\documentclass[12pt,reqno]{amsart}
\usepackage{latexsym, amsfonts, amsmath, amsthm, amssymb,amscd,color}
\usepackage{pstricks}
\usepackage{pst-optic}
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\setlength{\textwidth}{6.3in} \setlength{\textheight}{8.7in}
%\setlength{\topmargin}{0pt} \setlength{\headsep}{0pt}
%\setlength{\headheight}{0pt} \setlength{\oddsidemargin}{0pt}
%\setlength{\evensidemargin}{0pt} 
%\pagestyle{plain}
\allowdisplaybreaks

\newtheorem{thm}{Theorem}[section]
\newtheorem{prop}[thm]{Proposition}
\newtheorem{property}[thm]{Property}
\newtheorem{cor}[thm]{Corollary}
\newtheorem{defn}[thm]{Definition}
\newtheorem{lem}[thm]{Lemma}
\newtheorem{conj}[thm]{Conjecture}
\newtheorem{fact}[thm]{Fact}
\newtheorem{problem}[thm]{Problem}

\theoremstyle{remark}
\newtheorem{rmk}{Remark}[section]
\newtheorem{exam}{Example}[section]

\numberwithin{equation}{section}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
  \def\xI{0.866025} \def\xII{1.73205} \def\xIII{2.59808}
  \def\xIV{3.4641} \def\xV{4.33013} \def\xVI{5.19615}
  \def\xVII{6.062175} 
 \def\pav{\rput{60}(0,0){\psline(0,0)(0,3)}
 \rput{-60}(0,0){\psline(0,0)(0,4)}
 \rput{0}(-\xIII,1.5){\psline(0,0)(0,5)}
 \rput{0}(\xIV,2){\psline(0,0)(0,5)}
 \rput{60}(\xIV,7){\psline(0,0)(0,3)} 
 \rput{120}(0.866025,8.5){\psline(0,0)(0,4)}
 \rput{60}(\xI,0.5){\psline(0,0)(0,1)\psline(0,2)(0,4)}
 \rput{60}(\xII,1){\psline(0,1)(0,3)\psline(0,4)(0,5)}
 \rput{60}(\xIII,1.5){\psline(0,1)(0,3)\psline(0,5)(0,6)} 
 \rput{60}(\xIV,2){\psline(0,2)(0,4)\psline(0,6)(0,7)}
 \rput{60}(\xIV,3){\psline(0,1)(0,2)\psline(0,3)(0,4)\psline(0,5)(0,6)}
 \rput{60}(\xIV,4){\psline(0,1)(0,3)\psline(0,5)(0,6)} 
 \rput{60}(\xIV,5){\psline(0,1)(0,3)\psline(0,4)(0,5)}
 \rput{60}(\xIV,6){\psline(0,0)(0,2)\psline(0,3)(0,4)} 
 \rput{120}(\xIV,3){\psline(0,0)(0,3)\psline(0,4)(0,5)}
 \rput{120}(\xIV,4){\psline(0,0)(0,2)\psline(0,3)(0,5)}
 \rput{120}(\xIV,5){\psline(0,0)(0,1)\psline(0,2)(0,3)\psline(0,4)(0,6)} 
 \rput{120}(\xIV,6){\psline(0,0)(0,1)\psline(0,2)(0,3)\psline(0,4)(0,6)}
 \rput{120}(\xIV,7){\psline(0,1)(0,2)\psline(0,3)(0,6)}
 \rput{120}(\xIII,7.5){\psline(0,1)(0,5)} 
 \rput{120}(\xII,8){\psline(0,0)(0,3)\psline(0,4)(0,5)}
 \rput{0}(-\xII,1){\psline(0,0)(0,4)\psline(0,5)(0,6)}
 \rput{0}(-\xI,0.5){\psline(0,0)(0,1)\psline(0,2)(0,6)}
 \rput{0}(0,0){\psline(0,1)(0,2)\psline(0,3)(0,7)} 
 \rput{0}(\xI,0.5){\psline(0,0)(0,1)\psline(0,2)(0,4)\psline(0,5)(0,7)}
 \rput{0}(\xII,1){\psline(0,0)(0,2)\psline(0,3)(0,5)\psline(0,6)(0,7)}
 \rput{0}(\xIII,1.5){\psline(0,0)(0,4)\psline(0,5)(0,6)}} 

 \def\centpav{ \rput{-60}(0,0){\psline(0,0)(0,4)}
 \rput{0}(\xIV,2){\psline(0,0)(0,5)}
 \rput{0}(-\xIV,2){\psline(0,0)(0,5)}
 \rput{-60}(-\xIV,7){\psline(0,0)(0,4)} 
 \rput{60}(\xIV,7){\psline(0,0)(0,4)}
 \rput{-60}(-\xIV,7){\psline(0,0)(0,4)} 
 \rput{60}(0,0){\psline(0,0)(0,4)}
 \rput{60}(\xI,0.5){\psline(0,1)(0,5)}
 \rput{60}(\xII,1){\psline(0,2)(0,6)}
 \rput{60}(\xIII,1.5){\psline(0,2)(0,3)\psline(0,4)(0,7)} 
 \rput{60}(\xIV,2){\psline(0,2)(0,3)\psline(0,4)(0,5)\psline(0,6)(0,8)}
 \rput{60}(\xIV,3){\psline(0,1)(0,2)\psline(0,3)(0,4)\psline(0,5)(0,7)}
 \rput{60}(\xIV,4){\psline(0,1)(0,3)\psline(0,5)(0,7)} 
 \rput{60}(\xIV,5){\psline(0,1)(0,3)\psline(0,4)(0,6)}
 \rput{60}(\xIV,6){\psline(0,0)(0,2)\psline(0,3)(0,5)} 
 \rput{-60}(-\xI,0.5){\psline(0,1)(0,5)}
 \rput{-60}(-\xII,1){\psline(0,2)(0,6)}
 \rput{-60}(-\xIII,1.5){\psline(0,2)(0,5)\psline(0,6)(0,7)}
 \rput{-60}(-\xIV,2){\psline(0,3)(0,6)\psline(0,7)(0,8)} 
 \rput{-60}(-\xIV,3){\psline(0,2)(0,5)\psline(0,6)(0,7)}
 \rput{-60}(-\xIV,4){\psline(0,2)(0,6)}
 \rput{-60}(-\xIV,5){\psline(0,1)(0,2)\psline(0,3)(0,6)}
 \rput{-60}(-\xIV,6){\psline(0,0)(0,1)\psline(0,2)(0,5)} 
 \rput{0}(-\xIII,1.5){\psline(0,0)(0,4)\psline(0,5)(0,6)}
 \rput{0}(-\xII,1){\psline(0,0)(0,4)\psline(0,5)(0,6)}
 \rput{0}(-\xI,0.5){\psline(0,0)(0,3)\psline(0,4)(0,6)} 
 \rput{0}(0,0){\psline(0,0)(0,2)\psline(0,3)(0,4)\psline(0,5)(0,7)}
 \rput{0}(\xI,0.5){\psline(0,0)(0,2)\psline(0,3)(0,4)\psline(0,5)(0,7)}
 \rput{0}(\xII,1){\psline(0,0)(0,2)\psline(0,3)(0,5)\psline(0,6)(0,7)} 
 \rput{0}(\xIII,1.5){\psline(0,0)(0,4)\psline(0,5)(0,6)}
 \pspolygon[fillstyle=solid,fillcolor=lightgray](0,4)(-\xI,4.5)(0,5)(\xI,4.5)}

 \def\semipavf{ \rput{60}(0,0){\psline(0,0)(0,7)}
 \rput{60}(0,1){\psline(0,0)(0,3)\psline(0,4)(0,7)}
 \rput{60}(0,2){\psline(0,0)(0,1)\psline(0,2)(0,4)\psline(0,5)(0,7)}
 \rput{60}(0,3){\psline(0,0)(0,2)\psline(0,3)(0,6)} 
 \rput{60}(0,4){\psline(0,0)(0,1)\psline(0,3)(0,6)}
 \rput{60}(0,5){\psline(0,0)(0,1)\psline(0,2)(0,4)\psline(0,5)(0,6)}
 \rput{60}(0,6){\psline(0,0)(0,3)\psline(0,4)(0,5)}
 \rput{60}(0,7){\psline(0,1)(0,3)\psline(0,4)(0,5)} 
 \rput{60}(0,8){\psline(0,1)(0,4)}
 \rput{60}(0,9){\psline(0,0)(0,2)\psline(0,3)(0,4)}
 \rput{60}(0,10){\psline(0,0)(0,3)} \rput{60}(0,11){\psline(0,0)(0,2)}
 \rput{60}(0,12){\psline(0,0)(0,1)} 
 \rput{120}(0,1){\psline(0,0)(0,1)} \rput{120}(0,2){\psline(0,0)(0,2)}
 \rput{120}(0,3){\psline(0,1)(0,3)}
 \rput{120}(0,4){\psline(0,0)(0,2)\psline(0,3)(0,4)}
 \rput{120}(0,5){\psline(0,1)(0,4)} 
 \rput{120}(0,6){\psline(0,1)(0,3)\psline(0,4)(0,5)}
 \rput{120}(0,7){\psline(0,0)(0,3)\psline(0,4)(0,5)}
 \rput{120}(0,8){\psline(0,0)(0,1)\psline(0,2)(0,4)\psline(0,5)(0,6)} 
 \rput{120}(0,9){\psline(0,0)(0,1)\psline(0,3)(0,5)\psline(0,6)(0,7)}
 \rput{120}(0,10){\psline(0,1)(0,2)\psline(0,3)(0,5)\psline(0,6)(0,7)}
  \rput{120}(0,11){\psline(0,0)(0,1)\psline(0,2)(0,4)\psline(0,5)(0,7)} 
 \rput{120}(0,12){\psline(0,0)(0,3)\psline(0,4)(0,7)}
 \rput{120}(0,13){\psline(0,0)(0,7)} 
 \rput{0}(0,0){\psline(0,2)(0,3)\psline(0,4)(0,6)\psline(0,7)(0,10)}
 \rput{0}(-\xI,0.5){\psline(0,2)(0,3)\psline(0,4)(0,8)\psline(0,9)(0,10)} 
 \rput{0}(-\xII,1){\psline(0,1)(0,2)\psline(0,3)(0,5)\psline(0,6)(0,8)\psline(0,9)(0,10)}
 \rput{0}(-\xIII,1.5){\psline(0,1)(0,4)\psline(0,6)(0,9)} 
 \rput{0}(-\xIV,2){\psline(0,0)(0,1)\psline(0,2)(0,4)\psline(0,5)(0,7)\psline(0,8)(0,9)}
 \rput{0}(-\xV,2.5){\psline(0,0)(0,2)\psline(0,3)(0,5)\psline(0,6)(0,8)} 
 \rput{0}(-\xVI,3){\psline(0,0)(0,2)\psline(0,3)(0,7)}
 \rput{0}(-\xVII,3.5){\psline(0,0)(0,6)} 
 \def\completion{\rput{-60}(0,0){\psline(0,0)(0,1)}\rput{-120}(0,1){\psline(0,1)(0,0)}}
 \rput{0}(0,0){\completion} \rput{0}(0,1){\completion}
 \rput{0}(0,3){\completion} \rput{0}(0,6){\completion}
 \rput{0}(0,10){\completion} 
 \rput{0}(0,11){\completion} \rput{0}(0,12){\completion}
 \psline[linestyle=dotted](0,-0.5)(0,14) \uput[0](0,13.5){$\ell$}} 

\def\semipava{ \rput{60}(0,0){\psline(0,0)(0,7)}
 \rput{60}(0,1){\psline(0,0)(0,3)\psline(0,4)(0,7)}
 \rput{60}(0,2){\psline(0,0)(0,1)\psline(0,2)(0,4)\psline(0,5)(0,7)}
 \rput{60}(0,3){\psline(0,0)(0,2)\psline(0,3)(0,6)} 
 \rput{60}(0,4){\psline(0,0)(0,2)\psline(0,3)(0,6)}
 \rput{60}(0,5){\psline(0,0)(0,2)\psline(0,2)(0,4)\psline(0,5)(0,6)}
 \rput{60}(0,6){\psline(0,0)(0,3)\psline(0,4)(0,5)}
 \rput{60}(0,7){\psline(0,1)(0,3)\psline(0,4)(0,5)} 
 \rput{60}(0,8){\psline(0,1)(0,4)}
 \rput{60}(0,9){\psline(0,0)(0,2)\psline(0,3)(0,4)}
 \rput{60}(0,10){\psline(0,0)(0,3)} \rput{60}(0,11){\psline(0,0)(0,2)}
 \rput{60}(0,12){\psline(0,0)(0,1)} 
 \rput{120}(0,1){\psline(0,0)(0,1)} \rput{120}(0,2){\psline(0,0)(0,2)}
 \rput{120}(0,3){\psline(0,1)(0,3)}
 \rput{120}(0,4){\psline(0,1)(0,2)\psline(0,3)(0,4)}
 \rput{120}(0,5){\psline(0,2)(0,4)} 
 \rput{120}(0,6){\psline(0,2)(0,3)\psline(0,4)(0,5)}
 \rput{120}(0,7){\psline(0,0)(0,1)\psline(0,2)(0,3)\psline(0,4)(0,5)}
 \rput{120}(0,8){\psline(0,0)(0,1)\psline(0,2)(0,4)\psline(0,5)(0,6)} 
 \rput{120}(0,9){\psline(0,0)(0,1)\psline(0,3)(0,5)\psline(0,6)(0,7)}
 \rput{120}(0,10){\psline(0,1)(0,2)\psline(0,3)(0,5)\psline(0,6)(0,7)}
 \rput{120}(0,11){\psline(0,0)(0,1)\psline(0,2)(0,4)\psline(0,5)(0,7)} 
 \rput{120}(0,12){\psline(0,0)(0,3)\psline(0,4)(0,7)}
 \rput{120}(0,13){\psline(0,0)(0,7)} 
 \rput{0}(0,0){\psline(0,2)(0,6)\psline(0,7)(0,10)}
 \rput{0}(-\xI,0.5){\psline(0,2)(0,8)\psline(0,9)(0,10)}
 \rput{0}(-\xII,1){\psline(0,1)(0,2)\psline(0,3)(0,5)\psline(0,6)(0,8)\psline(0,9)(0,10)} 
 \rput{0}(-\xIII,1.5){\psline(0,1)(0,4)\psline(0,6)(0,9)}
 \rput{0}(-\xIV,2){\psline(0,0)(0,1)\psline(0,2)(0,4)\psline(0,5)(0,7)\psline(0,8)(0,9)}
 \rput{0}(-\xV,2.5){\psline(0,0)(0,2)\psline(0,3)(0,5)\psline(0,6)(0,8)} 
 \rput{0}(-\xVI,3){\psline(0,0)(0,2)\psline(0,3)(0,7)}
 \rput{0}(-\xVII,3.5){\psline(0,0)(0,6)} 
 \def\completion{\rput{-60}(0,0){\psline(0,0)(0,1)}\rput{-120}(0,1){\psline(0,1)(0,0)}}
 \rput{0}(0,0){\completion} \rput{0}(0,1){\completion}
 \rput{0}(0,6){\completion} \rput{0}(0,10){\completion} 
 \rput{0}(0,11){\completion} \rput{0}(0,12){\completion}
 \psline[linestyle=dotted](0,-0.5)(0,14) \uput[0](0,13.5){$\ell$} 
 \pspolygon[fillstyle=solid,fillcolor=lightgray](-\xII,6)(-\xIII,6.5)(-\xII, 7)}

