1]{$6$}; \draw (-0.5,6.5) node[scale=1]{$7$}; \draw (-0.5,7.5) node[scale=1]{$8$}; \draw (-0.5,8.5) node[scale=1]{$9$}; \draw (-0.5,9.5) node[scale=1]{$10$}; \draw (-0.5,10.5) node[scale=1]{$11$}; \draw (-0.5,11.5) node[scale=1]{$12$}; \draw (-0.5,12.5) node[scale=1]{$13$}; \draw (-0.5,13.5) node[scale=1]{$14$}; \draw (0.5,-0.5) node[scale=1]{$1$}; \draw (1.5,-0.5) node[scale=1]{$2$}; \draw (2.5,-0.5) node[scale=1]{$3$}; \draw (3.5,-0.5) node[scale=1]{$4$}; \draw (4.5,-0.5) node[scale=1]{$5$}; \draw (5.5,-0.5) node[scale=1]{$6$}; \draw (6.5,-0.5) node[scale=1]{$7$}; \draw (0,0) grid[step=1] (7,14); \draw (0,0) -- (7,7); \draw [dashed] (0,14) -- (7,7); \fill (0.5,0.5) circle (0.2); \fill (0.5,1.5) circle (0.2); \fill (2.5,2.5) circle (0.2); \fill (3.5,3.5) circle (0.2); \fill (4.5,4.5) circle (0.2); \fill (4.5,5.5) circle (0.2); \fill (1.5,6.5) circle (0.2); \fill (5.5,7.5) circle (0.2); \draw (5.5,8.5) node[scale=0.8]{$\bigstar$}; \draw (6.5,9.5) node[scale=0.8]{$\bigstar$}; \draw (6.5,10.5) node[scale=0.8]{$\bigstar$}; \draw (2.5,11.5) node[scale=0.8]{$\bigstar$}; \draw (1.5,12.5) node[scale=0.8]{$\bigstar$}; \draw (3.5,13.5) node[scale=0.8]{$\bigstar$}; \draw[color=red] (3.5,13.5) node[scale=0.8]{$\bigstar$}; \draw[color=red] [very thick] (3.5,13.5) [densely dashed] -- (3.5,9.5); \draw[color=red] [very thick] (3.5,9.5) [densely dashed] -- (6.5,9.5); \fill[color=blue] (1.5,6.5) circle (0.2); \draw[color=blue] [very thick] (1.5,6.5) [densely dashed] -- (1.5,1.5); \draw[color=blue] [very thick] (1.5,1.5) [densely dashed] -- (0.5,1.5); \draw[color=blue] [very thick] (0.5,1.5) [densely dashed] -- (0.5,12.5); \draw[color=blue] [very thick] (0.5,12.5) [densely dashed] -- (1.5,12.5); \draw[color=blue] [very thick] (1.5,12.5) [densely dashed] -- (1.5,11.5); \draw[color=blue] [very thick] (1.5,11.5) [densely dashed] -- (2.5,11.5); \draw[color=blue] (2.5,11.5) node[scale=0.8]{$\bigstar$}; \draw[color=blue] [very thick] (2.5,11.5) [densely dashed] -- (2.5,10.5); \draw[color=blue] [very thick] (2.5,10.5) [densely dashed] -- (6.5,10.5); \draw[color=ForestGreen] [very thick] (1.3,1.3) [densely dashed] -- (0.3,1.3); \draw[color=ForestGreen] [very thick] (0.3,1.3) [densely dashed] -- (0.3,12.7); \draw[color=ForestGreen] [very thick] (0.3,12.7) [densely dashed] -- (1.3,12.7); \draw[color=ForestGreen] (1.5,12.5) node[scale=0.8]{$\bigstar$}; \end{tikzpicture} \end{center} \caption{Extended Dellac configuration $T_1 \in \mathcal{T}_7^e$ such that $(\text{b}(T_1),\text{r}(T_1),\text{g}(T_1)) = (\textcolor{blue}{2},\textcolor{red}{1},\textcolor{ForestGreen}{1})$.} \label{fig:examplepaths} \end{figure} \begin{definition} \label{defi:TPath} Let $j \in [n]$. We consider a tableau $T$ made up of $n$ columns (each of which contains at most two points) and $2n$ rows (each of which contains at most one point), such that each of the first $j-1$ rows (from bottom to top) of $T$ contains exactly one point on or above the line $y = x$, and each of the first $j-1$ columns (from left to right) contains exactly two points (for example, any tableau $T \in \mathcal{T}_n^e$ satisfies this condition). Let $i \in [j,2n]$ be such that the first $j-1$ boxes of the row $L_i^T$ are empty (for example, any $i \in [j,2n]$ such that the box $(j:i)$ of $T$ contains a point). We define a sequence $\mathcal{I}_T(j:i) = (i_0,i_1,\dots)$ of elements of $[2n]$ as follows: first of all, let $i_0$ be the integer $i$, afterwards, if the integers $i_0,\dots,i_k$ are defined for some $k \geq 0$: \begin{enumerate} \item if $i_k \in [j,2n-j] \sqcup \{2n\}$, then $i_{k+1}$ is defined as $i_k$; \item if $i_k \in [2n-j+1,2n-1]$, let $j_k = 2n - i_k \in [j-1]$, then $i_{k+1}$ is defined by $(j_k:i_{k+1})$ being the upper point of $C_{j_k}^T$; \item otherwise $i_k \in [j-1]$, and $i_{k+1}$ is defined by $(i_k:i_{k+1})$ being the lower point of the $C_{i_k}^T$. \end{enumerate} \end{definition} For example, in Figure~\ref{fig:examplepaths}, one can read the sequences corresponding to the free points $p = (j:i)$ that appear in the red, blue or green path, by following the path in the opposite direction $(\dots,p_{k},p_{k-1},\dots)$. We have $\mathcal{I}_{T_1}(7:10) = (10,14,14,\dots),\mathcal{I}_{T_1}(7:11)=(11,12,13,2,7,7,\dots)$, etc. The green path, as a subpath of the blue path, also gives $\mathcal{I}_{T_1}(2:13) = (13,2,2,\dots)$. \begin{proposition} \label{prop:TPathstationnaire} In the context of Definition~\ref{defi:TPath}, the sequence $\mathcal{I}_T(j:i) = (i_0,i_1,\dots)$ becomes stationary, i.e., there exists an integer ${\rm root}_T(j:i) \in [j,2n-j] \sqcup \{2n\}$ (named the root of the box $(j:i)$ in $T$) and an integer $k_0 \geq 0$ such that $i_k = {\rm root}_T(j:i)$ for all $k \geq k_0$. Moreover, the map $i \mapsto {\rm root}_T(j:i)$ is a bijection from the set of integers $i \in [j,2n]$ such that the first $j-1$ boxes of $L_i^T$ are empty, to $[j,2n-j] \sqcup \{2n\}$. \end{proposition} \begin{proof} Since $[2n]$ is a finite set, there exist $0 \leq k_1 < k_2$ such that $i_{k_1} = i_{k_2}$. Suppose that $i_k \not\in [j,2n-j] \sqcup \{2n\}$ for all $k \geq 0$, i.e., the rule (i) of Definition~\ref{defi:TPath} is never applied. Then, the sequence $(i_k)_{k \geq 0}$ is reversible: for all $k > 0$, let $j_{k-1} \in [n]$ such that $p_{i_k}^T = (j_{k-1}:i_k)$; if $(j_{k-1}:i_k)$ is the upper point of its column, then $i_{k-1} = 2n-j_{k-1}$, otherwise $i_{k-1} = j_{k-1}$. This reversibility and the equality $i_{k_2} = i_{k_1}$ imply $i_{k_2-k_1} = i_{k_1-k_1} = i_0 = i$. Since $k_2 - k_1 > 0$ and, for all $k > 0$, the point $p_{i_k}^T$ is of the kind $(j_{k-1}:i_k)$ for some $j_{k-1} \in [j-1]$, then the point $p = (j:i)$ equals $(j_{k_2-k_1-1}:i_{k_2-k_1})$ for some $j_{k_2-k_1-1} < j$, which is absurd, so $i_k$ belongs to $[j,2n-j] \sqcup \{2n\}$ for $k$ big enough. Let $k_{\min}$ be the smallest integer $k \geq 0$ such that $i_k = \text{root}_T(j:i)$. As stated in the previous paragraph, the sequence $(i=i_0,i_1,\hdots,i_{k_{\min}})$ is reversible because it never involves the rule (i) of Definition~\ref{defi:TPath}, so the map $i \mapsto \text{root}_T(j:i)$ is injective. Finally, the number of integers $i \in [j,2n]$ such that the first $j-1$ boxes of $L_i^T$ are empty is exactly the cardinality $2(n-j+1)$ of the set $[j,2n-j] \sqcup \{2n\}$ (hence the map $i \mapsto \text{root}_T(j:i)$ is bijective): since the first $j-1$ rows of $T$ contain exactly $j-1$ points, and the first $j-1$ columns of $T$ contain exactly $2(j-1)$ points, then, among the $2n-j+1$ upper rows of $T$ (the rows $C_j^T,C_{j+1}^T,\dots,C_{2n}^T$), exactly $j-1$ of them contain a point in one of their first $j-1$ boxes, so $2n-j+1 - (j-1) = 2(n-j+1)$ of them have their first $j-1$ boxes empty. \end{proof} \begin{definition} \label{defi:green} We define $\mathcal{G}(T)$ as the set of the free points $p = (j:i)$ of $T$ such that: \begin{itemize} \item the point other than $p$ in $C_j^T$ is an element of $B(T)$; \item the integer $\text{root}_T(j:i)$ equals $j$. \end{itemize} The cardinality of $\mathcal{G}(T)$ is denoted by $\text{g}(T)$ and its elements are painted in green in the graphical representation of $T$. \end{definition} For example, in Figure~\ref{fig:examplepaths}, we have $\mathcal{G}(T_1) = \{(2:13)\}$. \begin{definition} Let $n \geq 1$ and $T \in \mathcal{T}_n^e$. We denote by $\text{Omax}(T)$ (whose cardinality is $\text{omax}(T)$) the set of the elements of $B(T),R(T)$ and $\mathcal{G}(T)$. Note that $\text{omax}(T) \geq 2$. The set $\text{Max}(T)$ (whose cardinality is $\max(T)$) is defined as $\text{Omax}(T)$ from which we removed the last element of $B(T)$ and the last element of $R(T)$. Hence $$\max(T) = b(T)+r(T)+g(T) = \omax(T)-2.$$ \end{definition} \begin{remark} \label{rem:evenmaximalfreepoints} For all $T \in \mathcal{T}_n^e$, the $\text{b}(T)$ last points of $B(T)$, the $\text{r}(T)+1$ points of $R(T)$, and the $\text{g}(T)$ points of $\mathcal{G}(T)$, are pairwise distinct free points of $T$, so $$\text{fr}(T) \geq \max(T)+1.$$ \end{remark} The first main result of this paper is the following. \begin{theorem} \label{theo:PnintermsofTne} For all $n \geq 1$ we have \begin{equation*} \label{eq:PnintermsofTne} P_n(x) = \sum_{T \in \mathcal{T}_n^e} 2^{{\rm fr}(T)-1-\max(T)} x^{\max(T)}. \end{equation*} \end{theorem} For example, the three elements of $\mathcal{T}_2^e$ shown in Figure~\ref{fig:T2ecolors} give $$ P_2(x) = 2^{2-1-\textcolor{blue}{1}-\textcolor{red}{0}-\textcolor{ForestGreen}{0}} x^{\textcolor{blue}{1}+\textcolor{red}{0}+\textcolor{ForestGreen}{0}} + 2^{1-1-\textcolor{blue}{0}-\textcolor{red}{0}-\textcolor{ForestGreen}{0}} x^{\textcolor{blue}{0}+\textcolor{red}{0}+\textcolor{ForestGreen}{0}}+2^{2-1-\textcolor{blue}{0}-\textcolor{red}{1}-\textcolor{ForestGreen}{0}} x^{\textcolor{blue}{0}+\textcolor{red}{1}+\textcolor{ForestGreen}{0}} = 2x+1. $$ \begin{figure}[!htbp] \begin{center} \begin{tikzpicture}[scale=0.4] \draw (0,0) grid[step=1] (2,4); \draw (0,0) -- (2,2); \draw [dashed] (0,4) -- (2,2); \fill (0.5,0.5) circle (0.2); \fill [color=blue] (0.5,1.5) circle (0.2); \draw (1.5,2.5) node[scale=0.8]{$\bigstar$}; \draw (1.5,3.5) node[scale=0.8]{$\bigstar$}; \draw[color=black] (1.5,3.5) node[scale=0.8]{$\bigstar$}; \draw[color=blue] [very thick] (0.5,1.5) [densely dashed] -- (0.5,2.5); \draw[color=blue] [very thick] (0.5,2.5) [densely dashed] -- (1.5,2.5); \draw[color=black] (1.5,2.5) node[scale=0.8]{$\bigstar$}; \begin{scope}[shift={(3,0)}] \draw (0,0) grid[step=1] (2,4); \draw (0,0) -- (2,2); \draw [dashed] (0,4) -- (2,2); \fill (0.5,0.5) circle (0.2); \fill (0.5,2.5) circle (0.2); \fill (1.5,1.5) circle (0.2); \draw (1.5,3.5) node[scale=0.8]{$\bigstar$}; \draw[color=black] (1.5,3.5) node[scale=0.8]{$\bigstar$}; \end{scope} \begin{scope}[shift={(6,0)}] \draw (0,0) grid[step=1] (2,4); \draw (0,0) -- (2,2); \draw [dashed] (0,4) -- (2,2); \fill (0.5,0.5) circle (0.2); \fill (1.5,1.5) circle (0.2); \draw (0.5,3.5) node[scale=0.8]{$\bigstar$}; \draw (1.5,2.5) node[scale=0.8]{$\bigstar$}; \draw[color=red] (0.5,3.5) node[scale=0.8]{$\bigstar$}; \draw[color=red] [very thick] (0.5,3.5) [densely dashed] -- (0.5,2.5); \draw[color=red] [very thick] (0.5,2.5) [densely dashed] -- (1.5,2.5); \draw[color=black] (1.5,2.5) node[scale=0.8]{$\bigstar$}; \end{scope} \end{tikzpicture} \end{center} \caption{The three elements of $\mathcal{T}_2^e$.} \label{fig:T2ecolors} \end{figure} The cardinality of $\SDC_{2n}$ being $r_n = 2P_n(2)$ is then obtained by setting $x = 2$ in Theorem~\ref{theo:PnintermsofTne} in view of Formula~\eqref{eq:TnegeneratesSpDC2n}. Section \ref{sec:PnintermsofTne} is dedicated to the proof of Theorem~\ref{theo:PnintermsofTne} (by induction, using the recursive formulas \eqref{eq:an0}, \eqref{eq:ank} and \eqref{eq:anmoins1}). \begin{remark} An independent proof of $|\SDC_{2n}| = r_n$ is given in a previous paper~\cite{B1}. The proof is based on a surjection $\varphi: \mathcal{T}_n^e \twoheadrightarrow \SP_n$ such that \begin{equation} \label{eq:orbitofapistol} 2^{n-\text{fd}(f)} = \sum_{T \in \varphi^{-1}(f)} 2^{\text{fr}(T)} \end{equation} for all $f \in \SP_n$, which proves $|\SDC_{2n}| = 2P_n(2) = r_n$ in view of Formula~\eqref{eq:PnintermsofSPn} and Formula~\eqref{eq:TnegeneratesSpDC2n}. Afterwards, one can check that for all $f \in \SP_n$ and $T \in \varphi^{-1}(f)$, \begin{enumerate} \item $\text{g}(T)$ is the number of integers $j \in \{2,4,\dots,2n-2\}$ such that $f(j-1) = f(j) = 2n$ and $\min\{j' k_0$; \item in Definition~\ref{defi:evenpath}, instead of $R(T)$ and $B(T)$, we define a single finite sequence $V(T) = S_T(2n+1)$ (possibly empty), whose number of elements is denoted by $\text{v}(T)$; \item the set $\mathcal{G}(T)$ is defined by replacing $B(T)$ with $V(T)$ in Definition~\ref{defi:green}. \end{itemize} For example, in Figure~\ref{fig:examplepaths2} we represent a tableau $T_3 \in \mathcal{T}_6^o$ such that $V(T_3)$ is the tuple $((2:13),(3:11),(4:10))$ and $\mathcal{G}(T_3) = \{(2:12)\}$. In general, when depicting a tableau $T \in \mathcal{T}_n^o$, the $\text{v}(T)$ elements of $V(T)$ and the $\text{g}(T)$ elements of $\mathcal{G}(T)$ are painted in purple and green respectively. \begin{figure}[!htbp] \begin{center} \begin{tikzpicture}[scale=0.5] \draw (-0.5,0.5) node[scale=1]{$1$}; \draw (-0.5,1.5) node[scale=1]{$2$}; \draw (-0.5,2.5) node[scale=1]{$3$}; \draw (-0.5,3.5) node[scale=1]{$4$}; \draw (-0.5,4.5) node[scale=1]{$5$}; \draw (-0.5,5.5) node[scale=1]{$6$}; \draw (-0.5,6.5) node[scale=1]{$7$}; \draw (-0.5,7.5) node[scale=1]{$8$}; \draw (-0.5,8.5) node[scale=1]{$9$}; \draw (-0.5,9.5) node[scale=1]{$10$}; \draw (-0.5,10.5) node[scale=1]{$11$}; \draw (-0.5,11.5) node[scale=1]{$12$}; \draw (-0.5,12.5) node[scale=1]{$13$}; \draw (0.5,-0.5) node[scale=1]{$1$}; \draw (1.5,-0.5) node[scale=1]{$2$}; \draw (2.5,-0.5) node[scale=1]{$3$}; \draw (3.5,-0.5) node[scale=1]{$4$}; \draw (4.5,-0.5) node[scale=1]{$5$}; \draw (5.5,-0.5) node[scale=1]{$6$}; \draw (0,0) grid[step=1] (6,13); \draw (0,0) -- (6,6); \draw [dashed] (0,13) -- (6,7); \fill (0.5,0.5) circle (0.2); \fill (0.5,1.5) circle (0.2); \fill (2.5,2.5) circle (0.2); \fill (3.5,3.5) circle (0.2); \fill (4.5,4.5) circle (0.2); \fill (4.5,5.5) circle (0.2); \fill (5.5,6.5) circle (0.2); \draw (1.5,12.5) node[scale=0.8]{$\bigstar$}; \draw (1.5,11.5) node[scale=0.8]{$\bigstar$}; \draw (2.5,10.5) node[scale=0.8]{$\bigstar$}; \draw (3.5,9.5) node[scale=0.8]{$\bigstar$}; \draw (5.5,7.5) node[scale=0.8]{$\bigstar$}; \draw[color=Purple] (1.5,12.5) node[scale=0.8]{$\bigstar$}; \draw[color=Purple] [very thick] (1.5,12.5) [densely dashed] -- (1.5,10.5); \draw[color=Purple] [very thick] (1.5,10.5) [densely dashed] -- (2.5,10.5); \draw[color=Purple] (2.5,10.5) node[scale=0.8]{$\bigstar$}; \draw[color=Purple] [very thick] (2.5,10.5) [densely dashed] -- (2.5,9.5); \draw[color=Purple] [very thick] (2.5,9.5) [densely dashed] -- (3.5,9.5); \draw[color=Purple] (3.5,9.5) node[scale=0.8]{$\bigstar$}; \draw[color=Purple] [very thick] (3.5,9.5) [densely dashed] -- (3.5,8.5); \draw[color=ForestGreen] [very thick] (1.5,1.5) [densely dashed] -- (0.5,1.5); \draw[color=ForestGreen] [very thick] (0.5,1.5) [densely dashed] -- (0.5,11.5); \draw[color=ForestGreen] [very thick] (0.5,11.5) [densely dashed] -- (1.5,11.5); \draw[color=ForestGreen] (1.5,11.5) node[scale=0.8]{$\bigstar$}; \end{tikzpicture} \end{center} \caption{Odd extended Dellac configuration $T_3 \in \mathcal{T}_6^o$ such that $(\text{v}(T_3),\text{g}(T_3)) = (\textcolor{Purple}{3},\textcolor{ForestGreen}{1})$.