 \def\semipavb{
 \rput{60}(0,0){\psline(0,0)(0,7)}
 \rput{60}(0,1){\psline(0,0)(0,3)\psline(0,4)(0,7)}
 \rput{60}(0,2){\psline(0,0)(0,1)\psline(0,2)(0,4)\psline(0,5)(0,7)}
 \rput{60}(0,3){\psline(0,0)(0,2)\psline(0,3)(0,6)} 
 \rput{60}(0,4){\psline(0,1)(0,2)\psline(0,3)(0,6)}
 \rput{60}(0,5){\psline(0,1)(0,2)\psline(0,3)(0,4)\psline(0,5)(0,6)}
 \rput{60}(0,6){\psline(0,1)(0,2)\psline(0,4)(0,5)}
 \rput{60}(0,7){\psline(0,1)(0,2)\psline(0,4)(0,5)} 
 \rput{60}(0,8){\psline(0,1)(0,2)\psline(0,3)(0,4)}
 \rput{60}(0,9){\psline(0,1)(0,2)\psline(0,3)(0,4)}
 \rput{60}(0,10){\psline(0,1)(0,3)} \rput{60}(0,11){\psline(0,0)(0,2)}
 \rput{60}(0,12){\psline(0,0)(0,1)} 
 \rput{120}(0,1){\psline(0,0)(0,1)} \rput{120}(0,2){\psline(0,0)(0,2)}
 \rput{120}(0,3){\psline(0,1)(0,3)}
 \rput{120}(0,4){\psline(0,0)(0,2)\psline(0,3)(0,4)}
 \rput{120}(0,5){\psline(0,0)(0,1)\psline(0,2)(0,4)} 
 \rput{120}(0,6){\psline(0,0)(0,1)\psline(0,2)(0,3)\psline(0,4)(0,5)}
 \rput{120}(0,7){\psline(0,0)(0,1)\psline(0,2)(0,3)\psline(0,4)(0,5)}
 \rput{120}(0,8){\psline(0,0)(0,1)\psline(0,2)(0,4)\psline(0,5)(0,6)} 
 \rput{120}(0,9){\psline(0,0)(0,1)\psline(0,2)(0,5)\psline(0,6)(0,7)}
 \rput{120}(0,10){\psline(0,0)(0,1)\psline(0,2)(0,5)\psline(0,6)(0,7)}
 \rput{120}(0,11){\psline(0,0)(0,1)\psline(0,2)(0,4)\psline(0,5)(0,7)} 
 \rput{120}(0,12){\psline(0,0)(0,3)\psline(0,4)(0,7)}
 \rput{120}(0,13){ \psline(0,0)(0,7)}
 \rput{0}(0,0){\psline(0,2)(0,3)\psline(0,4)(0,11)}
  \rput{0}(-\xI,0.5){\psline(0,2)(0,10)} 
 \rput{0}(-\xII,1){\psline(0,1)(0,2)\psline(0,3)(0,10)}
 \rput{0}(-\xIII,1.5){\psline(0,1)(0,4)\psline(0,6)(0,9)}
 \rput{0}(-\xIV,2){\psline(0,0)(0,1)\psline(0,2)(0,4)\psline(0,5)(0,7)\psline(0,8)(0,9)} 
 \rput{0}(-\xV,2.5){\psline(0,0)(0,2)\psline(0,3)(0,5)\psline(0,6)(0,8)}
 \rput{0}(-\xVI,3){\psline(0,0)(0,2)\psline(0,3)(0,7)}
 \rput{0}(-\xVII,3.5){\psline(0,0)(0,6)} 
 \def\completion{\rput{-60}(0,0){\psline(0,0)(0,1)}\rput{-120}(0,1){\psline(0,1)(0,0)}}
 \rput{0}(0,0){\completion} \rput{0}(0,1){\completion}
 \rput{0}(0,3){\completion}  \rput{0}(0,11){\completion} 
 \rput{0}(0,12){\completion} \psline[linestyle=dotted](0,-0.5)(0,14)
 \uput[0](0,13.5){$\ell$}
 \pspolygon[fillstyle=solid,fillcolor=lightgray](-\xIII,5.5)(-\xIII,7.5)(-\xV,
 6.5)} 

 \def\semipav{ \rput{60}(0,0){\psline(0,0)(0,7)}
 \rput{60}(0,1){\psline(0,0)(0,3)\psline(0,4)(0,7)}
 \rput{60}(0,2){\psline(0,0)(0,1)\psline(0,2)(0,4)\psline(0,5)(0,7)}
 \rput{60}(0,3){\psline(0,0)(0,2)\psline(0,3)(0,6)} 
 \rput{60}(0,4){\psline(0,0)(0,1)\psline(0,3)(0,6)}
 \rput{60}(0,5){\psline(0,0)(0,1)\psline(0,2)(0,4)\psline(0,5)(0,6)}
 \rput{60}(0,6){\psline(0,0)(0,3)\psline(0,4)(0,5)}
 \rput{60}(0,7){\psline(0,1)(0,3)\psline(0,4)(0,5)} 
 \rput{60}(0,8){\psline(0,1)(0,4)}
 \rput{60}(0,9){\psline(0,0)(0,2)\psline(0,3)(0,4)}
 \rput{60}(0,10){\psline(0,0)(0,3)} \rput{60}(0,11){\psline(0,0)(0,2)}
 \rput{60}(0,12){\psline(0,0)(0,1)} \rput{120}(0,1){\psline(0,0)(0,1)} 
 \rput{120}(0,2){\psline(0,0)(0,2)} \rput{120}(0,3){\psline(0,1)(0,3)}
 \rput{120}(0,4){\psline(0,0)(0,2)\psline(0,3)(0,4)}
 \rput{120}(0,5){\psline(0,1)(0,4)}
 \rput{120}(0,6){\psline(0,1)(0,3)\psline(0,4)(0,5)} 
 \rput{120}(0,7){\psline(0,0)(0,3)\psline(0,4)(0,5)}
 \rput{120}(0,8){\psline(0,0)(0,1)\psline(0,2)(0,4)\psline(0,5)(0,6)}
 \rput{120}(0,9){\psline(0,0)(0,1)\psline(0,3)(0,5)\psline(0,6)(0,7)} 
 \rput{120}(0,10){\psline(0,1)(0,2)\psline(0,3)(0,5)\psline(0,6)(0,7)}
 \rput{120}(0,11){\psline(0,0)(0,1)\psline(0,2)(0,4)\psline(0,5)(0,7)}
 \rput{120}(0,12){\psline(0,0)(0,3)\psline(0,4)(0,7)} 
 \rput{120}(0,13){\psline(0,0)(0,7)}
 \rput{0}(0,0){\psline(0,2)(0,3)\psline(0,4)(0,6)\psline(0,7)(0,10)}
 \rput{0}(-\xI,0.5){\psline(0,2)(0,3)\psline(0,4)(0,8)\psline(0,9)(0,10)} 
 \rput{0}(-\xII,1){\psline(0,1)(0,2)\psline(0,3)(0,5)\psline(0,6)(0,8)\psline(0,9)(0,10)}
 \rput{0}(-\xIII,1.5){\psline(0,1)(0,4)\psline(0,6)(0,9)} 
 \rput{0}(-\xIV,2){\psline(0,0)(0,1)\psline(0,2)(0,4)\psline(0,5)(0,7)\psline(0,8)(0,9)}
 \rput{0}(-\xV,2.5){\psline(0,0)(0,2)\psline(0,3)(0,5)\psline(0,6)(0,8)} 
 \rput{0}(-\xVI,3){\psline(0,0)(0,2)\psline(0,3)(0,7)}
 \rput{0}(-\xVII,3.5){\psline(0,0)(0,6)}
\def\completion{\rput{-60}(0,0){\psline(0,0)(0,1)}
 \rput{-120}(0,1){\psline(0,1)(0,0)}} 
 \rput{0}(0,0){\completion} \rput{0}(0,1){\completion}
 \rput{0}(0,3){\completion} \rput{0}(0,10){\completion}
 \rput{0}(0,11){\completion} \rput{0}(0,12){\completion}} 

 \def\pathsemipav{
 \semipav\pspolygon[fillstyle=solid,fillcolor=lightgray](0,6)(-\xI,6.5)(0,7) 
 \def\mI{-0.43301} \def\mII{-1.29903} \def\mIII{-2.165063}
 \def\mIV{-3.031088} \def\mV{-3.8971143} \def\mVI{-4.7631397}
 \def\mVII{-5.629165126} 
\psdots[dotsize=7pt](\mI,0.25)(\mII,0.75)(\mIII,1.25)(\mIV,1.75)%
  (\mV,2.25)(\mVI,2.75)(\mVII,3.25)\psdots[dotsize=7pt](\mI,1.25)%
  (\mI,3.25)(\mI,6.25)(\mI,10.25)(\mI,11.25)(\mI,12.25) 
\psset{linestyle=dotted,linewidth=2.5pt}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\def\Pf{\operatorname{Pf}}
\def\inv{\operatorname{inv}}
\def\sgn{\operatorname{sgn}}
\def\s{\sigma}
\def\l{\lambda}
\def\Acal{\mathcal{A}}
\def\Scal{\mathcal{S}}
\def\Ical{\mathcal{I}}
\def\T{\mathcal{T}}
\def\Tv{\mathcal{T}v}
\def\Th{\mathcal{T}h}
\def\ST{\mathcal{ST}}
\def\Mtil{\widetilde M}
\def\Atil{\widetilde A}
\def\Btil{\widetilde B}



\begin{document}

\title[centered symmetric  rhombus tilings of a hexagon]{Enumeration
of symmetric centered\\ rhombus tilings of a hexagon} 

\author[A. Kasraoui]{Anisse Kasraoui}
\author[C. Krattenthaler]{Christian Krattenthaler}
\address{Fakult\"at f\"ur Mathematik, Universit\"at Wien,
Nordbergstrasze 15,
A-1090 Vienna,
Austria}
%\email{anisse.kasraoui@univie.ac.at, christian.krattenthaler@univie.ac.at}
\thanks{Research supported by the Austrian
Science Foundation FWF, grants Z130-N13 and S50-N15,
the latter in the framework of the Special Research Program
``Algorithmic and Enumerative Combinatorics"}



\date{}


\begin{abstract}
A  rhombus tiling of a hexagon is said to be centered if it contains 
the central rhombus. We compute the number of vertically symmetric 
rhombus tilings of a hexagon with side lengths $a, b, a, a, b, a$ which 
are centered.  When $a$ is odd and $b$ is even, this shows that the 
probability  that a random vertically symmetric rhombus tiling of an 
$a, b, a, a, b, a$ hexagon is centered is exactly 
the same as the probability that a random rhombus tiling of an 
$a, b, a, a, b, a$ hexagon is centered. 
This also leads to a factorization theorem for the number of all rhombus 
tilings of a hexagon which are centered.
\end{abstract}

\maketitle



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%                              INTRODUCTION
%
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\section{Introduction}

The enumeration of plane partitions, equivalently of rhombus 
tilings of a hexagon,
was initiated by MacMahon in the early twentieth century. 
Let $a$, $b$ and $c$ be positive integers.
By an \textit{$(a,b,c)$-hexagon} we mean an equi-angular hexagon with 
side-lengths $a,b,c,a,b,c$. We
always draw such a hexagon with  the sides of lengths $a,b,c,a,b,c$ in 
clockwise order starting from
the southwestern side, so that the sides of length $b$ are vertical.
  From a classical result of MacMahon~\cite[Sec.~429, $q\to1$]{MacM}, 
we know that
the number of tilings of an $(a,b,c)$-hexagon by rhombi whose sides 
have length 1 and whose angles
are~60 and~120 degrees (equivalently, of plane partitions contained 
in an $a\times b \times c$ box)
is given by the product
\begin{equation}\label{eq:def P(a,b)}
 T(a,b,c)=\prod_{i=1}^a\prod_{j=1}^b\prod_{k=1}^c \frac{i+j+k-1}{i+j+k-2}.
\end{equation}
We call such tilings \textit{rhombus tilings}. The picture on the left
in Figure~\ref{fig:Til} shows a rhombus tiling of a
$(3,5,4)$-hexagon. 
MacMahon also conjectured that the number of \textit{vertically
symmetric} rhombus tilings of an $(a,b,a)$-hexagon (i.e, that are
invariant under reflection 
across the vertical symmetry axis of the hexagon; see the
picture on the right in Figure~\ref{fig:H-F} for an example) 
is given by the product 
\begin{equation}\label{eq:def N(a,b)}
ST(a,b,a)=\prod_{i=1}^a \frac{2i+b-1}{2i-1} \prod_{1\leq i<j\leq a}
\frac{i+j+b-1}{i+j-1}.
\end{equation}
This was first proved by Andrews~\cite{Andrews}. Other proofs, and
refinements, were later found by e.g.\ Gordon~\cite{Gordon}, 
Macdonald~\cite[pp.~83--85]{McDon}, Proctor~\cite[Prop.~7.3]{Proct},
Fischer~\cite{Ilse2}, and the second author of the present 
paper~\cite[Theorem~13]{Kratt}.
 \begin{figure}[h!]\label{fig:Til}
\psset{xunit=0.5,yunit=0.5}
\begin{pspicture}(0,0.5)(18,9)
\rput{0}(0,1){\pav}
\rput{0}(16,0.5){\centpav}
\end{pspicture}
\caption{\textit{Left}: a rhombus tiling of a $(3,5,4)$-hexagon.
\textit{Right}: a centered rhombus tiling of a $(4,5,4)$-hexagon where
the central rhombus is marked.} 
\end{figure}

During the last two decades, there has been an increasing interest
in enumerating rhombus tilings of planar regions with holes. One of
the first results in this direction concerned the enumeration of rhombus
tilings of an $(a,b,a)$-hexagon that contain the central rhombus.
(These tilings may be seen as the rhombus tilings of the region
obtained from an $(a,b,a)$-hexagon by removing its central rhombus.)
We call such tilings \textit{centered}. Note that an $(a,b,a)$-hexagon
has a central rhombus only if~$a$ and~$b$ have opposite parity. The
picture on the right in 
Figure~\ref{fig:Til} shows a centered tiling of a $(4,5,4)$-hexagon.
The corresponding enumeration result (see
Theorem~\ref{thm:Pcentered(2n+1,2x)} below) 
is independently due to Ciucu and the second author~\cite[Theorems~1
and 2 and Corollaries~3 and 4]{CiuKrat-centered} and to Gessel and
Helfgott~\cite[Theorems~15 and 17]{GeHel}.
For non-negative integers~$n$ and~$x$, define
\begin{equation}\label{eq:def Q(x,n)}
Q(n,x)=\frac{1}{2}\frac{(2n)!^2\,(2x)!\,(x+2n-1)!}
                    {n!^2\,(x)!\,(2x+4n-2)!}
       \left(\sum_{i=0}^{n-1} \frac{(-1)^{n-i-1}}{2n-2i-1}
       \frac{(x+n-i)_{2i}}{i!^2}\right), 
\end{equation}
where $(a)_k$ stands for the Pochhammer symbol, defined by
$(a)_k:=a(a+1)\cdots (a+k-1)$ for $k\geq 1$, 
and $(a)_0:=1$.

\begin{thm}\label{thm:Pcentered(2n+1,2x)}
Let $n$ and~$x$ be two non-negative integers.

\textit{(i)} For $x\geq1$, the number of centered rhombus tilings of
a $(2n+1,2x,2n+1)$-hexagon is \linebreak $Q(n+1,x)\cdot
T(2n+1,2x,2n+1),$ where $Q$ and $T$ are defined by~\eqref{eq:def
Q(x,n)} and~\eqref{eq:def P(a,b)}. Similarly, for $n\geq1$, the
number of centered rhombus tilings of a $(2n,2x+1,2n)$-hexagon is\break
$Q(n,x+1)\cdot T(2n,2x+1,2n)$.

\textit{(ii)} For $n\geq1$, exactly one third of the rhombus tilings of a 
$(2n+1,2n,2n+1)$-hexagon are centered. The same is true for a 
$(2n,2n+1,2n)$-hexagon.

\textit{(iii)} Let $a$ be a non-negative real number. For $x\sim an$, the
probability that a random rhombus tiling of a 
$(2n+1,2x,2n+1)$-hexagon is centered is $\sim \frac 2 \pi
\arcsin(1/(a+1))$ as $n$ tends 
to infinity. The same is true for a $(2n,2x+1,2n)$-hexagon.
\end{thm}

 Generalizations of the preceding result were later obtained by Fulmek
 and the second author~\cite{FulKrat-I,FulKrat-II},  
and by Fischer~\cite{Ilse}. For other results on the enumeration of rhombus 
tilings of hexagons of which central triangles 
are removed, see e.g.~\cite{Ciu,Ciu-Zare,CiuKrat-Dual,Ther,KrattOkada}. 
Another result which is particularly relevant to our paper is the
one by Ciucu and the second author in~\cite{CiuKrat-interaction},
where for the first time the number of rhombus tilings of a half
$(a,b,a)$-hexagon with a triangular hole of size two and a free
boundary was computed. Here, by a \textit{half $(a,b,a)$-hexagon} with a
\textit{free boundary} we mean the region --- denoted in the rest of
the paper by $F_{a,b,a}$ --- obtained from the left half of an
$(a,b,a)$-hexagon by regarding its boundary along~$\ell$, the
vertical symmetry axis of the hexagon, as free; i.e, rhombi in a
tiling of $F_{a,b,a}$ are allowed to protrude outward across~$\ell$
to its right. In Figure~\ref{fig:HalfTil}, the picture on the left 
shows a rhombus tiling of the half-hexagon $F_{7,6,7}$, the other two
 show tilings of $F_{7,6,7}$ with  
triangular holes of size one and two.
%%%%%%%%%%%%FIG 2
 \begin{figure}[h!]
\psset{xunit=0.5,yunit=0.5}
\begin{pspicture}(0,0)(22,15)
\rput{0}(3,0){\semipavf}
\rput{0}(13,0){\semipavb}
\rput{0}(23,0){\semipava }
\end{pspicture}
\caption{\textit{Left}: a rhombus tiling of the half-hexagon $F_{7,6,7}$.
\textit{Middle}: a tiling of $F_{7,6,7}$ with a triangular hole of size two.
\textit{Right}: a tiling of $F_{7,6,7}$ with a triangular hole of size
 one.}
\label{fig:HalfTil} 
\end{figure}