} \label{fig:examplepaths2} \end{figure} \begin{remark} \label{rem:oddmaximalfreepoints} For all $T \in \mathcal{T}_{n-1}^o$, the $\text{v}(T)$ elements of $V(T)$ and the $\text{g}(T)$ elements of $\mathcal{G}(T)$ are pairwise distinct free points, so $\text{fr}(T) \geq \text{v}(T) + \text{g}(T)$. \end{remark} The second main result of this paper is the following. \begin{theorem} \label{theo:PnintermsofTno} For all $n \geq 2$, \begin{equation*} \label{eq:PnintermsofTno} P_n(x) = \sum_{T \in \mathcal{T}_{n-1}^o} 2^{{\rm fr}(T)-\text{g}(T)} x^{{\rm v}(T)}(1+x)^{\text{g}(T)}. \end{equation*} \end{theorem} For example, the two elements of $\mathcal{T}_1^o$ shown in Figure~\ref{fig:T1o} give $$P_2(x) = 2^{0-\textcolor{ForestGreen}{0}} x^{\textcolor{Purple}{0}} (1+x)^{\textcolor{ForestGreen}{0}} + 2^{1-\textcolor{ForestGreen}{0}} x^{\textcolor{Purple}{1}} (1+x)^{\textcolor{ForestGreen}{0}} = 2x+1,$$ and the nine elements of $\mathcal{T}_2^o$ shown in Figure~\ref{fig:T2ocolors} give \begin{align*} P_3(x) &= 2^{1-\textcolor{ForestGreen}{0}} x^{\textcolor{Purple}{0}} (1+x)^{\textcolor{ForestGreen}{0}} + 2^{1-\textcolor{ForestGreen}{0}} x^{\textcolor{Purple}{1}} (1+x)^{\textcolor{ForestGreen}{0}} + 2^{2-\textcolor{ForestGreen}{1}} x^{\textcolor{Purple}{1}} (1+x)^{\textcolor{ForestGreen}{1}}\\ &+ 2^{1-\textcolor{ForestGreen}{0}} x^{\textcolor{Purple}{0}} (1+x)^{\textcolor{ForestGreen}{0}} + 2^{1-\textcolor{ForestGreen}{0}} x^{\textcolor{Purple}{1}} (1+x)^{\textcolor{ForestGreen}{0}} + 2^{1-\textcolor{ForestGreen}{0}} x^{\textcolor{Purple}{1}} (1+x)^{\textcolor{ForestGreen}{0}}\\ &+ 2^{0-\textcolor{ForestGreen}{0}} x^{\textcolor{Purple}{0}} (1+x)^{\textcolor{ForestGreen}{0}} + 2^{1-\textcolor{ForestGreen}{0}} x^{\textcolor{Purple}{1}} (1+x)^{\textcolor{ForestGreen}{0}} + 2^{2-\textcolor{ForestGreen}{0}} x^{\textcolor{Purple}{2}} (1+x)^{\textcolor{ForestGreen}{0}}\\ &= 2+2x+2x(1+x)+2+2x+2x+1+2x+4x^2\\ &= 6x^2+10x+5. \end{align*} \begin{figure}[!htbp] \begin{center} \begin{tikzpicture}[scale=0.4] \draw (0,0) grid[step=1] (1,3); \draw (0,0) -- (1,1); \draw [dashed] (0,3) -- (1,2); \fill (0.5,0.5) circle (0.2); \fill [color=black] (0.5,1.5) circle (0.2); \begin{scope}[shift={(2,0)}] \draw (0,0) grid[step=1] (1,3); \draw (0,0) -- (1,1); \draw [dashed] (0,3) -- (1,2); \fill (0.5,0.5) circle (0.2); \draw[color=Purple] (0.5,2.5) node[scale=0.8]{$\bigstar$}; \draw[color=Purple] [very thick] (0.5,2.5) [densely dashed] -- (0.5,1.5); \end{scope} \end{tikzpicture} \end{center} \caption{The two elements of $\mathcal{T}_1^o$.} \label{fig:T1o} \end{figure} \begin{figure}[!htbp] \begin{center} \begin{tikzpicture}[scale=0.4] \draw (0,0) grid[step=1] (2,5); \draw (0,0) -- (2,2); \draw [dashed] (0,5) -- (2,3); \fill (0.5,0.5) circle (0.2); \fill (0.5,1.5) circle (0.2); \fill (1.5,2.5) circle (0.2); \draw (1.5,3.5) node[scale=0.8]{$\bigstar$}; \begin{scope}[shift={(3,0)}] \draw (0,0) grid[step=1] (2,5); \draw (0,0) -- (2,2); \draw [dashed] (0,5) -- (2,3); \fill (0.5,0.5) circle (0.2); \fill (0.5,1.5) circle (0.2); \fill (1.5,2.5) circle (0.2); \draw[color=Purple] (1.5,4.5) node[scale=0.8]{$\bigstar$}; \draw[color=Purple] [very thick] (1.65,4.5) [densely dashed] -- (1.65,2.5); \draw[color=Purple] [very thick] (1.35,2.5) [densely dashed] -- (1.35,1.5); \draw[color=Purple] [very thick] (1.35,1.5) [densely dashed] -- (0.5,1.5); \draw[color=Purple] [very thick] (0.5,1.5) [densely dashed] -- (0.5,3.5); \end{scope} \begin{scope}[shift={(6,0)}] \draw (0,0) grid[step=1] (2,5); \draw (0,0) -- (2,2); \draw [dashed] (0,5) -- (2,3); \fill (0.5,0.5) circle (0.2); \fill (0.5,1.5) circle (0.2); \draw[color=Purple] (1.5,4.5) node[scale=0.8]{$\bigstar$}; \draw[color=Purple] [very thick] (1.5,4.5) [densely dashed] -- (1.5,2.5); \draw[color=ForestGreen] (1.5,3.5) node[scale=0.8]{$\bigstar$}; \draw[color=ForestGreen] [very thick] (1.5,3.5) [densely dashed] -- (0.5,3.5); \draw[color=ForestGreen] [very thick] (0.5,3.5) [densely dashed] -- (0.5,1.5); \draw[color=ForestGreen] [very thick] (0.5,1.5) [densely dashed] -- (1.5,1.5); \end{scope} \begin{scope}[shift={(9,0)}] \draw (0,0) grid[step=1] (2,5); \draw (0,0) -- (2,2); \draw [dashed] (0,5) -- (2,3); \fill (0.5,0.5) circle (0.2); \fill (0.5,2.5) circle (0.2); \fill (1.5,1.5) circle (0.2); \draw (1.5,3.5) node[scale=0.8]{$\bigstar$}; \begin{scope}[shift={(3,0)}] \draw (0,0) grid[step=1] (2,5); \draw (0,0) -- (2,2); \draw [dashed] (0,5) -- (2,3); \fill (0.5,0.5) circle (0.2); \fill (0.5,2.5) circle (0.2); \fill (1.5,1.5) circle (0.2); \draw[color=Purple] (1.5,4.5) node[scale=0.8]{$\bigstar$}; \draw[color=Purple] [very thick] (1.5,4.5) [densely dashed] -- (1.5,2.5); \draw[color=Purple] [very thick] (1.5,2.5) [densely dashed] -- (0.5,2.5); \draw[color=Purple] [very thick] (0.5,2.5) [densely dashed] -- (0.5,3.5); \end{scope} \begin{scope}[shift={(6,0)}] \draw (0,0) grid[step=1] (2,5); \draw (0,0) -- (2,2); \draw [dashed] (0,5) -- (2,3); \fill (0.5,0.5) circle (0.2); \fill (1.5,1.5) circle (0.2); \fill (0.5,3.5) circle (0.2); \draw[color=Purple] (1.5,4.5) node[scale=0.8]{$\bigstar$}; \draw[color=Purple] [very thick] (1.5,4.5) [densely dashed] -- (1.5,2.5); \end{scope} \end{scope} \begin{scope}[shift={(18,0)}] \draw (0,0) grid[step=1] (2,5); \draw (0,0) -- (2,2); \draw [dashed] (0,5) -- (2,3); \fill (0.5,0.5) circle (0.2); \fill (1.5,1.5) circle (0.2); \fill (0.5,3.5) circle (0.2); \fill (1.5,2.5) circle (0.2); \begin{scope}[shift={(3,0)}] \draw (0,0) grid[step=1] (2,5); \draw (0,0) -- (2,2); \draw [dashed] (0,5) -- (2,3); \fill (0.5,0.5) circle (0.2); \fill (1.5,1.5) circle (0.2); \draw[color=Purple] (0.5,4.5) node[scale=0.8]{$\bigstar$}; \draw[color=Purple] [very thick] (0.5,4.5) [densely dashed] -- (0.5,3.5); \fill (1.5,2.5) circle (0.2); \end{scope} \begin{scope}[shift={(6,0)}] \draw (0,0) grid[step=1] (2,5); \draw (0,0) -- (2,2); \draw [dashed] (0,5) -- (2,3); \fill (0.5,0.5) circle (0.2); \fill (1.5,1.5) circle (0.2); \draw[color=Purple] (0.5,4.5) node[scale=0.8]{$\bigstar$}; \draw[color=Purple] [very thick] (0.5,4.5) [densely dashed] -- (0.5,3.5); \draw[color=Purple] [very thick] (0.5,3.5) [densely dashed] -- (1.5,3.5); \draw[color=Purple] (1.5,3.5) node[scale=0.8]{$\bigstar$}; \draw[color=Purple] [very thick] (1.5,3.5) [densely dashed] -- (1.5,2.5); \end{scope} \end{scope} \end{tikzpicture} \end{center} \caption{The nine elements of $\mathcal{T}_2^o$.} \label{fig:T2ocolors} \end{figure} The cardinality of $\SDC_{2n-1}$ being $P_n(1) = l_n$ for all $n \geq 2$ is then obtained by setting $x=1$ in Theorem~\ref{theo:PnintermsofTno}, in view of Formula~\eqref{eq:TnogeneratesSpDC2nmoins1}. Section~\ref{sec:PnintermsofTno} is dedicated to the proof of Theorem~\ref{theo:PnintermsofTno}. \section{Proof of Theorem~\ref{theo:PnintermsofTne}} \label{sec:PnintermsofTne} For all $n \geq 1$, let \begin{align*} E_n(x) = \sum_{T \in \mathcal{T}_n^e} 2^{\text{fr}(T)-1-\max(T)} x^{\max(T)}. \end{align*} By Remark~\ref{rem:evenmaximalfreepoints} and the inequality $\text{fr}(T) \leq n$ for all $T \in \mathcal{T}_n^e$ (the free points of $T$ are of the kind $(j:i)$ such that $2n \geq i \geq 2n+1-j \geq n+1$, and hence are located in the $n$ top rows), the polynomial $E_n(x)$ has positive integers coefficients and its degree is at most $n-1$. Moreover, if we define $T_{n,\max} \in \mathcal{T}_n^e$ as the tableau whose points are $(j:j)$ and $(j:2n+1-j)$ for all $j \in [n]$ (see Figure~\ref{fig:T4max} for the case $n = 4$), then $\text{b}(T_{n,\max}) = 0,\text{r}(T_{n,\max}) = n-1$ and $\text{g}(T_{n,\max}) =0$, so $\max(T_{n,\max}) = n-1$, which makes the degree of $E_n(x)$ exactly $n-1$. \begin{figure}[!htbp] \begin{center} \begin{tikzpicture}[scale=0.5] \draw (-0.5,0.5) node[scale=1]{$1$}; \draw (-0.5,1.5) node[scale=1]{$2$}; \draw (-0.5,2.5) node[scale=1]{$3$}; \draw (-0.5,3.5) node[scale=1]{$4$}; \draw (-0.5,4.5) node[scale=1]{$5$}; \draw (-0.5,5.5) node[scale=1]{$6$}; \draw (-0.5,6.5) node[scale=1]{$7$}; \draw (-0.5,7.5) node[scale=1]{$8$}; \draw (0.5,-0.5) node[scale=1]{$1$}; \draw (1.5,-0.5) node[scale=1]{$2$}; \draw (2.5,-0.5) node[scale=1]{$3$}; \draw (3.5,-0.5) node[scale=1]{$4$}; \draw (0,0) grid[step=1] (4,8); \draw (0,0) -- (4,4); \draw [dashed] (0,8) -- (4,4); \fill (0.5,0.5) circle (0.2); \fill (1.5,1.5) circle (0.2); \fill (2.5,2.5) circle (0.2); \fill (3.5,3.5) circle (0.2); \draw[color=red] (0.5,7.5) node[scale=0.8]{$\bigstar$}; \draw[color=red] [very thick] (0.5,7.5) [densely dashed] -- (0.5,6.5); \draw[color=red] [very thick] (0.5,6.5) [densely dashed] -- (1.5,6.5); \draw[color=red] (1.5,6.5) node[scale=0.8]{$\bigstar$}; \draw[color=red] [very thick] (1.5,6.5) [densely dashed] -- (1.5,5.5); \draw[color=red] [very thick] (1.5,5.5) [densely dashed] -- (2.5,5.5); \draw[color=red] (2.5,5.5) node[scale=0.8]{$\bigstar$}; \draw[color=red] [very thick] (2.5,5.5) [densely dashed] -- (2.5,4.5); \draw[color=red] [very thick] (2.5,4.5) [densely dashed] -- (3.5,4.5); \draw[color=black] (3.5,4.5) node[scale=0.8]{$\bigstar$}; \end{tikzpicture} \end{center} \caption{The tableau $T_{4,\max} \in \mathcal{T}_4^e$.} \label{fig:T4max} \end{figure} For all $k \in [0,n-1]$, let $\mathcal{T}^e_{n,k}$ be the set of the tableaux $T \in \mathcal{T}_n^e$ such that $\max(T) = k$, and let $e_{n,k} = \sum\limits_{T \in \mathcal{T}_{n,k}^e} 2^{\text{fr}(T)-1-k}$, so that $$E_n(x) = \sum_{k = 0}^{n-1} e_{n,k} x^k.$$ Theorem~\ref{theo:PnintermsofTne} states that $E_n(x) = P_n(x)$, i.e., that $e_{n,k} = c_{n,k}$ for all $k \in [0,n-1]$. It is clear that $e_{1,0} = 1 = c_{1,0}$. We will prove the rest by checking that $(e_{n,k})_{0 \leq k \leq n}$ satisfies the recursive formulas \eqref{eq:an0} ,\eqref{eq:ank},\eqref{eq:anmoins1}. To do so, we first define in the next section a surjection from $\mathcal{T}_n^e$ to $\mathcal{T}_{n-1}^e$. \subsection{A surjection $\Pi: \mathcal{T}_n^e \twoheadrightarrow \mathcal{T}_{n-1}^e (n \geq 2)$} Consider $T \in \mathcal{T}_n^e$ and the integers $(j_{n-1},j_{n+1}) \in [n]^2$ such that the points of $L_{n-1}^T$ and $L_{n+1}^T$ are located in $C_{j_{n-1}}^T$ and $C_{j_{n+1}}^T$ respectively. We define $(i_{\min},i_{\max})$ as $(n-1,n+1)$ if $j_{n-1} \leq j_{n+1}$, as $(n+1,n-1)$ otherwise. We define two new paths \begin{align*} (B'(T),R'(T)) = \begin{cases} (S_T(i_{\min}),S_T(i_{\max})), &\text{if $C_{n-1}^T$ contains an element of $R(T)$;}\\ (S_T(i_{\max}),S_T(i_{\min})), &\text{if $C_{n-1}^T$ contains an element of $\mathcal{G}(T)$;}\\ (S_T(n+1),S_T(n-1)), &\text{otherwise and if $p_{n+1}^T$ is free;}\\ (S_T(n-1),S_T(n+1)), &\text{otherwise.