The present paper was motivated by an attempt to find similar
results to those obtained in~\cite{CiuKrat-interaction} for a
triangular hole of size one. While we did not succeed for a general
position of the hole, we have been able to obtain a formula
for the number of rhombus tilings of the region $F_{a,b,a}^*$
obtained from the half-hexagon $F_{a,b,a}$ by removing a triangular
hole of size one pointing to the left such that the centre of its
right side coincides with the centre of the free boundary. (This region
is only defined if~$a$ and~$b$ have opposite parity.)
The picture on the left in Figure~\ref{fig:H-F} shows a tiling of
$F_{7,6,7}^*$. As illustrated in Figure~\ref{fig:H-F}, by reflecting
the tilings of $F_{a,b,a}^*$ across the free boundary, it is easily seen that
these are equinumerous with the centered {\it vertically symmetric}
rhombus tilings of an $(a,b,a)$-hexagon.
%%%%%%%%%%%%FIG 2
 \begin{figure}[h!]
\begin{center}
\psset{xunit=0.5,yunit=0.5}
\begin{pspicture}(0,0)(24,15)
\rput{0}(5,0){\semipav  \psline[linestyle=dotted](0,-0.5)(0,14)
\pspolygon[fillstyle=solid, fillcolor=lightgray] (0,6)(0,7)(-0.866,
6.5)} 
\def\pavcomplet{ \semipav
\pnode(0,-1){V1} \pnode(0,10){V2} \symPlan(V1)(V2){\semipav}}
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fillcolor=lightgray] (0,6)(0.866, 6.5)(0,7)(-0.866, 6.5)}
\psline{<->}(7.5,6.5)(12,6.5) 
\end{pspicture}
\caption{The correspondence between tilings of $F^*_{2n+1,2x,2n+1}$
and centered vertically symmetric tilings of a
$(2n+1,2x,2n+1)$-hexagon.}
\label{fig:H-F} 
\end{center}
\end{figure}
 The next two theorems, which may be seen as a ``symmetrization'' of
Theorem~\ref{thm:Pcentered(2n+1,2x)}(i), are our main results.

\begin{thm}\label{thm:Ncentered(2n+1,2x)}
Let $n$ and $x$ be non-negative integers. For $x\geq 1$, the number of\break 
centered
vertically symmetric rhombus tilings of a $(2n+1,2x,2n+1)$-hexagon is\break
$Q(n+1,x)\cdot ST(2n+1,2x,2n+1)$, 
where $Q$ and $ST$ are defined by~\eqref{eq:def Q(x,n)}
and~\eqref{eq:def N(a,b)}. 
\end{thm}

For a positive integer $n$ and a non-negative integer $x$, define
\begin{multline}
\label{eq:def-Ux}
U_n(x)=\sum_{i=1}^n
\big((2n-1)!!+(-1)^{i+1}(2n)!!\big)
\frac{\left(\frac{3}{2}-i\right)_{2n-1}}{(i-1)!\,(2n-i)!}\\
\times
\big((x+1)_{i-1}(x+i+1)_{2n-i}-(x+1)_{2n-i}(x+2n+2-i)_{i-1}\big),
\end{multline}
and
\begin{equation}\label{eq:def R(x,n)}
R(n,x)=2^{3n-2}\frac{(2x+2)!\,(x+2n)!}{n!\,(x+1)!\,(2x+4n)!}\,U_n(x).
\end{equation}
Here, as usual, $a!!$ stands for the double factorial, i.e., for
$a\cdot (a-2)\cdot (a-4)\cdots 2$ if $a$ is even, and for
$a\cdot (a-2)\cdot (a-4)\cdots 1$ if $a$ is odd.

\begin{thm}\label{thm:Ncentered(2n,2x+1)}
Let $n$ and $x$ be non-negative integers. For $n\geq 1$, the number of\break
centered 
vertically symmetric rhombus tilings of a $(2n,2x+1,2n)$-hexagon is\break
$R(n,x)\cdot ST(2n,2x+1,2n)$, 
where $R$ and $ST$ are defined by~\eqref{eq:def R(x,n)}
and~\eqref{eq:def N(a,b)}. 
\end{thm}

The next result, which is rather striking and deserves further
investigation, is immediate  
from Theorems~\ref{thm:Ncentered(2n+1,2x)} and~\ref{thm:Pcentered(2n+1,2x)}.

\begin{cor}\label{cor:Pcentered(2n+1,2x)-Ncentered(2n+1,2x)}
Let $n$ and $x$ be  non-negative integers. The probability that a
random vertically symmetric rhombus tiling of a 
$(2n+1,2x,2n+1)$-hexagon is centered is exactly the same as 
the probability that a
random rhombus tiling of a $(2n+1,2x,2n+1)$-hexagon is centered.
\end{cor}

 We should note here that the preceding result leads to an interesting
 (and intriguing) factorization 
 for the number of all centered tilings of a hexagon. Given a planar
 region $R$ which is symmetric with respect to a 
vertical axis and to a horizontal axis, let $\T(R)$ be the set of
all rhombus tilings of $R$. We also let $\T^{(|)}(R)$ (respectively $\T^{(-)}(R)$)
be the set of those tilings in $\T(R)$ that are vertically symmetric
(respectively horizontally symmetric). 
Let $H_{a,2b,a}$ be an $(a,2b,a)$-hexagon. 
Ciucu and the second author~\cite{CiuKrat-factor} observed the factorization 
\begin{equation}\label{eq:FactorizationTil}
\# \T(H_{a,2b,a})= \# \T^{(|)}(H_{a,2b,a}) \cdot \# \T^{(-)}(H_{a,2b,a}).
\end{equation}
This can be proved by combining~\eqref{eq:def P(a,b)}
 and~\eqref{eq:def N(a,b)} with a formula of Proctor~\cite{Proct2} for
 $\# \T^{(-)}(H_{a,2b,a})$  
(equivalently, the number of transpose complementary plane partitions
 in a $2b\times a\times a$ box); see also~\cite{CiuKrat-factor}, where
 the above relation  
was put in a more general context. 
 Now, suppose that $a$ is odd, and let $H^*_{a,2b,a}$ denote the region
 obtained  by 
removing the central rhombus from $H_{a,2b,a}$. Recall that a centered
 tiling of $H_{a,2b,a}$ is obviously  
equivalent to a tiling of $H^*_{a,2b,a}$. Then, combining
 Corollary~\ref{cor:Pcentered(2n+1,2x)-Ncentered(2n+1,2x)}
 with~\eqref{eq:FactorizationTil}  
(and noting that any horizontally symmetric tiling of $H_{a,2b,a}$ is
 centered), we arrive at 
the following factorization for the number of centered tilings.

\begin{cor}\label{cor:FactorizationCentTil}
For any non-negative integers $n$ and $x$, we have
\begin{equation}\label{eq:FactorizationCentTil}
\# \T(H^*_{2n+1,2x,2n+1})= \# \T^{(|)}(H^*_{2n+1,2x,2n+1}) \cdot \#
\T^{(-)}(H^*_{2n+1,2x,2n+1}).
\end{equation}
\end{cor}
%This is in line with a more general conjecture of Ciucu which,
%roughly speaking, asserts that the above factorization a
%is valid  when $H^*_{2n+1,2x,2n+1}$ is replaced by any region
%obtained from a $(2n+1,2x,2n+1)$-hexagon by removing two unit
%triangular 
%situated on the horizontal axis.
Note that the above factorization is very similar to the one
in~\eqref{eq:FactorizationTil}. 

Another result which is immediate from
Corollary~\ref{cor:Pcentered(2n+1,2x)-Ncentered(2n+1,2x)} 
and Theorem~\ref{thm:Pcentered(2n+1,2x)} is the following.

\begin{cor}\label{cor:Ncentered(2n+1,2x)}
$(i)$ For $n\geq 1$, the probability that a random vertically
symmetric rhombus tiling of 
a $(2n+1,2n,2n+1)$-hexagon is centered is $1/3$.

$(ii)$ Let $a$ be a non-negative real number. For $x\sim an$, the
probability that a random vertically symmetric rhombus tiling of
a $(2n+1,2x,2n+1)$-hexagon is centered  is $\sim \frac 2 \pi
\arcsin(1/(a+1))$, as $n$ tends to infinity. 
\end{cor}

As one can expect, the second part of the preceding proposition is
still valid for a $(2n,2x+1,2n)$-hexagon.

\begin{cor}\label{cor:Ncentered(2n,2x+1)}
 Let $a$ be a non-negative real number.  Then, for $x\sim an$, the
 probability 
that a random vertically symmetric rhombus tiling of a
$(2n,2x+1,2n)$-hexagon is centered is $\sim \frac 2 \pi
\arcsin(1/(a+1))$, as $n$ tends to infinity.
\end{cor}

The rest of this paper is devoted to the proof of the above corollary
 and Theorems~\ref{thm:Ncentered(2n+1,2x)} 
 and~\ref{thm:Ncentered(2n,2x+1)}. As in many previous papers, our
 approach to proving 
 Theorems~\ref{thm:Ncentered(2n+1,2x)}
 and~\ref{thm:Ncentered(2n,2x+1)} is to first translate the centered
 vertically symmetric rhombus 
 tilings into  families of non-intersecting lattice paths. Then, to
 enumerate these non-intersecting 
 lattice paths, we use a slight extension of a theorem of
 Stembridge~\cite{Stem} to obtain Pfaffians for the numbers 
 we are interested in.  
Here, due to the fact that we are considering an odd-size hole in
our half-hexagon, we encounter a complication which has not appeared
earlier (respectively has been circumvented or avoided so far):
a naive direct application of the relevant formulas does not lead
to formulas for the number of families of non-intersecting lattice
paths but rather to ``weighted" enumerations in which each family
is assigned a weight of $+1$ or $-1$, depending on the sign of the permutation
with respect to which starting points are connected with end points. 
There is a trick which is used 
here for the first time to cope with this difficulty: it consists
of introducing appropriate edge weights to ``correct" this sign, so 
that each family 
of non-intersecting lattice paths is indeed counted by $+1$ in
the relevant formulas. All this is explained in
Section~\ref{sec:2}. 

The evaluation of the resulting Pfaffians occupies
Sections~\ref{sec:3} and~\ref{sec:4}, 
with some auxiliary results proved in Section~\ref{sec:5}. It
is based on the ``exhaustion/identification 
of factors'' method (see e.g.~\cite[Sec.~2.4]{Krattt}) and turns
out to be particularly demanding. In particular, we need a 
Pfaffian factorization due to Ciucu and the second  author, an
evaluation of a perturbed Mehta--Wang Pfaffian, and evaluations of 
very intricate combinatorial sums.   In the final section, 
Section~\ref{sec:6},
we perform the asymptotic calculation needed 
to derive Corollary~\ref{cor:Ncentered(2n,2x+1)} from
Theorem~\ref{thm:Ncentered(2n,2x+1)}. 




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%
%  rhombus tilings, non-intersecting lattice paths and Pfaffians
%
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%\newpage
\section{Centered vertically symmetric tilings, non-intersecting
lattice paths and Pfaffians} 
\label{sec:2}


\subsection{Centered vertically symmetric tilings and non-intersecting
lattice paths}\label{sect:tilings-paths} 
As explained in the introduction (see Figure~\ref{fig:H-F}), centered
vertically symmetric 
tilings of an $(a,b,a)$-hexagon may be seen as tilings of the region
$F^*_{a,b,a}$. Throughout this section, the term \textit{lattice path}
will always 
refer to a path in the lattice $\mathbb{Z}^2$ consisting of unit
horizontal and vertical steps in the positive direction. 
 There is a well known bijection between rhombus tilings of lattice regions
and families of non-intersecting lattice paths. An illustration in
our situation is given in Figure~\ref{fig:tilings-paths}.
%%%%%%%%%%%%FIG 3
 \begin{figure}[h!]
\begin{center}
\psset{xunit=0.7,yunit=0.7}
\begin{pspicture}(0,0)(20,17)
\psset{linewidth=1.5pt,griddots=10}
 \rput{0}(6,3){\pathsemipav}
\rput{0}(19,3){\path}
\end{pspicture}
\end{center}
\caption{The correspondence between tilings of $F^*_{n,x,n}$ and
 non-intersecting lattice paths.}\label{fig:tilings-paths}
\end{figure}
 By this bijection, tilings of $F^*_{2n+1,2x,2n+1}$ are seen to be
 equinumerous with families $(P_1, P_2, \ldots, P_{2n+1})$ of 
non-intersecting lattice paths, where for $i = 1, 2,\ldots , 2n+1$,
$P_i$ runs from~$(-i,i)$ to some point from the set $I=\{(-1,j):\, 
1\leq j\leq 2x+2n+1\}$ with the additional condition that $(-1,x+n+1)$
{\it must\/} be an ending point of {\it some} path. (We should note 
that an empty path, 
i.e., a path with no steps, may occur as in Figure~\ref{fig:tilings-paths},
where the bottommost path is the empty path from $(-1,1)$ to
$(-1,1)$.)
Similarly, tilings of $F^*_{2n,2x+1,2n}$ are equinumerous with
families $(P_1, P_2, \ldots, P_{2n})$ of non-intersecting lattice 
paths, where $P_i$ runs from $(-i, i)$ to some point from the set $I$
with the same additional condition. 
To enumerate these non-intersecting lattice paths, we shall use a
slight extension of 
a theorem of Stembridge~\cite[Theorem~3.2]{Stem}, see 
Theorem~\ref{thm:Paths-Pfaffians} below.

\subsection{Non-intersecting lattice paths and
Pfaffians}\label{sect:paths-Pfaffian} 
We first set up some terminology.  The \textit{Pfaffian} of a
skew-symmetric matrix $A=(A_{i,j})_{1\le i,j\le 2n}$ 
will be denoted by $\Pf A $. It is well-known
(see e.g.~\cite[Props.~2.2 and 2.3]{Stem}) 
that the Pfaffian satisfies the following three properties:
\begin{align}\label{eq:pf-det}
(\Pf A)^2&=\det A,\\
\label{eq:PfBAB}
\Pf (B\,A\,B^t)&=(\det B)(\Pf A),\\
\label{eq:xPf}
\underset{1\le i,j\le 2n}\Pf (x_ix_jA_{i,j})&=
x_1x_2\cdots x_{2n}\Pf A,
\end{align}
for some $2n\times 2n$ matrix $B$.
The \textit{sign} of a permutation $\pi\in\mathfrak{S}_n$,
where $\mathfrak{S}_n$ is the symmetric group of rank $n$, will
be denoted by $\sgn\pi$. Recall that
\begin{equation}\label{eq:sign}
\sgn\pi=(-1)^{\inv \pi},
\end{equation}
where $\inv \pi$ is the number of \textit{inversions} of $\pi$; i.e.,
the number of pairs $(i,j)$ with $1\leq i<j\leq n$ and $\pi(i)>\pi(j)$.

Given a weight function $w$ that assigns values in the field of
complex numbers to each edge of the integer lattice
$\mathbb{Z}^2$, we extend $w$ multiplicatively to multisets of
edges, so that $w(M)=\prod_{e\in M} w(e)$ for any such multiset $M$.
The weight of a lattice path or $r$-tuple of lattice paths
is defined as the weight of the underlying multiset of edges. Given
any family $\mathcal{F}$ of edge multisets, we will write
$GF[\mathcal{F}]$ for the generating function for $\mathcal F$ 
with respect to the weight $w$;
 i.e., $GF[\mathcal{F}]=\sum_{M\in \mathcal{F}} w(M)$. In particular,
 let us define 
\begin{equation}\label{eq:def h(A,B)}
h(A,B)=GF[\mathcal{P}(A\rightarrow B)]=\sum_{P\in
\mathcal{P}(A\rightarrow B)} w(P),
\end{equation}
where $\mathcal{P}(A\rightarrow B)$ stands for the set of lattice paths
from $A$ to $B$.

\begin{thm}\label{thm:Paths-Pfaffians}
Let $\Acal=(A_1,A_2,\ldots ,A_p)$, $\Scal=(S_1, S_2, \ldots , S_q)$
and $\Ical= 
(I_1, I_2,\ldots)$ be finite lists of points in the lattice
$\mathbb{Z}^2$, with $p+q$ even. 
For $\pi\in \mathfrak{S}_p$, let $\mathcal{P}^{nonint}
(\Acal_{\pi}\rightarrow (\Scal,\Ical))$ denote the set of 
families $(P_1, P_2,\ldots, P_p)$ of non-intersecting lattice paths
with $P_k$ running from 
$A_{\pi(k)}$ to $S_k$, for $k = 1, 2, \ldots, q$, and to $I_{j_k}$,
for $k = q + 1, q + 2,\ldots, p$, the indices being required to
satisfy $j_{q+1} < j_{q+2} <\ldots < j_p$. Then
\begin{equation}\label{eq:Pfaffian-nonintersecting}
\Pf\left(
    \begin{array}{cc}
      Q & H \\
      -H^t & 0 \\
    \end{array}
  \right)=(-1)^{{\binom q 2}} \sum_{\pi\in \mathfrak{S}_p} (\sgn
\pi)\; GF[\mathcal{P}^{nonint} (\Acal_{\pi}\rightarrow (\Scal,\Ical))],
\end{equation}
where the matrix $Q = (Q_{i,j})_{1\leq i,j\leq p}$ is defined by
\begin{equation}\label{eq:defQij}
Q_{i,j}=\sum_{1\leq s<t} \big(h(A_i,I_s)\, h(A_j,I_t) - h(A_j,I_s)\,
h(A_i,I_t)\big),
\end{equation}
and the matrix $H=(H_{i,j})_{1\leq i\leq p, 1\leq j\leq q}$ by
\begin{equation}\label{eq:defHij}
H_{i,j} = h(A_i,S_j),
\end{equation}
with $h(A,B)$ defined by~\eqref{eq:def h(A,B)}.
\end{thm}

The preceding theorem is a slight extension of a result of
Stembridge~\cite[Theorem~3.2]{Stem} and is just 
a weighted version of~Theorem~5 in~\cite{CiuKrat-interaction}. Since
the proof of Stembridge's theorem (alternatively, we may use the
minor summation formula of Ishikawa and
Wakayama~\cite[Theorem~2]{Ishikawa}) is easily adapted with only a
little extra effort to the preceding result, the proof details are
omitted.