} \end{cases} \end{align*} We also define a new set $\mathcal{G}'(T)$: the set of the free points of $T$ of the kind $p = (j:i)$ such that the other point of $C_j^T$ is an element of $B'(T)$, and such that $\text{root}_T(j:i) =j$. Their respective numbers of elements located in $C_1^T,C_2^T,\dots,C_{n-1}^T$ are denoted by $\text{b}'(T)$, $\text{r}'(T)$ and $\text{g}'(T)$. Note that $S_T(n+1) = ((n:n+1))$ for all tableau $T$ in which $p_{n+1}^T$ is free, so $r'(T)$ or $b'(T)$ may equal $0$. However, the first element of $S_T(n-1)$, the point $p_{n-1}^T$, is always of the kind $(j:n-1)$ for some $jn+1$.} \end{cases}$$ For example, for the tableau $T_1 \in \mathcal{T}_7^e$ depicted in Figure~\ref{fig:newpaths}, we obtain the tableau $\Pi_0(T_1)$ of Figure~\ref{fig:Pi0T1}. \begin{figure}[!htbp] \begin{center} \begin{tikzpicture}[scale=0.5] \draw (-0.5,0.5) node[scale=1]{$1$}; \draw (-0.5,1.5) node[scale=1]{$2$}; \draw (-0.5,2.5) node[scale=1]{$3$}; \draw (-0.5,3.5) node[scale=1]{$4$}; \draw (-0.5,4.5) node[scale=1]{$5$}; \draw (-0.5,5.5) node[scale=1]{$6$}; \draw (-0.5,6.5) node[scale=1]{$7$}; \draw (-0.5,7.5) node[scale=1]{$8$}; \draw (-0.5,8.5) node[scale=1]{$9$}; \draw (-0.5,9.5) node[scale=1]{$10$}; \draw (-0.5,10.5) node[scale=1]{$11$}; \draw (-0.5,11.5) node[scale=1]{$12$}; \draw (0.5,-0.5) node[scale=1]{$1$}; \draw (1.5,-0.5) node[scale=1]{$2$}; \draw (2.5,-0.5) node[scale=1]{$3$}; \draw (3.5,-0.5) node[scale=1]{$4$}; \draw (4.5,-0.5) node[scale=1]{$5$}; \draw (5.5,-0.5) node[scale=1]{$6$}; \draw (0,0) grid[step=1] (6,12); \draw (0,0) -- (6,6); \draw [dashed] (0,12) -- (6,6); \fill (0.5,0.5) circle (0.2); \fill (0.5,1.5) circle (0.2); \fill (2.5,2.5) circle (0.2); \fill (3.5,3.5) circle (0.2); \fill (4.5,4.5) circle (0.2); \end{tikzpicture} \end{center} \caption{The tableau $\Pi_0(T_1)$.} \label{fig:Pi0T1} \end{figure} Afterwards, let $X_T$ be the empty set. For $j$ from $1$ to $n-1$, if $C_j^T$ contains a point $p$ labeled with $\beta$ (respectively $\rho,\gamma$), we insert a point in the box $(j:n-1)$ of $\Pi_0(T)$ (respectively the box $(j:2n-2)$, the box $(j:j)$), following Definition~\ref{defi:insertion}. We say that \textit{we have inserted $p$ in $\Pi_0(T)$}. Also, if $p \in \text{Max}(T)$, let $p_{\text{new}}$ be the point that we just plotted in $\Pi_0(T)$ following the latter insertion, then we add $p_{\text{new}}$ to $X_T$. Let $\Pi(T)$ be the tableau obtained at the end of this algorithm. It is straightforward by Definition~\ref{defi:insertion} that $\Pi(T)$ is an element of $\mathcal{T}_{n-1}^e$, and that $X_T \subsetneq \text{Omax}(\Pi(T))$ (it is not equal because the point $p_{n-1}^{\Pi(T)}$ is an element of $\text{Omax}(\Pi(T))$ that results from the insertion in $\Pi_0(T)$ of the point $p_{n-1}^T$, but $p_{n-1}^T$ belongs to $\text{Max}'(T)$, not to $\text{Max}(T)$. Hence $p_{n-1}^{\Pi(T)} \not\in X_T$). The set $X_T$ will be useful to reconstruct $T$ from $\Pi(T)$ (see Proposition~\ref{prop:Pimoins1}). For example, the tableau $\Pi(T_1) \in \mathcal{T}_6^e$ is as depicted in Figure~\ref{fig:PiT1}, and $X_{T_1}$ is the tuple $\{(2:6),(2:11),(3:10),(4:12)\}$. \begin{figure}[!htbp] \begin{center} \begin{tikzpicture}[scale=0.5] \draw (-0.5,0.5) node[scale=1]{$1$}; \draw (-0.5,1.5) node[scale=1]{$2$}; \draw (-0.5,2.5) node[scale=1]{$3$}; \draw (-0.5,3.5) node[scale=1]{$4$}; \draw (-0.5,4.5) node[scale=1]{$5$}; \draw (-0.5,5.5) node[scale=1]{$6$}; \draw (-0.5,6.5) node[scale=1]{$7$}; \draw (-0.5,7.5) node[scale=1]{$8$}; \draw (-0.5,8.5) node[scale=1]{$9$}; \draw (-0.5,9.5) node[scale=1]{$10$}; \draw (-0.5,10.5) node[scale=1]{$11$}; \draw (-0.5,11.5) node[scale=1]{$12$}; \draw (0.5,-0.5) node[scale=1]{$1$}; \draw (1.5,-0.5) node[scale=1]{$2$}; \draw (2.5,-0.5) node[scale=1]{$3$}; \draw (3.5,-0.5) node[scale=1]{$4$}; \draw (4.5,-0.5) node[scale=1]{$5$}; \draw (5.5,-0.5) node[scale=1]{$6$}; \draw (0,0) grid[step=1] (6,12); \draw (0,0) -- (6,6); \draw [dashed] (0,12) -- (6,6); \fill (0.5,0.5) circle (0.2); \fill (0.5,1.5) circle (0.2); \fill (2.5,2.5) circle (0.2); \fill (3.5,3.5) circle (0.2); \fill (4.5,4.5) circle (0.2); \draw[color=red] (3.5,11.5) node[scale=0.8]{$\bigstar$}; \draw[color=red] [very thick] (3.5,11.5) [densely dashed] -- (3.5,7.5); \draw[color=red] [very thick] (3.5,7.5) [densely dashed] -- (5.5,7.5); \draw[color=Peach] (5.5,7.5) node[scale=0.8]{$\bigstar$}; \fill[color=blue] (1.5,5.5) circle (0.2); \draw[color=blue] [very thick] (1.5,5.5) [densely dashed] -- (1.5,1.5); \draw[color=blue] [very thick] (1.5,1.5) [densely dashed] -- (0.5,1.5); \draw[color=blue] [very thick] (0.5,1.5) [densely dashed] -- (0.5,10.5); \draw[color=blue] [very thick] (0.5,10.5) [densely dashed] -- (1.5,10.5); \draw[color=blue] [very thick] (1.5,10.5) [densely dashed] -- (1.5,9.5); \draw[color=blue] [very thick] (1.5,9.5) [densely dashed] -- (2.5,9.5); \draw[color=blue] (2.5,9.5) node[scale=0.8]{$\bigstar$}; \draw[color=blue] [very thick] (2.5,9.5) [densely dashed] -- (2.5,8.5); \draw[color=blue] [very thick] (2.5,8.5) [densely dashed] -- (4.5,8.5); \draw[color=Turquoise] (4.5,8.5) node[scale=0.8]{$\bigstar$}; \draw[color=blue] [very thick] (4.5,8.5) [densely dashed] -- (4.5,6.5); \draw[color=blue] [very thick] (4.5,6.5) [densely dashed] -- (5.5,6.5); \draw[color=Turquoise] (5.5,6.5) node[scale=0.8]{$\bigstar$}; \draw[color=ForestGreen] [very thick] (1.3,1.3) [densely dashed] -- (0.3,1.3); \draw[color=ForestGreen] [very thick] (0.3,1.3) [densely dashed] -- (0.3,10.7); \draw[color=ForestGreen] [very thick] (0.3,10.7) [densely dashed] -- (1.3,10.7); \draw[color=ForestGreen] (1.5,10.5) node[scale=0.8]{$\bigstar$}; \end{tikzpicture} \end{center} \caption{The tableau $\Pi(T_1) \in \mathcal{T}_6^e$.} \label{fig:PiT1} \end{figure} Note that $(\text{b}(\Pi(T_1)),\text{r}(\Pi(T_1)),\text{g}(\Pi(T_1))) = (3,1,1)$. \end{definition} \begin{remark} \label{rem:newstat} For all $T \in \mathcal{T}_n^e$, we have \begin{align*} \max(\Pi(T)) = \max(T) + \text{max}'(T)-2. \end{align*} Indeed, the insertions in $\Pi_0(T)$ of the two points of $C_{n-1}^T$, which both belong to $\text{Max}(T)$ or $\text{Max}'(T)$ (because their roots belong to $\{n-1,n,n+1,2n\}$), produce the two points of $C_{n-1}^{\Pi(T)}$, in other words, the last element of $B(\Pi(T))$ and the last element of $R(\Pi(T))$, which are not counted by the statistics $\text{b}$ and $\text{r}$. In particular, since $\max'(T) \geq 1$ in general (the set $\text{Max}'(T)$ always contains $p_{n-1}^T$), we have $$\max(\Pi(T)) \in [\max(T)-1,n-2].$$ For example, for the tableau $T_1 \in \mathcal{T}_7^e$, we obtain $\max(\Pi(T_1)) = (2+1+1)+(2+1)-2 = 5$. \end{remark} We now intend to construct $\Pi^{-1}(T_0)$ for all $T_0 \in \mathcal{T}_{n-1}^e$. To do so, we first need to define labels for the elements of $\text{Omax}(T_0)$, given a subset $X \subsetneq \Omax(T_0)$. \begin{definition} \label{defi:newlabels} For a tableau $T_0 \in \mathcal{T}_{n-1}^e$ and a subset $X \subsetneq \text{Omax}(T_0)$, let us define labels of the elements of $\text{Omax}(T_0)$ as follows. \begin{itemize} \item For all points $p \in \mathcal{G}(T_0)$, let $q \neq p$ be the other point of the column of $p$ (the point $q$ is then an element of $B(T_0)$). \begin{itemize} \item If neither $p$ nor $q$ belong to $X$, we label $p$ and $q$ with the respective letters $\mathfrak{g}'$ and $\mathfrak{b}'$. \item If only $q$ belongs to $X$, we label $p$ and $q$ with the respective letters $\mathfrak{b}'$ and $\mathfrak{b}$. \item If only $p$ belongs to $X$, we label $p$ and $q$ with the respective letters $\mathfrak{r}$ and $\mathfrak{r}'$. \item If both $p$ and $q$ belong to $X$, we label $p$ and $q$ with the respective letters $\mathfrak{g}$ and $\mathfrak{b}$. \end{itemize} \item For all points $p$ among the first $b_0(T)$ elements of $B(T_0)$ (respectively among the first $r(T_0)$ elements of $R(T_0)$) that are not covered by the first case, if $p \not\in X$, then we label $p$ with $\mathfrak{b}'$ (respectively $\mathfrak{r}'$), otherwise we label $p$ with $\mathfrak{b}$ (respectively $\mathfrak{r}$). \item Let $p^b$ (respectively $p^r$) be the last element of $B(T_0)$ (respectively of $R(T_0)$). Recall that $p^b$ and $p^r$ are the two points of $C_{n-1}^{T_0}$. \begin{itemize} \item If $p^r \not\in X$, we label $p^r$ with the letter $\mathfrak{r}'$. Afterwards, we label $p^b$ with the letter $\mathfrak{b}'$ if $p^b \not\in X$, with the letter $\mathfrak{b}$ otherwise. \item Otherwise, let $j_{\min}$ be the smallest integer $j \in [n-1]$ such that $C_j^{T_0}$ contains an element of $\text{Omax}(T_0) \backslash X$. If $j_{\min} = n-1$, then $\Omax(T_0) = \{p^b,p^r\}$ and hence $X = \{p^r\}$, and we label $p^b$ and $p^r$ with the respective letters $\mathfrak{b}'$ and $\mathfrak{r}$. If $j_{\min} < n-1$, all the elements of $\Omax(T_0)$ that appear in $C_{j_{\min}}^{T_0}$ have already been labeled. We then have two cases: \begin{itemize} \item If $C_{j_{\min}}^{T_0}$ contains an element labeled with $\mathfrak{b}'$, then we label $p^r$ with the letter $\mathfrak{r}$. Afterwards, we label $p^b$ with the letter $\mathfrak{b}'$ if $p^b \not\in X$, with the letter $\mathfrak{b}$ otherwise. \item Otherwise, we label $p^b$ and $p^r$ with the respective letters $\mathfrak{r}'$ and $\mathfrak{r}$ if $p^b \not\in X$, with the respective letters $\mathfrak{b}$ and $\mathfrak{g}$ otherwise. \end{itemize} \end{itemize} \end{itemize} \end{definition} \begin{remark} \label{rem:labelsPi} In the context of Definition~\ref{defi:newlabels}, the elements of $\text{Omax}(T_0)$ that are labeled with $\mathfrak{b},\mathfrak{r}$ or $\mathfrak{g}$ are exactly the elements of $X$. \end{remark} \begin{definition} \label{defi:Pimoins1} Let $i \in [0,n-2]$, and $T_0 \in \mathcal{T}_{n-1,i}^e$. We consider a subset $X \subsetneq \text{Omax}(T_0)$, whose cardinality $|X|$ is denoted by $k$. Note that $\text{omax}(T_0) = i+2$ hence $k \in [0,i+1]$. We intend