%\newpage
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Centered vertically symmetric tilings and Pfaffians}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Combining the non-intersecting lattice path interpretation of
centered vertically symmetric 
rhombus tilings with Theorem~\ref{thm:Paths-Pfaffians}, we obtain the
following counting formulas in terms of Pfaffians.

\begin{lem}\label{lem:tilings-Pfaffian_casA}
Let $n$ be a fixed non-negative integer. Then, for any non-negative integer $x$,
the number of centered vertically symmetric rhombus tilings of a 
$(2n+1,2x,2n+1)$-hexagon
is equal to the Pfaffian of the skew-symmetric
matrix $M(x)= (M_{i,j}(x))_{1\leq i,j\leq 2n+2}$ defined by
\begin{alignat}2
\label{eq:def-Mij}
M_{i,j}(x)&=R_{i,j}(x)+T_{i,j}(x)-T_{j,i}(x),\quad &&1\leq i,j\leq 2n+1,\\
M_{i,2n+2}(x)&=\binom {x+n}{i-1},\quad &&1\leq i\leq 2n+1,\label{eq:def-Mij-Lr}
\end{alignat}
where $R_{i,j}(x)$ and $T_{i,j}(x)$ are defined by
\begin{align}
R_{i,j}(x)&:=\sum_{t=1}^{2x+2n+1}\frac{j-i}{t}{\binom t i}{\binom t
j}=\sum_{t=1}^{2x+2n+1}\left({\binom t i}{\binom {t-1}{ j-1}}
-{\binom {t-1}{i-1}}{\binom t j}\right),
\label{eq:def-Rij}\\
T_{i,j}(x)&:={\binom {2x+2n+1} i}\left({\binom {x+n} j}+{\binom {x+n+1} j}
\right).\label{eq:def-Tij}
\end{align}
\end{lem}

\begin{lem}\label{lem:tilings-Pfaffian_casB}
Let $n$ be a fixed positive integer. Then, for any non-negative
integer $x$,  the number of centered vertically symmetric rhombus
tilings of a $(2n,2x+1,2n)$-hexagon  is equal to the Pfaffian of the
skew-symmetric matrix $N(x)= (N_{i,j}(x))_{1\leq i,j\leq 2n+2}$
defined by
\begin{alignat}2
N_{i,j}(x)&=M_{i,j}(x)=R_{i,j}(x)+T_{i,j}(x)-T_{j,i}(x),\quad &&1\leq
i,j\leq 2n+1,\label{eq:def-Nij}\\ 
N_{i,2n+1}(x)&={\binom {2n+2x+1 } i}-{\binom {n+x } i}-{\binom {n+x+1
} i},\quad &&1\leq i\leq 2n,\label{eq:Qij_B_lastcolumn}\\ 
N_{i,2n+2}(x)&=M_{i,2n+2}(x)=\binom {x+n}{i-1},\quad &&1\leq i\leq
2n,
\label{eq:def-Nij-Lr}
\\ 
N_{2n+1,2n+2}(x)&=0,
\nonumber
\end{alignat}
where $M_{i,j}(x)$, $R_{i,j}(x)$ and $T_{i,j}(x)$ are defined as in
Lemma~{\em\ref{lem:tilings-Pfaffian_casA}}. 
\end{lem}

\begin{rmk}\label{rmk:Mn vs Nn}
 The matrices $M(x)$ and $N(x)$ differ only in the $(2n+1)$-st row and
 $(2n+1)$-st column. 
\end{rmk}

Before we turn to the proof of these two lemmas, we
provide an alternative expression for $R_{i,j}(x)$ which has the
advantage to be polynomial in $x$:
\begin{equation}\label{eq:def-Rij-bis}
R_{i,j}(x)=\sum_{\ell=0}^{i-1}\frac{j-i}{i}{\binom {j-1}{ i-1-\ell}}
{\binom {\ell+j} \ell}{\binom {2x+2n+2}{ \ell+j+1}}.
\end{equation}
The proof that the above relation is equivalent to~\eqref{eq:def-Rij}
(for $x\geq 0$) amounts to a routine computation 
(see e.g.\ the proof of Equation~(4.13) in~\cite{CiuKrat-interaction}).
We should also notice that the equivalence of these two relations
also holds for $x<0$ once we interpret sums by
\begin{equation}
 \sum_{k=m}^{n-1}\textrm{Expr}(k)=
           \begin{cases}
             \sum_{k=m}^{n-1}\textrm{Expr}(k), & \text{if $n>m$,} \\
             0, & \hbox{if $n=m$,}\\
             -\sum_{k=n}^{m-1}\textrm{Expr}(k), & \text{if $n<m$,}
           \end{cases}
\end{equation}
as in~\cite[Sec.~5]{CiuKrat-interaction}. With what we have said,
the expression~\eqref{eq:def-Rij} for $R_{i,j}(x)$ makes sense for
negative integers $x$ also and is equal to the expression
in~\eqref{eq:def-Rij-bis}. We will make use of this fact later in
this paper.

\begin{proof}[Proof of Lemma~\ref{lem:tilings-Pfaffian_casA}]
 Let $T(F^*_{2n+1,2x,2n+1})$ denote the number of rhombus tilings of
 the region $F^*_{2n+1,2x,2n+1}$. 
We apply Theorem~\ref{thm:Paths-Pfaffians}.
Let $\Acal$, $\Scal$ and $\Ical$ be the list of points in
 $\mathbb{Z}^2$ defined by
\begin{equation}\label{eq:def-SandI}
\begin{split}
 &\Acal=(A_1,A_2,\ldots,A_{2n+1})\quad\text{with $A_i=(-i, i)$},\\
 &\Scal=(I_{n+x+1})\quad\text{and}\quad\Ical=(I_1,
 I_2,\ldots,I_{2x+2n+1})\setminus(I_{n+x+1})\quad\text{with
 $I_j=(-1,j)$}. 
\end{split}
\end{equation}
It follows from what we have said in Section~\ref{sect:tilings-paths} that
\begin{equation}\label{eq:T(Hnx)}
T(F^*_{2n+1,2x,2n+1})=\sum_{\pi\in \mathfrak{S}_{2n+1}}
\#\mathcal{P}^{nonint} (\Acal_{\pi}\rightarrow (\Scal,\Ical)),
\end{equation}
where $\mathcal{P}^{nonint} (\Acal_{\pi}\rightarrow (\Scal,\Ical))$ is
 defined 
as in Theorem~\ref{thm:Paths-Pfaffians}. Therefore,
by~\eqref{eq:Pfaffian-nonintersecting}, in order to express
$T(H_{2n+1,2x})$ in terms of a Pfaffian, it suffices to find a
 weight function that assigns $\sgn\pi$ to each family of paths
 in 
$\mathcal{P}^{nonint} (\Acal_{\pi}\rightarrow (\Scal,\Ical))$ for
$\pi\in\mathfrak{S}_{2n+1}$. To this end, it is convenient to
slightly modify the list $\Acal$ (because of the possible occurrence
of empty paths, 
as in Figure~\ref{fig:tilings-paths}, which always have weight 1).
Let $\widetilde\Acal$ be defined by
\begin{equation}\label{eq:def-Amod}
\widetilde\Acal=(\widetilde A_1,\widetilde A_2,\ldots,\widetilde
A_{2n+1})\quad 
\text{with $\widetilde A_1=(-1, 0)$ and $\widetilde A_i=A_i=(-i, i)$
for $i\geq 2$}. 
\end{equation}
It is easily checked that
$\#\mathcal{P}^{nonint} (\Acal_{\pi}\rightarrow
(\Scal,\Ical))=\#\mathcal{P}^{nonint}
(\widetilde\Acal_{\pi}\rightarrow (\Scal,\Ical))$ 
for any $\pi$ in~$\mathfrak{S}_{2n+1}$. So, by~\eqref{eq:T(Hnx)}, we
have 
\begin{equation}\label{eq:T(Hnx)-mod}
T(F^*_{2n+1,2x,2n+1})=\sum_{\pi\in \mathfrak{S}_{2n+1}}
\#\mathcal{P}^{nonint} (\widetilde\Acal_{\pi}\rightarrow (\Scal,\Ical)).
\end{equation}
Let $\pi=\pi_1\pi_2\ldots\pi_{2n+1}$  be a permutation in
$\mathfrak{S}_{2n+1}$, written in one-line notation, such that
$\mathcal{P}^{nonint}(\widetilde\Acal_{\pi}\rightarrow
(\Scal,\Ical))$ is non-empty. From the definition of the $\widetilde
A_i$'s and $I_j$'s, it is not difficult to see that the sequence
$\pi_2\pi_3\cdots\pi_{2n+1}$ must be increasing, and thus the number
of inversions of $\pi$, $\inv\pi$, is equal to $\pi_1-1$. This
implies that, for any family $\left(P_1, P_2,\ldots, P_{2n+1}\right)$ in
$\mathcal{P}^{nonint}(\widetilde\Acal_{\pi}\rightarrow
(\Scal,\Ical))$,  $\inv\pi$ is the number of paths $P_i$ that
end in a point $I_j$ with $j<n+x+1$. So, if $w$ is a weight function
on the edges of the lattice $\mathbb{Z}^2$ such that the
weight of a path $P$ from $\widetilde A_i$ to $I_j$ satisfies
\begin{equation}
  w(P)=
  \begin{cases}
    +1, & \text{if $j\geq n+x+1$,} \\
    -1, & \text{if $j<n+x+1$,}
  \end{cases}
\label{eq:def_weight1}
\end{equation}
we would obtain from~\eqref{eq:T(Hnx)-mod},~\eqref{eq:sign} and what
we have said that
\begin{equation}\label{eq:T(Hnx)-bis}
T(F^*_{2n+1,2x,2n+1})=\sum_{\pi\in \mathfrak{S}_{2n+1}} (\sgn\pi)\;
GF[\mathcal{P}^{nonint} (\widetilde\Acal_{\pi}\rightarrow (\Scal,\Ical)],
\end{equation}
where $GF$ is the generating function with respect to the weight $w$,
as defined in Section~\ref{sect:paths-Pfaffian}. Note that the
construction of such a weight function $w$ is easy: for instance, we
may assign to each edge of the lattice~$\mathbb{Z}^2$ the value $1$
except for the edges $\{(-1,0),(-1,1)\}$, $\{(-2,j),(-1,j)\}$ for
$2\leq j<n+x+1$, and $\{(-1,n+x),(-1,n+x+1)\}$, to which we assign
the value $-1$. So, we may assume that~\eqref{eq:T(Hnx)-bis} is true
for some weight function $w$ satisfying~\eqref{eq:def_weight1}.

Let $h(\widetilde A_i,I_j)$ be defined as in~\eqref{eq:def h(A,B)}.
Then, 
by~\eqref{eq:def_weight1}, we have 
$h(\widetilde
A_i,I_j)=\epsilon \#\mathcal P(\widetilde A_i\rightarrow I_j)$,
where $\epsilon=-1$ if $j<n+x+1$ and $\epsilon=1$ otherwise,
and where $\mathcal P(A\rightarrow B)$ 
denotes the set of lattice paths from $A$ to $B$, as earlier. 
In combination with the well-known 
counting formula for lattice paths with given starting and end point, 
this implies that
\begin{equation}\label{eq:NumbLatticePaths}
h(\widetilde A_i, I_j)=(-1)^{\chi\left(j<  n+x+1\right)}\binom {j-1}{i-1},
\end{equation}
where, as usual, for an assertion $\mathcal X$, 
$\chi(\mathcal X)=1$ if $\mathcal X$ is true and
$\chi(\mathcal X)=0$ otherwise. 
Combining~\eqref{eq:T(Hnx)-bis} with
Theorem~\ref{thm:Paths-Pfaffians}, we see that 
\begin{equation}\label{eq:def-MatTil1}
T(F^*_{2n+1,2x,2n+1})=\Pf\left(
    \begin{array}{cc}
      Q& H \\
      -H^t & 0 \\
    \end{array}
  \right),
\end{equation}
where $Q= (Q_{i,j})_{1\leq i,j\leq 2n+1}$ is the matrix defined by
\begin{equation}\label{eq:def_Qij}
Q_{i,j}=\underset{s,t\neq n+x+1}{\sum_{1\leq s<t\leq 2x+2n+1}}
h(\widetilde
A_i,I_s)\, h(\widetilde A_j,I_t) - h(\widetilde A_j,I_s)\, h(\widetilde
A_i,I_t),
\end{equation}
and $H=(H_{i,1})_{1\leq i\leq 2n+1}$ is the column vector defined by
\begin{equation}\label{eq:def_Hi1}
H_{i,1}= h(\widetilde A_i,I_{x+n+1})=\binom
{x+n}{i-1},
\end{equation}
where the last equality follows from~\eqref{eq:NumbLatticePaths}. So,
to complete the proof of the lemma, 
it suffices to show that the matrix entry $Q_{i,j}$
in~\eqref{eq:def_Qij} is equal 
to the right-hand side of\eqref{eq:def-Mij}.

Let us define
\begin{equation}\label{eq:def_Pij}
P_{i,j}(s,t) ={\binom {s-1}{ i-1}}{\binom {t-1}{ j-1}}-{\binom {s-1}{
j-1}}{\binom {t-1}{ i-1}}.
\end{equation}
Then, combining~\eqref{eq:def_Qij} with~\eqref{eq:NumbLatticePaths},
we obtain 
\begin{equation}\label{eq:decompo-Qij}
Q_{i,j}=\sum_{1\leq s<t\leq x+n}P_{i,j}(s,t)+\sum_{x+n+2\leq s<t\leq
2x+2n+1}P_{i,j}(s,t)-\sum_{s=1}^{n+x}\sum_{t=n+x+2}^{2n+2x+1}P_{i,j}(s,t).
\end{equation}
By~\eqref{eq:def_Pij}, the last sum in the preceding relation can be
evaluated as follows: 
\begin{align}
\sum_{s=1}^{n+x}\sum_{t=n+x+2}^{2n+2x+1}P_{i,j}(s,t)
&=\sum_{s=1}^{n+x}{\binom {s-1}{ i-1}}
\sum_{t=n+x+2}^{2n+2x+1}{\binom {t-1}{ j-1}}-\sum_{s=1}^{n+x}
{\binom {s-1}{ j-1}}\sum_{t=n+x+2}^{2n+2x+1} {\binom {t-1}{ i-1}}
\nonumber\\
\begin{split}\label{eq:SumPij eq1}
&={\binom {n+x}{ i}}\left({\binom {2n+2x+1}{ j}}-{\binom {n+x+1}{
j}}\right)\\ 
&\quad- {\binom {n+x}{ j}}\left({\binom {2n+2x+1}{ i}}-{\binom
{n+x+1}{ i}}\right)
\end{split}\\
\begin{split}\label{eq:SumPij eq2}
&={\binom {n+x}{ i}}{\binom {2n+2x+1}{ j}}-{\binom {n+x}{ j}}{\binom
{2n+2x+1}{ i}}\\ 
&\quad+\frac{i-j}{n+x+1}{\binom {n+x+1}{ i}}{\binom {n+x+1}{ j}},
\end{split}
\end{align}
where we used the well-known identity
\begin{equation}\label{eq:BinomialSum}
\sum_{\ell=m}^{n}{\binom {\ell}{ i}}={\binom {n+1}{ i+1}}-{\binom {m}{ i+1}}
\end{equation}
to obtain~\eqref{eq:SumPij eq1}. Similarly,
using~\eqref{eq:BinomialSum}, we can prove (for details, see
e.g.~\cite[Eq.~(4.10)]{CiuKrat-interaction}, where the case
$M=1$ is treated) that
\begin{align}
\sum_{M\leq s<t\leq N}P_{i,j}(s,t)&=\sum_{M\leq s<t\leq N}\left(
{\binom {s-1}{ i-1}}{\binom {t-1}{ j-1}}- {\binom {s-1}{ j-1}}{\binom {t-1}{
i-1}}\right)\nonumber\\
&=\sum_{t=M}^N \frac{j-i}{t}{\binom {t}{ i}}{\binom {t}{ j}}\;+{\binom {N}{
i}}{\binom {M-1}{ j}}- {\binom {N}{ j}}{\binom {M-1}{ i}}.
\label{eq:Pij from M to N}
\end{align}
Finally, after substituting~\eqref{eq:SumPij eq2} and~\eqref{eq:Pij from M
to N} in~\eqref{eq:decompo-Qij}, 
it is easily checked that the matrix entry $Q_{i,j}$ is equal to the
right-hand side of 
the relation~\eqref{eq:def-Mij}. This finishes the proof. 
\end{proof}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\subsection{Centered rhombus tilings of $F_{2n,2x+1}$ and Pfaffians}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{proof}[Proof of Lemma~\ref{lem:tilings-Pfaffian_casB}]
Using the correspondence illustrated in
Figure~\ref{fig:tilings-paths} and the same reasoning as in the
proof of~\eqref{eq:T(Hnx)-mod}, we see that $T(F^*_{2n,2x+1,2n})$,  the
number of tilings of the region $F^*_{2n,2x+1,2n}$, is equal to the
number of families $(P_1, 
P_2, \ldots, P_{2n})$ of non-intersecting lattice paths, where for
$i= 1, 2,\ldots,2n$, the path $P_i$ runs from $\widetilde A_i$ to some $I_j$,
with the additional condition that $I_{n+x+1}$ must be an ending
point of some path. Here, the $\widetilde A_i$'s and $I_j$'s have
the same meaning as in~\eqref{eq:def-Amod} and~\eqref{eq:def-SandI}.
To be able to use Theorem~\ref{thm:Paths-Pfaffians}, we shall need
to introduce a phantom vertex (see the remark after Theorem~3.1
in~\cite{Stem}) for the parity condition on $p+q$ in
Theorem~\ref{thm:Paths-Pfaffians} to be satisfied.

Let $\Acal'=(A'_1,A'_2,\ldots,A'_{2n+1})$, $\Scal'=(I_{n+x+1}')$ and
$\Ical'=(I_1', I_2',\ldots,I_{2x+2n+2}')\setminus(I_{n+x+1}')$ 
be the list of points in $\mathbb{Z}^2$ defined by
\begin{equation}\label{eq:def-S'andI'}
\begin{split}
 A'_1&=\widetilde A_1=(-1, 0),\quad   A'_i=\widetilde A_i=(-i,
 i)\quad\text{for $1\leq i\leq 2n$},\quad A'_{2n+1}=(0,-1),\\ 
 I'_j&=I_j=(-1,j)\quad\text{for $1\leq j\leq 2x+2n+1$},\quad
 I'_{2x+2n+2}=A'_{2n+1}=(0,-1). 
\end{split}
\end{equation}
Then, from what we have said in the
preceding paragraph it is not hard to show that
\begin{equation}\label{eq:T(Hnx)'}
T(F^*_{2n,2x+1,2n})=\sum_{\pi\in \mathfrak{S}_{2n+1}}
\#\mathcal{P}^{nonint} (\Acal'_{\pi}\rightarrow (\Scal',\Ical')).
\end{equation}
Let $\pi=\pi_1\pi_2\ldots\pi_{2n+1}$ be a permutation in
$\mathfrak{S}_{2n+1}$  such that
$\mathcal{P}^{nonint}(\Acal'_{\pi}\rightarrow (\Scal',\Ical'))$ 
is non-empty. Then, from the definition of the $A_i'$'s and $I_j'$'s,
it is not difficult to 
see that the sequence $\pi_2\pi_3\cdots\pi_{2n+1}$ must be increasing
and $\pi_{2n+1}=2n+1$ (for, if $\left(P_1,\ldots, 
P_{2n+1}\right)\in P^{nonint}(\Acal'_{\pi}\rightarrow
(\Scal',\Ical'))$, then 
$P_{2n+1}$ must be the empty path from $A'_{2n+1}$ to
$I'_{2x+2n+2}$). Consequently, $\inv\pi$ is equal to $\pi_1-1$.
So, for any family $\left(P_1, 
P_2,\ldots, P_{2n+1}\right)$ in $P^{nonint}(\Acal'_{\pi}\rightarrow
(\Scal',\Ical'))$, it is also the 
number of paths $P_i$ such that $P_i$ ends at a point $I'_j$ with
$1\leq j<n+x+1$. In a way similar to the one in the proof of
Lemma~\ref{lem:tilings-Pfaffian_casA}, one can find a
weight function 
$w'$ on the edges of $\mathbb{Z}^2$ such that the weight of a path $P$
from $A'_i$ to $I'_j$ satisfies 
\begin{equation}
 w'(P)=
  \begin{cases}
    +1, & \text{if $j\geq n+x+1$,} \\
    -1, & \text{if $1\leq j<n+x+1$.}
  \end{cases}
\label{eq:def_weight2}
\end{equation}
We omit the details. It follows
from~\eqref{eq:T(Hnx)'},~\eqref{eq:sign} and  
what we have said that
\begin{equation}\label{eq:T(Hnx)-bis'}
T(F^*_{2n,2x+1,2n})=\sum_{\pi\in \mathfrak{S}_{2n+1}} (\sgn\pi)\;
GF[\mathcal{P}^{nonint} (\Acal'_{\pi}\rightarrow (\Scal',\Ical'))].
\end{equation}

Similarly to~\eqref{eq:NumbLatticePaths},
we have $h(A_i',I_j')=\epsilon \#\mathcal P(A_i'\rightarrow I_j')$, 
where $\epsilon=-1$ if
$j< n+x+1$ and $\epsilon=1$ otherwise, whence
\begin{equation}\label{eq:NumbLatticePaths-b}
h(A_i', I_j')=\begin{cases}
     (-1)^{\chi\left(j<  n+x+1\right)}\binom {j-1}{i-1}, & \text{if
 $i\neq 2n+1$ and $j\neq 2x+2n+2$,} \\ 
     1, & \text{if $(i,j)=(2n+1,2x+2n+2)$,} \\
     0, & \text{otherwise.}
  \end{cases}
\end{equation}
Combining~\eqref{eq:T(Hnx)-bis'} with
Theorem~\ref{thm:Paths-Pfaffians} and~\eqref{eq:NumbLatticePaths-b}, 
we obtain 
\begin{equation}\label{eq:defMat1}
T(F^*_{2n,2x+1,2n})=\Pf\left(
    \begin{array}{cc}
      Q& H \\
      -H^t & 0 \\
    \end{array}
  \right),
\end{equation}
where $Q= (Q_{i,j})_{1\leq i,j\leq 2n+1}$ is the matrix defined by
\begin{equation}\label{eq:def_Qij-b}
Q_{i,j}=\underset { s,t\neq n+x+1}{\sum_{1\leq s<t\leq 2x+2n+2}}
h(A_i',I_s')\, h(A_j',I_t') - h(A_j',I_s')\, h(A_i',I_t'),
\end{equation}
and $H=(H_{i,1})_{1\leq i\leq 2n+1}$ is the column vector defined by
\begin{equation}\label{eq:def_Hi1-b}
H_{i,1}= h(A_i',I_{x+n+1}')=\chi(i\neq 2n+1)\binom {x+n}{i-1},
\end{equation}
where the last equality follows from~\eqref{eq:NumbLatticePaths-b}.

Comparing~\eqref{eq:NumbLatticePaths-b}
with~\eqref{eq:NumbLatticePaths}, it is not hard to see that, for
$1\leq i,j\leq 2n$, the right-hand side of~\eqref{eq:def_Qij-b} is
equal to the right-hand side of~\eqref{eq:def_Qij} (which was shown
to be the right-hand side of $\eqref{eq:def-Mij}$). 
Note that this can also be directly proved from the combinatorial
interpretation of~\eqref{eq:def_Qij} and~\eqref{eq:def_Qij-b} 
(see e.g.\ Equation~(3.1) in~\cite{Stem}).
On the other hand, when $1\leq i\leq 2n$ and $j=2n+1$,
from~\eqref{eq:def_Qij-b} 
and~\eqref{eq:NumbLatticePaths-b} we infer
\begin{equation}
Q_{i,2n+1}=\underset { t\neq n+x+1}{\sum_{t=1}^{2x+2n+1}}
h(A'_i,I'_t)
=-\sum_{t=1}^{n+x}{\binom {t-1 }{
i-1}}+\sum_{t=n+x+2}^{2n+2x+1}{\binom {t-1}{ i-1}},
\end{equation}
which, by~\eqref{eq:BinomialSum}, simplifies to the right-hand
side of~\eqref{eq:Qij_B_lastcolumn}. Summarizing, we have proved
that the matrix 
in~\eqref{eq:defMat1} is equal to the matrix $N(x)$ described in the lemma.
This finishes the proof.
\end{proof}

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%
%                         PREUVE THEOREME  1
%
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%\newpage
\section{Proof of Theorem~\ref{thm:Ncentered(2n+1,2x)}}
\label{sec:3}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%                 Plan de la preuve                          %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Throughout this section, we assume that $n$ is a (fixed) non-negative
integer. 
From~\eqref{eq:def N(a,b)}, it is easily checked that
$ST(2n+1,2x,2n+1)$ can be written in the form 
\begin{equation*}%\label{eq:N_2(2n+1,2x)}
ST(2n+1,2x,2n+1)=\frac{(x+\frac{1}{2})_{2n+1}}{(\frac{1}{2})_{2n+1}}
               \prod_{s=1}^{n}\frac{(2x+2s)_{4n-4s+3}}{(2s)_{4n-4s+3}}.
\end{equation*}
This, combined with~Lemma~\ref{lem:tilings-Pfaffian_casA}, implies
that Theorem~\ref{thm:Ncentered(2n+1,2x)} is equivalent to the
equation
\begin{multline}
\label{eq:Theorem1_bis}
 \Pf M(x)=
\frac{1}{4}\frac{(2n+2)!^2\,(2n)!}{(n+1)!^2\,(4n+1)!}\,
\prod_{s=1}^{n}\frac{(2x+2s)_{4n-4s+3}}{(2s)_{4n-4s+3}}\\
\times
\left(\sum_{i=0}^n\frac{(-1)^{n-i}}{2n+1-2i}
\frac{(x+n+1-i)_{2i}}{(i!)^2}\right),
\end{multline}
where $M(x)$ is the matrix defined in
Lemma~\ref{lem:tilings-Pfaffian_casA}. 
We shall prove the latter formula.

An outline of the proof is as follows.
In Section~\ref{sec:PfM-1}, we prove that
$\Pf\,M(x)$ is a polynomial in $x$ of degree at most $2n^2+3n$.
Subsequently, in Section~\ref{sec:PfM-2}, we show that
\begin{equation}\label{eq:inv_shift-Pfaffian}
\Pf\, M(x)=(-1)^n\Pf\,M(-2n-1-x).
\end{equation}
Then, in Section~\ref{sec:PfM-3}, we argue that
$\prod_{s=1}^n(x+s)^{s}$ divides $\Pf\, M(x)$ as a
polynomial in $x$, while in Section~\ref{sec:PfM-4} we
do the same for the product $\prod_{s=1}^n(x+s+\frac{1}{2})^{s}$.
By~\eqref{eq:inv_shift-Pfaffian}, this implies that
\begin{equation*}
\left(x+n+\frac{1}{2}\right)^{n}
\prod_{s=1}^{n-1}\left(x+s+\frac{1}{2}\right)^{s}
\left(x+2n+1-s-\frac{1}{2}\right)^{s}
\prod_{s=1}^n(x+s)^{s}(x+2n+1-s)^{s},
\end{equation*}
which simplifies to
\begin{equation}
2^{-2n^2-n}\,\prod_{s=1}^{n}\left(2x+2s\right)_{4n-4s+3}
\end{equation}
and is a polynomial of degree $2n^2+n$, divides $\Pf M(x)$ as a
polynomial in $x$. 
Altogether, this entails that
\begin{equation}\label{eq:def-Px}
\Pf M(x)=P(x)\,\prod_{s=1}^{n}\left(2x+2s\right)_{4n-4s+3}
\end{equation}
for some polynomial $P(x)$ in $x$ of degree at most $2n$. Therefore,
in order to prove~\eqref{eq:Theorem1_bis}, 
it remains to show that $P(x)$ is equal to the polynomial $K(x)$
defined by 
\begin{equation}\label{eq:def-Kx}
K(x)=
\frac{1}{4}\frac{(2n+2)!^2\,(2n)!}{(n+1)!^2\,(4n+1)!}
\prod_{s=1}^{n}\frac{1}{(2s)_{4n-4s+3}}\,
\left(\sum_{i=0}^n\frac{(-1)^{n-i}}{2n+1-2i}
\frac{(x+n+1-i)_{2i}}{(i!)^2}\right).
\end{equation}
Since $P(x)$ and $K(x)$ are polynomials in $x$ of degree at most
$2n$, it suffices to show that~$P(x)=K(x)$ holds for 
$2n+1$ distinct values of $x$. In Section~\ref{sec:PfM-5}, we
determine the evaluations of 
$P(x)$ and $K(x)$ at $x=0,-1,\dots,-n$, which shows that
$P(x)=K(x)$ at $x=0,-1,\ldots,-n$. 
Since, by~\eqref{eq:def-Px} and~\eqref{eq:inv_shift-Pfaffian},
$P(x)=P(-2n-1-x)$, and, by~\eqref{eq:def-Kx}, $K(x)=K(-2n-1-x )$ 
for any $x$, 
this shows at the same time that $P(x)=K(x)$ for
$x=-2n-1,-2n,\dots,-n-1$. Altogether, we have found $2n+2$ values 
of $x$ for which $P(x)=K(x)$. This would complete
the proof of~\eqref{eq:Theorem1_bis}. 


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%                 Step 1                                     %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsection{ $\Pf\,M(x)$ is a polynomial in $x$ of degree at most
$2n^2+3n$}\label{sec:PfM-1} 

By Lemma~\ref{lem:tilings-Pfaffian_casA}
and~\eqref{eq:def-Rij-bis}, the $(i,j)$-entry of 
$M(x)$ is a polynomial in $x$ of degree $i+j$ if $1\leq i, j\leq
2n+1$ and $i\neq j$, 
and of degree $i-1$ if $1\leq i\leq 2n+1$ and $j=2n+2$. Moreover,
$M_{i,j}(x)=0$ if $i=j$. 
By the formal definition of a Pfaffian (see
e.g.\ \cite[Sec.~2]{Stem}), this immediately shows that
$\Pf\,M(x)$ is 
a polynomial in $x$. This also implies that in the defining
expansion of the determinant $\det M(x)$ each nonzero term 
is a polynomial of degree
\begin{equation*}
\left(\sum_{i=1}^{2n+1}i\right)-1
+\left(\sum_{j=1}^{2n+1}j\right)-1=4n^2+6n.
\end{equation*}
Consequently, $\det M(x)$ is a polynomial of degree at most
$4n^2+6n$. 
The claim then follows from~\eqref{eq:pf-det}.


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%                 Step 2                                    %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{$\Pf M(x)=(-1)^n \Pf M(-2n-1-x)$}\label{sec:PfM-2}
We shall transform, up to sign, $M(x)$ into
$M(-2n-1-x)$ by a sequence of elementary row and column operations.
More precisely, let $B=\left(B_{i,j}\right)_{1\leq i,j\leq 2n+2}$ be
the lower triangular matrix of size $2n+2$ defined by
\begin{equation}
B=\left(
    \begin{array}{ccc|c}
       & & & 0 \\
      &\left(B_{i,j}\right)_{\substack{1\leq i\leq 2n+1 \\ 1\leq j\leq
      2n+1}} && \vdots \\ 
      &&& 0 \\\hline
      0&\cdots&0&-1
    \end{array}
  \right),
  \quad\text{with $B_{i,j}=\binom{i-1}{j-1}$ for $1\leq i,j\leq 2n+1$,}
\end{equation}
and let $M^{(1)}=\left(M^{(1)}_{i,j}\right)_{1\leq i,j\leq 2n+2}$ be
the skew-symmetric matrix (of size $2n+2$) defined by $M^{(1)}=B\,
M(x)\,B^t$. 
We claim that the $(i,j)$-entry in $M^{(1)}$ is, up to the sign
$(-1)^{i+j}$, the $(i,j)$-entry in $M(-2n-1-x)$. 
Since $\det B=-1$, by \eqref{eq:PfBAB} we would obtain
$$\Pf M^{(1)}=(\det B)(\Pf M(x))=-\Pf M(x),$$
and, on the other hand, by \eqref{eq:xPf},
$$\Pf M^{(1)}=(-1)^{1+2+\dots+(2n+2)}\Pf M(-2n-1-x)
=(-1)^{n+1}\Pf M(-2n-1-x).$$
If we combine these two relations, this 
would lead to~\eqref{eq:inv_shift-Pfaffian}, as desired. 