to define a set $\mathfrak{T}(T_0,X) \subset \mathcal{T}_{n}^e$ whose cardinality is either $1$ or $2$ following three situations described below. Consider the labels introduced in Definition~\ref{defi:newlabels}. \begin{enumerate}[label=\arabic*.] \item Let $\mathcal{S}_1$ be the situation where $k \leq i$ and no point of the last column $C_{n-1}^{T_0}$ of $T_0$ is labeled with $\mathfrak{r}$ or $\mathfrak{g}$. \item Let $\mathcal{S}_2$ be the situation where: \begin{itemize} \item either $k = i+1$; \item or one of the points of $C_{n-1}^{T_0}$ is labeled with $\mathfrak{r}$ or $\mathfrak{g}$, and there are no two elements of $\Omax(T_0) \backslash X$ labeled with $\mathfrak{b}'$ and $\mathfrak{r}'$. \end{itemize} \item Let $\mathcal{S}_3$ be the remaining situation, i.e., where $k \leq i$, one of the points of $C_{n-1}^{T_0}$ is labeled with $\mathfrak{r}$ or $\mathfrak{g}$, and there exist two elements of $\Omax(T_0) \backslash X$ labeled with $\mathfrak{b}'$ and $\mathfrak{r}'$. \end{enumerate} In any situation of the three above situations, first consider a $n \times 2n$ tableau $\mathfrak{T}_0(T_0)$ containing points only in its boxes $\left( j:\overline{i} \right)$ for all points $(j:i)$ of $T_0$ that does not belong to $\text{Omax}(T_0)$, where $\overline{i}$ is defined as $i$ if $i < n$, as $i+2$ otherwise. Let us set $T^1 = T^2 = T^{3/} = T^{3\backslash} = \mathfrak{T}_0(T_0)$. For $j$ from $1$ to $n-1$, assume that $C_j^{T_0}$ contains an element $p \in \text{Omax}(T_0)$ (otherwise $T^1,T^2,T^{3/},T^{3\backslash}$ already contain two points in their $j$-th columns and we do nothing). \begin{itemize} \item If the label of $p$ is $\mathfrak{b}$ (respectively $\mathfrak{r},\mathfrak{g},\mathfrak{g}'$), we insert a point in the box $(j:n)$ (respectively $(j:2n),(j:j),(j:j)$) of $T^1,T^2,T^{3/}$ and $T^{3\backslash}$. \item If the label of $p$ is $\mathfrak{b}'$ (respectively $\mathfrak{r}'$), \begin{itemize} \item in the situation $\mathcal{S}_1$, we insert a point in the box $(j:n-1)$ (respectively $(j:n+1)$) of $T^1$; \item in the situation $\mathcal{S}_2$, we insert a point in the box $(j:n-1)$ of $T^2$; \item in the situation $\mathcal{S}_3$, \begin{itemize} \item if the points of the last column of $T_0$ have the labels $\mathfrak{r}'$ and $\mathfrak{r}$, or $\mathfrak{b}$ and $\mathfrak{g}$, then we insert a point in the box $(j:n+1)$ (respectively $(j:n-1)$) of $T^{3/}$, and we insert a point in the box $(j:n-1)$ (respectively $(j:n+1)$) of $T^{3\backslash}$; \item otherwise, we insert a point in the box $(j:n-1)$ (respectively $(j:n+1)$) of $T^{3/}$, and we insert a point in the box $(j:n+1)$ (respectively $(j:n-1)$) of $T^{3\backslash}$. \end{itemize} \end{itemize} \end{itemize} We say that \textit{we inserted $p$ in $T^1$} (respectively $T^2$, $T^{3/}$ and $T^{3 \backslash}$) in the situation $\mathcal{S}_1$ (respectively $\mathcal{S}_2,\mathcal{S}_3$). At the end of this algorithm, we finally insert two points in the boxes $(n:n)$ and $(n:2n)$ of $T^1$ (respectively $T^2$, $T^{3/}$ and $T^{3 \backslash}$) in the situation $\mathcal{S}_1$ (respectively $\mathcal{S}_2,\mathcal{S}_3$). The choice of $T_0$ and $X \subsetneq \text{Omax}(T_0)$ results in one of the three situations $\mathcal{S}_1,\mathcal{S}_2,\mathcal{S}_3$, depending on whether $T^1$, or $T^2$, or both $T^{3/}$ and $T^{3\backslash}$ belong to $\mathcal{T}_{n}^e$ at the end of the above algorithm. Let $\mathfrak{T}(T_0,X) \subset \mathcal{T}_n^e$ be defined as $\{T^1\}$ (respectively $\{T^2\},\{T^{3/},T^{3 \backslash}\}$) in the situation $\mathcal{S}_1$ (respectively $\mathcal{S}_2,\mathcal{S}_3$). \end{definition} \begin{remark} \label{rem:maximalPimoins1} If $T \in \mathfrak{T}(T_0,X)$, then by construction the cardinality of $\text{Max}(T)$ is the number of points of $\text{Omax}(T_0)$ labeled with $\mathfrak{b},\mathfrak{r}$ or $\mathfrak{g}$, in other words it equals the cardinality $k$ of $X$ by Remark~\ref{rem:labelsPi}, and $\mathfrak{T}(T_0,X) \subset \mathcal{T}_{n,k}^e$. \end{remark} \begin{remark} \label{rem:freepointsofantecedents} In the context of Definition~\ref{defi:Pimoins1}, if a tableau is defined in the situation $\mathcal{S}_1$ or $\mathcal{S}_3$, then the point $p_{n+1}^T$ is not free, in other words the box $(n:n+1)$ of $T$ is empty, whereas in the situation $\mathcal{S}_2$ the point $p_{n+1}^T$ is $(n:n+1)$ hence is free. Indeed, in the situation $\mathcal{S}_1$, the points of $C_{n-1}^{T_0}$ have the labels $\mathfrak{b}'$ and $\mathfrak{r}'$, so the points of $C_{n-1}^T$ are obtained by inserting two points in the boxes $(n-1:n-1)$ and $(n-1:n+1)$ of $T^1$, which implies that $p_{n+1}^T$ is located in the first $n-1$ columns of $T$ and $p_{n+1}^T$ is then not free; in the situation $\mathcal{S}_3$, since there exists two elements of $\Omax(T_0) \backslash X$ with the labels $\mathfrak{b}'$ and $\mathfrak{r}'$, by construction we insert a point in boxes $(j_1:n-1)$ and $(j_2:n+1)$ for some $j_1,j_2