We now turn to the proof of the claim that
$M^{(1)}_{i,j}=(-1)^{i+j}M_{i,j}(-2n-1-x)$. 
By definition of the matrices $M^{(1)}$ and $B$, we have
\begin{align}
M^{(1)}_{i,j}&=\sum_{a = 1}^{i} \sum_{b = 1}^{j}{\binom {i-1}{a-1}}
{\binom {j-1}{b-1}}  M_{a,b}(x),\quad\text{$1\leq i,j\leq 2n+1$,}
\label{eq:def-Aij}\\
M^{(1)}_{i,2n+2}&=-\sum_{a = 1}^{i} {\binom {i-1 }{a-1}} M_{a,2n+2}(x),
\quad\text{$1\leq i\leq 2n+1$}.\label{eq:def-Aij-Lr}
\end{align}

Recall that, for $1\leq a,b\leq 2n+1$, we have $M_{a,b}(x)=R_{a,b}(x)
+T_{a,b}(x)-T_{b,a}(x)$, where the $R_{i,j}(x)$'s and $T_{i,j}(x)$'s
are given by~\eqref{eq:def-Rij} and~\eqref{eq:def-Tij}. It was
already shown in~\cite[Sec.~5, Step~1,
Equations~(5.5)--(5.10)]{CiuKrat-interaction} that 
\begin{equation}
\sum_{a = 1}^{i} \sum_{b = 1}^{j}{\binom {i-1}{a-1}} {\binom {j-1}
{b-1}}R_{a,b}(x)=
(-1)^{i+j}R_{i,j}(-2n-1-x).\label{eq:invarianceRij}
\end{equation}
On the other hand, using the expression~\eqref{eq:def-Tij}, after a
routine calculation we obtain
\begin{align}
&\sum_{a = 1}^{i} \sum_{b = 1}^{j}{\binom {i-1}{a-1}} {\binom {j-1}
{b-1}}\,T_{a,b}(x)\nonumber\\
&\quad=\sum_{a = 1}^{i} {\binom {i-1}{a-1}}{\binom {2x+2n+1}{
a}}\sum_{b = 1}^{j}{\binom {j-1}{b-1}}\left({\binom {x+n}{ b}}+{\binom
{x+n+1}{ b}}\right)\nonumber\\ 
&\quad=(-1)^{i+j} {\binom {-2x-2n-1 }{ i}}\left({\binom {-n-x}{
j}}+{\binom {-n-x-1}{ 
j}}\right)\label{eq:invarianceTij-bis}\\
&\quad=(-1)^{i+j}\,T_{i,j}(-2n-1-x),\label{eq:invarianceTij}
\end{align}
where, to obtain~\eqref{eq:invarianceTij-bis}, we used the identity
\begin{equation}\label{eq:Chu-Van}
\sum_{k\geq 0}\binom {L}{k-\epsilon}\binom {M}{k}= \binom {L+M}{L+\epsilon}
=(-1)^{L+\epsilon}\binom {-M-1+\epsilon}{L+\epsilon},
\end{equation}
which follows from the Chu--Vandermonde summation formula. After
substituting~\eqref{eq:def-Mij} in~\eqref{eq:def-Aij}, it is immediate 
from~\eqref{eq:invarianceRij} and~\eqref{eq:invarianceTij} that
$M^{(1)}_{i,j}=(-1)^{i+j}M_{i,j}(-2n-1-x)$ for $1\leq i,j\leq 2n+1$. 
Similarly, substituting~\eqref{eq:def-Mij-Lr} in~\eqref{eq:def-Aij-Lr},
we obtain for $1\leq i\leq 2n+1$ that
\begin{equation*}
M^{(1)}_{i,2n+2}= -\sum_{a = 1}^{i} {\binom {i-1 }{a-1}} \binom
{x+n}{a-1}=(-1)^{i}\binom{-x-n-1}{i-1}=(-1)^{i} M_{i,2n+2}(-2n-1-x),
\end{equation*}
where the second equality follows again from~\eqref{eq:Chu-Van}
(specialized to $\epsilon=0$).

Summarizing, we have shown that
$M^{(1)}=\left((-1)^{i+j}M_{i,j}(-2n-1-x)\right)_{1\leq i,j\leq 2n+2}$, as
desired. This finishes the proof of our claim.


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%                 Step 3                                     %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{ $\prod_{s=1}^n(x+s)^{s}$ divides $\Pf
M(x)$}\label{sec:PfM-3} 

Let $s$ be an integer with $1\leq s\leq n$. It is easily checked
that,
\begin{equation}\label{eq:comblineaire-s}
\textrm{for $2n-2s+2\leq a\leq 2n+1$, the $a$-th row of the
matrix $M(-s)$ vanishes.}
\end{equation}
Indeed, by Lemma~\ref{lem:tilings-Pfaffian_casA}, the entry in
row $a$ and column $j$ of $M(-s)$ equals
\begin{multline*}
M_{a,j}(-s)=\sum_{t=1}^{2n-2s+1}\frac{j-a}{t}{\binom {t}{ a}}{\binom {t}{
j}}+{\binom {2n-2s+1}{ a}}\left({\binom {n-s}{ j}}+{\binom {n-s+1}{
j}}\right)\\ 
-{\binom {2n-2s+1}{ j}}\left({\binom {n-s}{ a}}+{\binom
{n-s+1}{ a}}\right)
\end{multline*}
for $1\leq j\leq 2n+1$,
and $M_{a,2n+2}(-s)={\binom {n-s}{ a-1}}$. These entries are clearly
zero if $a>2n-2s+1$ since, for $t=1,\ldots,2n-2s+1$, 
the binomial coefficients ${\binom {n-s}{
a-1}}$ and ${\binom {t}{ a}}$ vanish.


By~\eqref{eq:comblineaire-s}, we have
$2s$ linear combinations of the rows of the matrix~$M(x)$
that are linearly independent and vanish at $x=-s$. This,
combined with the next lemma, yields the divisibility of $\Pf M(x)$
by~$(x+s)^{s}$, as desired. 

\begin{lem}[{\cite[Sec.~2]{Krat}}]
\label{lem:pf-factor}
Let $A(x)$ be a skew-symmetric matrix with polynomial entries. A
sufficient condition for $(x-b)^m$ to divide $\Pf A(x)$ is that
the dimension of the kernel of the matrix $A(b)$ is at least $2m$.
\end{lem}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%                 Step 3                                     %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{$\prod_{s=1}^n(x+s+1/2)^{s}$ divides $\Pf
M(x)$}\label{sec:PfM-4} 

We first notice that some coefficients of the matrix $M(x)$, when
specialized to $x=-s-1/2$, 
$1\leq s\leq n$, have a relatively simple form. Namely, for
$2n-2s+1\leq i\leq 2n+1$ and $j\leq 2n+1$, we have
\begin{equation}
M_{i,j}(-s-1/2)=-M_{j,i}(-s-1/2)=-{\binom {2n-2s}{
j}}\left({\binom {n-s-\frac{1}{2}}{ i}}+{\binom {n-s+\frac{1}{2}} i}
\right).
\label{eq:Qij-s-1/2_simplification_i>}
\end{equation}
Thus,
\begin{equation}\label{eq:Qij-s-1/2_annulation}
M_{i,j}(-s-1/2)=0\quad \textrm{for  $1\leq s\leq n$ and
$2n-2s+1\leq i,j\leq 2n+1$}.
\end{equation}
Indeed, by~\eqref{eq:def-Mij}, we have
\begin{multline*}
M_{i,j}(-s-1/2)=\sum_{t=1}^{2n-2s}\frac{j-i}{t}{\binom {t}{
i}}{\binom {t}{ j}}+{\binom {2n-2s}{ i}}\left({\binom {n-s-\frac{1}{2}}
j}+{\binom {n-s+\frac{1}{2}}{
j}}\right)\\
-{\binom {2n-2s}{ j}}\left({\binom {n-s-\frac{1}{2}}
i}+{\binom {n-s+\frac{1}{2}} i} \right),
\end{multline*}
and the binomial coefficients
${\binom {t}{ i}}$ vanish for $t=1,\ldots,2n-2s$ and $i\geq 2n-2s+1$.


\subsubsection{The case $1\leq s\leq n-1$} 
Suppose $1\leq s\leq n-1$. We shall show that 
\begin{multline}
(-1)^{a}{\binom {n+s-\frac{1}{2}}{ a+1}}\cdot\left( \textrm{row
$(2n-a)$ of $M(-s-1/2)$ } \right)\\ 
-\left(n+s-\frac{1}{2}\right) {\binom {2n}{ a}}\cdot \left(
\textrm{row $2n$ of $M(-s-1/2)$ } \right)\label{eq:comblineaire-s-1/2_a}\\
-a {\binom {2n+1}{ a+1}}\cdot\left( \textrm{row $(2n+1)$ of
$M(-s-1/2)$ } \right)=0,
\end{multline}
for $a=1,\ldots,2s-1$, and that
\begin{multline}
\sum_{i=1}^{2n-2s}(-2)^{i-1}\cdot\left( \textrm{row $i$ of
$M(-s-1/2)$ } \right)\\ 
-4^{n-s}
\frac{(2n-2s+2)_{2s}}{(n-s+\frac{1}{2})_{2s-1}}\cdot\left(
\textrm{row $2n$ of $M(-s-1/2)$ } \right)\label{eq:comblineaire-s-1/2_b}\\
-2n\cdot 4^{n-s}
\frac{(2n-2s+2)_{2s}}{(n-s+\frac{1}{2})_{2s}}\cdot\left( \textrm{row
$(2n+1)$ of $M(-s-1/2)$ } \right)=0.
\end{multline}
As these are $2s$ linear combinations of the rows of the matrix
$M(x)$ that are linearly independent and vanish at $x=-s-1/2$, 
by Lemma~\ref{lem:pf-factor} this
would lead to the divisibility of $\Pf
M(x)$ by~$(x+s+\frac{1}{2})^{s}$. 

\begin{proof}[Proof of~\eqref{eq:comblineaire-s-1/2_a}]
Suppose $1\leq a\leq 2s-1$. We have to show that, for $1\leq j\leq
2n+2$, we have
\begin{multline}
\label{eq:comblineaire-s-1/2_a_firstcolumns}
(-1)^{a}{\binom {n+s-\frac{1}{2}}{ a+1}} M_{2n-a,j}(-s-1/2) -
\left(n+s-\frac{1}{2}\right) {\binom {2n}{ a}} M_{2n,j}(-s-1/2)\\
- a {\binom {2n+1}{ a+1}}  M_{2n+1,j}(-s-1/2)=0.
\end{multline}

\medskip
(a) If $2n-2s+1\leq j\leq 2n+1$, all the matrix entries in
\eqref{eq:comblineaire-s-1/2_a_firstcolumns} are zero
by~\eqref{eq:Qij-s-1/2_annulation}. So, the identity is clearly
true.

\medskip
(b) Suppose $1\leq j\leq 2n-2s$. By
substituting~\eqref{eq:Qij-s-1/2_simplification_i>}
in~\eqref{eq:comblineaire-s-1/2_a_firstcolumns}, 
we obtain the relation
\begin{multline}\label{eq:comblineaire-s-1/2_a_eq1}
(-1)^{a}{\binom {n+s-\frac{1}{2}}{ a+1}}\left({\binom {n-s-\frac{1}{2}}{
2n-a}}+{\binom {n-s+\frac{1}{2}}{
2n-a}}\right)\\
-\left(n+s-\frac{1}{2}\right) {\binom {2n}{
a}}\left({\binom {n-s-\frac{1}{2}}{ 2n}}+{\binom {n-s+\frac{1}{2}}{
2n}}\right)\\
- a {\binom {2n+1}{
a+1}}\left({\binom {n-s-\frac{1}{2}}{ 2n+1}}+{\binom {n-s+\frac{1}{2}}{
2n+1}}\right)=0,
\end{multline}
as a routine verification shows.

\medskip
(c) Suppose $j=2n+2$. By~\eqref{eq:def-Mij-Lr},
\eqref{eq:comblineaire-s-1/2_a_firstcolumns} reduces to the relation
\begin{multline}
(-1)^{a}{\binom {n+s-\frac{1}{2}}{ a+1}}{\binom {n-s-\frac{1}{2}}{
2n-a-1}} 
-\left(n+s-\frac{1}{2}\right) {\binom {2n}{ a}}{\binom
{n-s-\frac{1}{2}}{ 2n-1}}\\ 
- a {\binom {2n+1}{ a+1}}{\binom {n-s-\frac{1}{2}}{
2n}}=0,
\label{eq:comblineaire-s-1/2_a_eq2}
\end{multline}
which is again seen by a routine verification. 

\medskip
This finishes the proof of~\eqref{eq:comblineaire-s-1/2_a}.
\end{proof}

\begin{proof}[Proof of~\eqref{eq:comblineaire-s-1/2_b}] 
We have to show that, for $1\leq j\leq 2n+2$, we have
\begin{multline}
\label{eq:comblineaire-s-1/2_b_eq}
\sum_{i=1}^{2n-2s}(-2)^{i-1} M_{i,j}(-s-1/2)=4^{n-s}
\frac{(2n-2s+2)_{2s}}{(n-s+\frac{1}{2})_{2s-1}}M_{2n,j}(-s-1/2)\\
+2n\cdot 4^{n-s}
\frac{(2n-2s+2)_{2s}}{(n-s+\frac{1}{2})_{2s}}M_{2n+1,j}(-s-1/2).
\end{multline}

\medskip
(a) Suppose $j=2n+2$. Then, using the
expression~\eqref{eq:def-Mij-Lr}, it is easily checked that the
right-hand 
side of~\eqref{eq:comblineaire-s-1/2_b_eq} is zero (when specialized
to $j=2n+2$), and thus~\eqref{eq:comblineaire-s-1/2_b_eq} 
reduces to the relation
\begin{equation}\label{eq:comblineaire-s-1/2_b_eq1_hypidentity}
 \sum_{i=0}^{2n-2s-1}(-2)^{i} {\binom {n-s-\frac{1}{2}}{
i}}=0.
\end{equation}
Reversing the order of summation over $i$ (that is, we replace $i$ by
$2n-2s-1-i$), the left-hand side
of~\eqref{eq:comblineaire-s-1/2_b_eq1_hypidentity} can be written
using the standard hypergeometric notation 
\begin{equation*}
{}_pF_q\!\left[\begin{matrix} {a_1,\ldots,a_p}\\
{b_1,\ldots,b_q}\end{matrix};z\right]
=\sum_{n\geq0}\frac{(a_1)_n\cdots(a_p)_n}{(b_1)_n
\cdots(b_q)_n}\frac{z^n}{n!}
\end{equation*}
as
\begin{equation*}
(-2)^{2n-2s-1} {\binom {n-s-\frac{1}{2}}{ 2n-2s-1}}\,
{}_2F_1\!\left[\begin{matrix} {1,-2n+2s+1}\\
{-n+s+\frac{3}{2}}\end{matrix};\frac{1}{2}\right].
\end{equation*}
This is indeed zero (recall that $1\leq s\leq n-1$) by means of
Gau{\ss}' second ${}_2F_1$-summation
(cf.\ \cite[(1.7.1.9); Appendix (III.6)]{SlatAC})
\begin{equation*}
{}_2F_1\!\left[\begin{matrix} {a,-N}\\
{\frac{1}{2}+\frac{a}{2}-\frac{N}{2}}\end{matrix};\frac{1}{2}\right] 
=
\begin{cases}
0, & \text{if $N$ is an odd non-negative integer,}\\[0.25cm]
\frac{\left(\frac{1}{2}\right)_{N/2}}{\left(\frac{1-a}{2}\right)_{N/2}},
& \text{if $N$ is an even non-negative integer.}
 \end{cases}
\end{equation*}

\medskip
(b) Suppose $2n-2s+1\leq j\leq 2n+1$. Then, using the
expressions~\eqref{eq:Qij-s-1/2_simplification_i>}
and~\eqref{eq:Qij-s-1/2_annulation} for the matrix 
entries in the sum in~\eqref{eq:comblineaire-s-1/2_b_eq},
it is easily checked that the right-hand side
of~\eqref{eq:comblineaire-s-1/2_b_eq} vanishes so
that \eqref{eq:comblineaire-s-1/2_b_eq} reduces to the identity
\begin{equation*}
 \sum_{i=1}^{2n-2s}(-2)^{i-1}\left({\binom {n-s-\frac{1}{2}}
j}+{\binom {n-s+\frac{1}{2}} j}\right){\binom {2n-2s}{ i}}=0,
\end{equation*}
which is an immediate consequence of the binomial theorem.

\medskip
(c) Suppose $1\leq j\leq 2n-2s$. Then, by~\eqref{eq:def-Mij}, the
left-hand side of~\eqref{eq:comblineaire-s-1/2_b_eq} is
\begin{multline}
\sum_{i=1}^{2n-2s}(-2)^{i-1}
M_{i,j}(-s-1/2)=\sum_{i=1}^{2n-2s}(-2)^{i-1}
\sum_{t=1}^{2n-2s}\frac{j-i}{t}{\binom {t}{
i}}{\binom {t}{ j}}\label{eq:sumQi}\\
+\left({\binom {n-s-\frac{1}{2}}
j}+{\binom {n-s+\frac{1}{2}}
j}\right)\sum_{i=1}^{2n-2s}(-2)^{i-1}{\binom {2n-2s}{ i}}\\ 
-{\binom {2n-2s}{ j}}\sum_{i=1}^{2n-2s}
(-2)^{i-1}\left({\binom {n-s-\frac{1}{2}} i}+{\binom {n-s+\frac{1}{2}}
i}\right).
\end{multline}
The first sum in~\eqref{eq:sumQi} is equal to ${\binom {2n-2s}{ j}}$.
This can be deduced from the formula
\begin{equation}\label{eq:sumbinom}
\sum_{i=1}^{N}(-2)^{i-1}\sum_{t=1}^{N}\frac{j-i}{t}{\binom {t}{
i}}{\binom {t}{ j}} =\frac{1+(-1)^N}{2}\binom{N}{j},
\end{equation}
which is valid for any non-negative integers $N$ and $j$. For $N$
even, \eqref{eq:sumbinom} was proved in~\cite[proof of
(5.21)]{CiuKrat-interaction}. The proof for odd $N$ is
similar to the case of even $N$ and involves
only elementary manipulations. It is thus left to the reader.

By the binomial theorem, the second sum in~\eqref{eq:sumQi} vanishes.
Moreover, the last sum in~\eqref{eq:sumQi} simplifies to
\begin{equation}\label{eq:sumQi_last1}
\sum_{i=1}^{2n-2s} (-2)^{i-1}\left({\binom {n-s-\frac{1}{2}}
i}+{\binom {n-s+\frac{1}{2}}
i}\right)=1-4^{n-s}{\binom {n-s-\frac{1}{2}}{ 2n-2s}}.
\end{equation}
This immediately follows from the telescoping equation
\begin{equation*}
(-2)^{i-1}\left({\binom {n-s-\frac{1}{2}} i}+{\binom {n-s+\frac{1}{2}}
i}\right)=G(i+1)-G(i),
\end{equation*}
with $G(i)=-(-2)^{i-1}{\binom {n-s-\frac{1}{2}}{ i-1}}$.
Finally, replacing the sums in~\eqref{eq:sumQi} by their evaluated forms,
we arrive at
\begin{equation}\label{eq:sumQi_simplification}
\sum_{i=1}^{2n-2s}(-2)^{i-1} M_{i,j}(-s-1/2)=4^{n-s}{\binom {2n-2s}{
j}}{\binom {n-s-\frac{1}{2}}{ 2n-2s}}.
\end{equation}
On the other hand, by~\eqref{eq:Qij-s-1/2_simplification_i>}, the
right-hand side of~\eqref{eq:comblineaire-s-1/2_b_eq} is equal to
\begin{multline*}
-4^{n-s}\frac{(2n-2s+2)_{2s}}{(n-s+\frac{1}{2})_{2s-1}}\binom{2n-2s}{j}
\left({\binom {n-s-\frac{1}{2}}{
2n}}+{\binom {n-s+\frac{1}{2}}{ 2n}}\right)\\
-2n 4^{n-s}
\frac{(2n-2s+2)_{2s}}{(n-s+\frac{1}{2})_{2s}}\binom{2n-2s}{j}
\left({\binom {n-s-\frac{1}{2}}{
2n+1}}+{\binom {n-s+\frac{1}{2}}{ 2n+1}}\right).
\end{multline*}
Therefore, to prove that~\eqref{eq:comblineaire-s-1/2_b_eq} holds
for $1\leq j\leq 2n-2s$, it suffices to show that the above
expression is equal to the right-hand side
of~\eqref{eq:sumQi_simplification}. This amounts to a routine
verification. 

\medskip
This concludes the proof of~\eqref{eq:comblineaire-s-1/2_b}.
\end{proof}

\subsubsection{The case $s=n$} 
From~\eqref{eq:Qij-s-1/2_annulation},
we see that all the coefficients of the $i$-th row, $1\leq i\leq
2n+1$, of the matrix $M(-n-1/2)$ are zero except its last
coefficient $M_{i,2n+2}(-n-1/2)$ (which is,
by~\eqref{eq:def-Mij-Lr}, equal to ${\binom {-1/2}{ i-1}}$). It is then
immediate to find $2n$ linear combinations of the rows of the matrix
$M(-n-1/2)$ that are linearly independent and vanish. For instance,
for $a=1,\ldots,2n$, we have
\begin{equation*}
{}\left( \textrm{row $a$ of $M(-n-1/2)$ } \right)-
\frac{M_{a,2n+2}(-n-1/2)}{M_{a+1,2n+2}(-n-\frac{1}{2})}\cdot\left(
\textrm{row $(a+1)$ of $M(-n-1/2)$} \right)=0.
\end{equation*}
This implies divisibility of $\Pf M(x)$ by~$(x+n+\frac{1}{2})^{n}$.


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%         Step 5                   %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{$P(x)=K(x)$ at $x=0,-1,\ldots,-n$}\label{sec:PfM-5}

Let $\s$ be a given integer with $0\leq \s \leq n$. It is not too
hard to evaluate the polynomial $K(x)$, defined by~\eqref{eq:def-Kx},
at~$x=-\s$. Indeed, using the 
Pfaff--Saalsch\"utz summation
(cf.\ \cite[(2.3.1.3); Appendix~(III.2)]{SlatAC}),
\begin{equation} \label{eq:PfSa}
{}_{3} F_{2} \! \left[ \begin{matrix} { a, b, -N} \\ { c, 1 + a + b
- c - N} \end{matrix} ; {\displaystyle 1} \right ] =
{\frac{({\textstyle c-a})_{N} \, ({\textstyle c-b})
_{N}}{({\textstyle c})_{N} \, ({\textstyle c-a-b})_{N}}},
\end{equation}
where $N$ is a non-negative integer, it is easily checked 
(see~\cite[Eq.~(5.26)]{CiuKrat-centered}) that
\begin{equation}
\sum_{i=0}^n\frac{(-1)^{n-i}}{2n+1-2i}\frac{(n-\s+1-i)_{2i}}{i!^2}
=(-1)^{n+1}\frac{\left(n+\frac{3}{2}\right)_{n-\s}}
{2\left(-n-\frac{1}{2}\right)_{n+1-\s}},
\quad\text{$0\leq \sigma\leq n$}.
\end{equation}
By inserting the latter identity in~\eqref{eq:def-Kx}, after
some simplification we arrive at
\begin{equation}\label{eq:Kx at -s}
K(-\s)=(-1)^{\s}\frac{(4n-2\s+1)!\,(2\s)!\,(2n)!}{(2n-\s)!\,\s!(4n+1)!}
\prod_{s=1}^n\frac{(2s-1)!}{(4n+2-2s)!},
\quad\text{$0\leq \sigma\leq n$}.
\end{equation}

The evaluation of $P(x)$ at $x=-\s$ is much more delicate. The
polynomial $P(x)$ is defined by means of \eqref{eq:def-Px}. Since the
product on 
the right-hand side of~\eqref{eq:def-Px} is zero for $x=-\s$,
$1\le \s\le n$, we should write~\eqref{eq:def-Px} in the form
\begin{multline*}
P(x)=\frac{2^{-\sigma}}{(x+\sigma)^{\sigma}} \Pf M(x)
\;\underset { s\neq \sigma}{\prod_{s=1}^n}(2x+2s)^{-s}\;
\prod_{s=1}^n(2x+4n+2-2s)^{-s} \\
\times\prod_{s=1}^{n}\left(2x+2s+1\right)^{-s}\;
\prod_{s=1}^{n-1}\left(2x+4n+1-2s\right)^{-s},
\end{multline*}
and subsequently specialize $x=-\sigma$. After some manipulation,
we arrive at
\begin{equation}\label{eq:Px at -s}
P(-\s)=\frac{2^{-\sigma}\prod _{s=1}^{n-\s} (2s-1)!}{\prod _{s=1}^{\s-1}
(2s)!\;\prod_{s=n-\s+1}^{2n-\s} (2s)!}
\left(\frac{1}{(x+\sigma)^{\sigma}} \Pf
M(x)\right)\bigg\vert_{x=-\sigma}.
\end{equation}
To evaluate $(x+\s)^{-\s} \Pf M(x)$ at $x=-\s$, we shall use the
following lemma due to Ciucu and the second author who used it 
in a similar context.
\begin{lem}[{\cite[Lemma~11]{CiuKrat-interaction}}]
\label{lem:FactorizationPfaffian}
Let $N,a,b$ be non-negative integers with $a< b\le N$, where $N$ and
$b-a$ are even. Let $A=(A_{i,j})_{1\le i,j\le N}$ be a
skew-symmetric matrix with the following properties:
\begin{enumerate}
\item The entries of $A$ are polynomials in $x$.
\item The entries in rows $a+1,a+2,\dots,b$ {\rm(}and, hence, also in
the corresponding columns{\rm)} are divisible by $x+s$.
\end{enumerate}
Then
$$\left(\frac {1} {(x+s)^{(b-a)/2}}\Pf A\right)\bigg\vert_{x=-s}=\Pf
\widetilde A\,\cdot\,\Pf S,$$
where $\widetilde A$ is the matrix which arises from
$A$ by deleting rows and columns $a+1,a+2,\dots,b$ and subsequently
specializing $x=-s$, and
$$S=\left(
\left(\frac {1} {x+s}A_{i,j}\right)\bigg\vert_{x=-s}\right) _{a+1\le
i,j\le b}.$$
\end{lem}

Let $\s$ be a given integer with $1\leq \s \leq n$. It follows
from~\eqref{eq:comblineaire-s} that 
for $2n-2\s+2 \leq i \leq 2n+1$ the coefficients of the $i$-th
row of the matrix $M(x)$ are divisible by
$(x+\s)$. Applying Lemma~\ref{lem:FactorizationPfaffian}, 
we obtain
\begin{equation}\label{eq:factorization_Pfaffian}
\left(\frac {1} {(x+\s)^{\s}}\Pf M(x)\right)\bigg\vert_{x=-\s}
=\Pf \widetilde M\, \cdot\, \Pf S,
\end{equation}
where $\Mtil= (\Mtil_{i,j})_{1\leq i,j\leq 2n-2\s+2}$ is the
skew-symmetric matrix of size $2n-2\s+2$ defined by 
\begin{equation}
\begin{split}\label{eq:def-Mtil}
\Mtil_{i,j}&=M_{i,j}(-\s),\quad 1\leq i,j\leq 2n-2\s+1,\\
\Mtil_{i,2n-2\s+2}&=M_{i,2n+2}(-\s),\quad 1\leq i\leq 2n-2\s+1,
\end{split}
\end{equation}
and $S= (S_{i,j})_{1\leq i,j\leq 2\s}$ is the skew-symmetric matrix
of size $2\s$ defined by 
\begin{equation}\label{eq:def-MatS}
S=\left( \left(\,\frac {1}
{x+\s}M_{i,j}(x)\right)\bigg\vert_{x=-\s}\,\right)_{2n-2\s+2\le
i,j\le 2n+1}.
\end{equation}
We point out that~\eqref{eq:factorization_Pfaffian} also holds for
$\s=0$ once we interpret the Pfaffian of an empty matrix (i.e., the 
Pfaffian of $S$) as $1$. In particular, under this convention, the
arguments below can be used for $0\le \s\le n$, that is, 
{\it including} $\s=0$.

We shall prove that $\Pf \Mtil=1$ and
\begin{equation}
\Pf S=(-1)^{\s} 2^{\s} \left(\prod_{i=1}^{2\s}(2n-2\s+i)! \right)
\left( \prod _{i=1} ^{\s}\frac {(2i-1)!}
{(4n-2\s+2i+1)!}\right).
\label{eq:evaluation_PfS}
\end{equation}
If we substitute these values for $\Pf \widetilde M$ and $\Pf S$
in~\eqref{eq:factorization_Pfaffian}, and then insert the
resulting equation in~\eqref{eq:Px at -s}, after
some simplification we would arrive at
\begin{equation}\label{eq:Pn at -s}
P(-\s)=(-1)^{\s}\frac{ \prod_{s=2n-2\s+1}^{2n}s!\; \prod _{s=1} ^{\s}
(2s-1)!\;\prod _{s=1} ^{n-\s} (2s-1)!} 
{ \prod _{s=1} ^{\s-1} (2s)!\;\prod _{s=n+1} ^{2n-\s} (2s)!\;\prod
_{s=0} ^{\s-1}(4n+1-2s)!}, \quad\text{$0\leq \sigma\leq n$}.
\end{equation}
Then, a routine comparison of~\eqref{eq:Pn at -s} with~\eqref{eq:Kx at -s}
would show that $P(x)=K(x)$ at $x=0,-1,\ldots,-n$, as desired. So, to
complete our proof, it remains to establish the claimed evaluations
of $\Pf \Mtil$ and $\Pf S$.

\medskip
\noindent\emph{Evaluation of $\Pf \Mtil$.} If we
compare the matrix $\Mtil$ defined by~\eqref{eq:def-Mtil} with the matrix
$M(x)$ in Lemma~\ref{lem:tilings-Pfaffian_casA}, then we see that
$\widetilde M$ is equal to the matrix $M(x)$ in
Lemma~\ref{lem:tilings-Pfaffian_casA} with $n$ replaced by $n-\s$
and with $x=0$. By Lemma~\ref{lem:tilings-Pfaffian_casA},
this implies that $\Pf \Mtil$ is equal to the number of centered
vertically symmetric tilings of a
$(2(n-\s)+1,0,2(n-\s)+1)$-hexagon. 
In a tiling of such a hexagon, all rhombi are forced, and trivially
the unique tiling of such a hexagon is centered and 
symmetric. Consequently, we have $\Pf \Mtil=1$.

\medskip
\noindent\emph{Evaluation of $\Pf S$.} By~\eqref{eq:def-MatS}
and~\eqref{eq:def-Mij}, the $(i,j)$-entry of the matrix $S$
satisfies
\begin{equation*}
S_{i,j}=\frac{R_{2N+1+i,2N+1+j}(x)}{x+\s}+\frac{T_{2N+1+i,2N+1+j}(x)}{x+\s}
-\frac{T_{2N+1+j,2N+1+i}(x)}{x+\s}\bigg\vert_{x=-\s},
\end{equation*}
where we have set $N=n-\s$. Using expression~\eqref{eq:def-Rij-bis}
for the $R_{i,j}$'s and~\eqref{eq:def-Tij} for the $T_{i,j}$'s,
after a routine calculation we arrive at
\begin{multline}
\label{eq:ValSij}
S_{i,j}= 2\cdot \sum _{\ell=0} ^{i+2n-2\s}(-1)^{\ell+j+1} \frac
{j-i} {i+2n-2\s+1}\binom {j+2n-2\s}{i+2n-2\s-\ell}\\
\cdot\binom{\ell+j+2n-2\s+1}{\ell} \frac {(2n-2\s+2)!\,(\ell+j-1)!}
{(\ell+j+2n-2\s+2)!}.
\end{multline}
To derive the above equation, we used the following
relations whose proof is left to the reader: for $r,t$ with $r\geq1$ and
$t\geq 0$, and $\epsilon=1,2$, we have
\begin{align*} %\label{eq:limit1-binom at -s}
\lim_{x\to -\s} \frac{1}{x+\s}\binom
{2x+2n+t}{r+2n-2\s+t}&=(-1)^{r-1}2\,\frac{(2n-2\s+t)!\,(r-1)!}{(r+2n-2\s+t)!},\\
\lim_{x\to -\s} \binom {x+n+\epsilon}{r+2n-2\s+1}&=0.
\end{align*}

\begin{lem}\label{lem:evaluationPfS} 
For integers $a,i,j$ with $a\ge-1$ and $i,j\ge1$, we have 
\begin{multline*}
\sum _{\ell=0} ^{i+a}(-1)^{\ell+j+1} \frac {j-i} {i+a+1}\binom
{j+a}{i+a-\ell}\binom {\ell+j+a+1}{\ell} \frac
{(a+2)!\,(\ell+j-1)!} {(\ell+j+a+2)!}\\
={{\left( -1 \right) }^{i + j }}\,{\frac {
   ( j-i ) \,(a+i)!\,(a+j)! }
   {(2a+i+j+2)!}}.
\end{multline*}
\end{lem}

The above result was proved in the particular case $a=2n-2\s-1$
in~\cite[p.~277]{CiuKrat-interaction}. 
Since the arguments in~\cite[p.~277]{CiuKrat-interaction} can be
used in the same way to prove Lemma~\ref{lem:evaluationPfS}, 
we omit the proof details. Combining the preceding lemma
with~\eqref{eq:ValSij}, we arrive at 
\begin{equation}\label{eq:ValSij-simplification}
S_{i,j}= {\left( -1 \right) }^{i + j }\,{\frac {2\,
   ( j-i ) \,(2n-2\s+i)!\,(2n-2\s+j)! }
   {(4n-4\s+i+j+2)!}}.
\end{equation}
Consequently, we have
\begin{equation*}
\underset {1\le i,j\le 2\s}\Pf(S_{i,j})=(-1)^{\s} 2^{\s}
\left(\prod_{i=1}^{2\s}(2n-2\s+i)! \right) \underset {1\le i,j\le
2\s}\Pf\left(\frac{j-i}{(4n-4\s+i+j+2)!}\right),
\end{equation*}
which simplifies to~\eqref{eq:evaluation_PfS}, as desired, by
the Pfaffian evaluation
\begin{equation}\label{eq:Mehta-Wang}
\underset {0\leq i,j\leq 2k-1}\Pf\left(\frac {(j-i )}
{(b+i+j)!}\right) =\prod_{i=0}^{k-1}
\frac{(2i+1)!}{(b+2k+2i-1)!},\quad k\geq 1.
\end{equation}
Note that the above equation is a slight variation of a Pfaffian
evaluation due to Mehta and Wang
(see~\cite[Cor.~10]{CiuKrat-interaction}). 
This completes the proof of~\eqref{eq:evaluation_PfS}. 






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%
%           PROOF OF THEOREM 2
%
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\newpage
\section{Proof of Theorem~\ref{thm:Ncentered(2n,2x+1)}}
\label{sec:4}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%         Plan de la preuve             %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Again, throughout this section, we assume that $n$ is a (fixed) positive
integer. 
From \eqref{eq:def N(a,b)}, it is easily checked that $ST(2n,2x+1,2n)$
can be written in the form
\begin{equation*}%\label{eq:N_2(2n+1,2x)}
ST(2n,2x+1,2n)=\frac{(x+1)_{2n}}{(\frac{1}{2})_{2n}}
        \prod_{s=1}^{n}\frac{(2x+1+2s)_{4n-4s+1}}{(2s)_{4n-4s+1}}.
\end{equation*}
This, combined with Lemma~\ref{lem:tilings-Pfaffian_casB}, leads to
the following reformulation of Theorem~\ref{thm:Ncentered(2n,2x+1)}:
\begin{equation}\label{eq:Theorem2_bis}
\Pf N(x)= \frac{2^{5n-1}}{n!\,(4n)!}\, (x+1)_{2n} \,
\prod_{s=2}^{n}\frac{(2x+2s)_{4n-4s+3}}{(2s)_{4n-4s+3}}\;U_{n}(x),
\end{equation}
where $N(x)$ is the matrix defined in
Lemma~\ref{lem:tilings-Pfaffian_casB} and 
$U_n(x)$ is defined by~\eqref{eq:def-Ux}.
Our proof of~\eqref{eq:Theorem2_bis} is, in spirit, quite similar to
the proof of~\eqref{eq:Theorem1_bis}. 

First, with exactly the same kind of reasoning as used in
Section~\ref{sec:PfM-1}, 
it is easily seen that $\Pf N(x)$ is a polynomial in $x$ of degree at
most $2n^2+n-1$. We omit the details. In Section~\ref{sec:PfN-2}, we
prove that 
\begin{equation}\label{eq:inv_shift-Pfaffian_N}
\Pf N(x)=(-1)^{n+1}\Pf N(-2n-1-x).
\end{equation}
In Sections~\ref{sec:PfN-3} and~\ref{sec:PfN-4}, we show that
$\prod_{s=1}^n(x+s)^{s}$ and 
$\prod_{s=1}^{n-1}(x+s+\frac{3}{2})^{s}$ divide $\Pf N(x)$ as a
polynomial in $x$. This, combined
with~\eqref{eq:inv_shift-Pfaffian_N}, implies that 
\begin{equation*}
\left(x+n+\frac{1}{2}\right)^{n-1}
\prod_{s=1}^{n-2}\left(x+s+\frac{3}{2}\right)^{s}
\left(x+2n+1-s-\frac{3}{2}\right)^{s}
\prod_{s=1}^n(x+s)^{s}(x+2n+1-s)^{s},
\end{equation*}
which is equal to
\begin{equation}
2^{-2n^2+3n-1}\,(x+1)_{2n}\,\prod_{s=2}^{n}\left(2x+2s\right)_{4n-4s+3}
\end{equation}
and is a polynomial of degree $2n^2-n+1$, divides $\Pf N(x)$ as a
polynomial in $x$. 
Altogether, we are led to the formula
\begin{equation}\label{eq:def-Tx}
\Pf N(x)=T(x)\,(x+1)_{2n}\,\prod_{s=2}^{n}\left(2x+2s\right)_{4n-4s+3},
\end{equation}
for some polynomial $T(x)$ in $x$ of degree at most
$2n-2$. Therefore, in order to prove~\eqref{eq:Theorem2_bis}, 
it remains to show that $T(x)$ is equal to the polynomial $L(x)$
defined by 
\begin{equation}\label{eq:def-Lx}
L(x)=\frac{2^{5n-1}}{n!\,(4n)!}\,\prod_{s=2}^{n}
\frac{1}{(2s)_{4n-4s+3}}\;U_{n}(x),
\end{equation}
with $U_n(x)$ given by~\eqref{eq:def-Ux}. Since $T(x)$ and $L(x)$ are
polynomials in $x$ of degree at most $2n-2$, 
it suffices to show that~$T(x)=L(x)$ holds for $2n-1$ distinct values
of $x$. In Section~\ref{sec:PfN-5}, 
we determine the evaluations of 
$T(x)$ and $L(x)$ at $x=-1,\dots,-n$, which shows that
$T(x)=L(x)$ at $x=-1,\ldots,-n$. 
Since, by~\eqref{eq:def-Tx} and~\eqref{eq:inv_shift-Pfaffian_N},
$T(x)=T(-2n-1-x)$ and, by~\eqref{eq:def-Lx}, $L(x)=L(-2n-1-x )$ for
any $x$, 
this shows at the same time that $T(x)=L(x)$ for
$x=-2n,-2n+1,\dots,-n-1$. Altogether, this yields $2n$ values of~$x$
for which $T(x)=L(x)$. This would complete the
proof of~\eqref{eq:Theorem2_bis}. 


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%         Step 1                  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsection{ $\Pf N(x)=(-1)^{n+1} \Pf N(-2n-1-x)$}\label{sec:PfN-2}
The proof is quite similar to the proof
of~\eqref{eq:inv_shift-Pfaffian} and requires only slight changes. 
Let $\Btil$ be the lower triangular matrix of size $2n+2$ defined by
\begin{equation}
\Btil=\left(
  \begin{array}{ccc|rr}
    & & & & \\
   &\left(B_{i,j}\right)_{\substack{1\leq i\leq 2n \\ 1\leq j\leq
 2n}} &&0& 0 \\ 
   &&&& \\\hline
   &0&&-1&0\\
   &0&&0&-1
  \end{array}
 \right),
 \quad\text{with $B_{i,j}=\binom{i-1}{j-1}$ for $1\leq i,j\leq 2n$,}
\end{equation}
and let $N^{(1)}$ be the skew-symmetric matrix (of size $2n+2$)
defined by 
$N^{(1)}=\Btil\, N(x)\,{\Btil}^t$.

We claim that the $(i,j)$-entry in $N^{(1)}$ is, up to the sign
$(-1)^{i+j}$, the $(i,j)$-entry in $N(-2n-1-x)$.
First, from Remark~\ref{rmk:Mn vs Nn}, the definition of $N^{(1)}$ and
what we have proved in Section~\ref{sec:PfM-2}, we have
\begin{equation*}
N^{(1)}_{i,j}=M^{(1)}_{i,j}=(-1)^{i+j}M_{i,j}(-2n-1-x)
=(-1)^{i+j}N_{i,j}(-2n-1-x)\quad\text{if
$i,j\neq 2n+1$},
\end{equation*}
where $M^{(1)}$ is the matrix defined in
Section~\ref{sec:PfM-2}. Furthermore, by definition of the 
matrices $N^{(1)}$ and $\Btil$ and~\eqref{eq:Qij_B_lastcolumn}, 
for $1\leq i\leq 2n$ we have
\begin{align*}
N^{(1)}_{i,2n+1}&=-\sum_{a = 1}^{i} {\binom {i-1 }{a-1}} N_{a,2n+2}(x)\\
&=\sum_{a = 1}^{i} {\binom {i-1 }{a-1}} \left( \binom {x+n}{a}+ \binom
{x+n+1}{a}-\binom {2x+2n+1}{a} \right),
\end{align*}
which simplifies, by Chu--Vandermonde summation (see
e.g.~\eqref{eq:Chu-Van}) and~\eqref{eq:Qij_B_lastcolumn}, to 
\begin{align*}
N^{(1)}_{i,2n+1}&=(-1)^i\left(\binom {-x-n}{i}+
\binom{-x-n-1}{i}-\binom {-2x-2n-1}{i}\right)\\
&=(-1)^{i+1}N_{i,2n+1}(-2n-1-x).
\end{align*}

Summarizing, we have shown that
$N^{(1)}_{i,j}=(-1)^{i+j}N_{i,j}(-2n-1-x)$ for all $i,j$. 
With exactly the same reasoning used in Section~\ref{sec:PfM-2}, this
leads to~\eqref{eq:inv_shift-Pfaffian_N}. 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%         Step 2                  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{ $\prod_{s=1}^n(x+s)^{s}$ divides $\Pf
N(x)$}\label{sec:PfN-3} 
Let $s$ be an integer with $1\leq s\leq n$. We claim that,
\begin{equation}\label{eq:comblineaire-s_N}
\textrm{for $2n-2s+2\leq a\leq 2n$, the $a$-th row of the
matrix $N(-s)$ is zero,}
\end{equation}
and that
\begin{equation}\label{eq:comblineaire-s_2_N}
\sum_{i=1}^{2n-2s+1}(-1)^{i-1}\left(2^{i-1}-\binom{i-1}{n-s}\right)
\cdot\left( \textrm{row $i$ of $N(-s)$} \right) -\left( \textrm{row
$(2n+1)$ of $N(-s)$ } \right)=0.
\end{equation}
As these are $2s$ linear combinations of the rows of the matrix
$N(x)$ that are linearly independent and vanish at $x=-s$,
this implies divisibility of $\Pf N(x)$ by~$(x+s)^{s}$.

\begin{proof}[Proof of~\eqref{eq:comblineaire-s_N}]
Suppose
$2n-2s+2\leq a\leq 2n$. It follows from Remark~\ref{rmk:Mn vs Nn}
and~\eqref{eq:comblineaire-s} that $N_{a,j}(-s)=0$ if $1\leq j\leq
2n+2$ and $j\neq 2n+1$. Moreover, by~\eqref{eq:Qij_B_lastcolumn}, we
have
$$
N_{a,2n+1}(-s)=\binom {2n-2s+1}{a}-\binom {n-s}{a}-\binom
{n-s+1}{a},
$$
which is clearly zero since $a> 2n-2s+1\geq n-s+1$. 
\end{proof}

\begin{proof}[Proof of~\eqref{eq:comblineaire-s_2_N}]
We have to
show that, for $1\leq j\leq 2n+2$, there holds
\begin{equation}\label{eq:comblineaire-s_2_N-bis}
\sum_{i=1}^{2n-2s+1}(-2)^{i-1}N_{i,j}(-s)
-\sum_{i=1}^{2n-2s+1}(-1)^{i-1}\binom{i-1}{n-s}N_{i,j}(-s)-N_{2n+1,j}(-s)=0.
\end{equation}

\medskip
(a) Suppose $j=2n+2$. By~\eqref{eq:def-Nij-Lr}, we have
$N_{i,2n+2}(-s)=\binom{n-s}{i-1}$ if $i\leq n-s+1$, and it equals
$0$ otherwise.
Therefore, when $j=2n+2$, the left-hand side
of~\eqref{eq:comblineaire-s_2_N-bis} simplifies to
\begin{equation*}
\sum_{i=1}^{n-s+1}
(-2)^{i-1}\binom{n-s}{i-1}-\sum_{i=1}^{n-s+1}(-1)^{i-1}
\binom{i-1}{n-s}\binom{n-s}{i-1}=(-1)^{n-s}-(-1)^{n-s}=0,
\end{equation*}
as desired. That the first (respectively second) sum in the above relation
is equal to $(-1)^{n-s}$ is immediate from the binomial theorem
(respectively from the fact that the term with index $i=n-s+1$ is the only
term which is nonzero).

\medskip
(b) Suppose $j=2n+1$. Using the expression~\eqref{eq:Qij_B_lastcolumn}
for the matrix entry 
$N_{i,2n+1}(-s)$, and noticing that $N_{2n+1,2n+1}(-s)=0$, we see
that~\eqref{eq:comblineaire-s_2_N-bis} reduces to
\begin{equation*}
\sum_{i=1}^{2n-2s+1}(-1)^{i-1}\left(2^{i-1}
-\binom{i-1}{n-s}\right)\left(\binom{2n-2s+1}{i}-{\binom {n-s}{
i}}-{\binom {n-s+1}{ i}}\right) =0,
\end{equation*}
which is immediate from the summations
\begin{align}
&\sum_{i=1}^{2n-2s+1} (-2)^{i-1} {\binom {2n-2s+1}{ i}}
=\sum_{i=1}^{2n-2s+1} (-2)^{i-1}\left({\binom {n-s}{ i}}+{\binom {n-s+1}{
i}}\right) =1,\label{eq:comb-Sa}\\
 &\sum_{i=1}^{2n-2s+1} (-1)^{i-1}\binom{i-1}{n-s}
\left({\binom {n-s}{ i}}+{\binom {n-s+1}{ i}}\right)
=(-1)^{n-s},\label{eq:comb-Sb}\\
&\sum_{i=1}^{2n-2s+1} (-1)^{i-1}\binom{i-1}{n-s} {\binom {2n-2s+1}{ i}}
 =(-1)^{n-s}.\label{eq:comb-Sc}
\end{align}
Indeed, the two sums in~\eqref{eq:comb-Sa} are easily evaluated by the
binomial theorem. 
The sum in~\eqref{eq:comb-Sb} has only one term which is nonzero, the
term with index $i=n-s+1$, which is equal to 
$(-1)^{n-s}$, whence~\eqref{eq:comb-Sb}. In the sum
in~\eqref{eq:comb-Sc}, the terms with index~$i$ between~$1$ and 
$n-s$ vanish. By shifting the summation index $i$ by $n-s+1$
(that is, we replace $i$ by $i+n-s+1$) and then using the relation 
$(-1)^{i}\binom{i+n-s}{n-s}=\binom{-n+s-1}{i}$, this sum becomes
\begin{equation*}
(-1)^{n-s}\sum_{i=0}^{n-s} \binom{-n+s-1}{i}\binom {2n-2s+1}{n-s-i},
\end{equation*}
which, by the Chu--Vandermonde summation formula, is equal to
$(-1)^{n-s}\binom{n-s}{n-s}=(-1)^{n-s}$. 

\medskip
(c) Suppose $2n-2s+2\leq j\leq 2n$. From~\eqref{eq:comblineaire-s_N}
and the skew-symmetry of $N(-s)$,
the $j$-th column of $N(-s)$ is zero, and thus the
relation~\eqref{eq:comblineaire-s_2_N-bis} is clearly true. 

\medskip
(d) Suppose $1\leq j\leq 2n-2s+1$. Using the
expressions~\eqref{eq:def-Nij} and~\eqref{eq:Qij_B_lastcolumn} 
for the corresponding matrix entries, we see
that~\eqref{eq:comblineaire-s_2_N-bis} reduces to 
\begin{multline}
\label{eq:reduction_comblineaire-s_2_N/j<=2n}
\sum_{i=1}^{2n-2s+1}
 (-1)^{i-1}\left(2^{i-1}-\binom{i-1}{n-s}\right)
\left(R_{i,j}(-s)+T_{i,j}(-s)-T_{j,i}(-s)\right)\\ 
+{\binom {2n-2s+1}{ j}}-{\binom {n-s}{ j}}-{\binom {n-s+1}{ j}} =0,
\end{multline}
where $R_{i,j}(x)$ and $T_{i,j}(x)$ are defined as
in~\eqref{eq:def-Rij} and~\eqref{eq:def-Tij}. 

From the expression~\eqref{eq:def-Tij} for $T_{i,j}(x)$
and~\eqref{eq:comb-Sa}--\eqref{eq:comb-Sc}, we have 
\begin{multline}
\label{eq:comb-Sd}
\sum_{i=1}^{2n-2s+1}
(-1)^{i-1}\left(2^{i-1}-\binom{i-1}{n-s}\right)
\left(T_{i,j}(-s)-T_{j,i}(-s)\right)\\
=\left(1-(-1)^{n-s}\right)\bigg{(} {\binom {n-s}{
j}}+{\binom {n-s+1}{ j}} -{\binom {2n-2s+1}{ j}} \bigg{)}.
\end{multline}
From~\eqref{eq:sumbinom}, we have
\begin{equation}\label{eq:comb-Se}
\sum_{i=1}^{2n-2s+1} (-2)^{i-1}R_{i,j}(-s)=\sum_{i=1}^{2n-2s+1}
(-2)^{i-1} \sum_{t=1}^{2n-2s+1}\frac{j-i}{t}{\binom {t}{ j}}{\binom {t}{
i}}=0.
\end{equation}
From the expression~\eqref{eq:def-Rij} for $R_{i,j}(x)$, we have
\begin{align*}
&\sum_{i=1}^{2n-2s+1}(-1)^{i-1} \binom{i-1}{n-s}R_{i,j}(-s)\\
&=\sum_{i=1}^{2n-2s+1}(-1)^{i-1}\binom{i-1}{n-s}
\sum_{t=1}^{2n-2s+1}\binom{t-1}{j-1}\binom{t}{i}
-\binom{t}{j}\binom{t-1}{i-1}\\
&=\sum_{t=1}^{2n-2s+1}{\binom {t-1}{ j-1}}
\sum_{i=1}^{t}(-1)^{i-1} \binom{t} {i} \binom{i-1}{n-s}
-\sum_{t=1}^{2n-2s+1}\binom{t}{j}\sum_{i=1}^{t} (-1)^{i-1}
\binom{t-1}{i-1}\binom{i-1}{n-s},
\end{align*}
where the last equality follows by interchanging the sums over $i$
and $t$. 
In the latter expression, the first inner sum is zero if $t\leq n-s$
(because of the binomial coefficient 
$\binom{i-1}{n-s}$), and it is equal to $(-1)^{n-s}$ if $t>n-s$
(see~\eqref{eq:comb-Sc} and its proof). The second inner sum is 
$(-1)^{n-s}$ if $t=n-s+1$, and it equals $0$ otherwise, 
as can be checked by using the relation 
$$\binom{t-1}{i-1}\binom{i-1}{n-s}=\binom{t-1}{n-s}
\binom{t-1-(n-s)}{i-1-(n-s)}$$ 
and the binomial theorem.

Altogether, this implies that
\begin{multline*}
\sum_{i=1}^{2n-2s+1}(-1)^{i-1}
\binom{i-1}{n-s}R_{i,j}(-s)=(-1)^{n-s}\sum_{t=n-s+1}^{2n-2s+1}{\binom {t-1}{
j-1}}-(-1)^{n-s} \binom{n-s+1}{j}\\
=(-1)^{n-s}\left({\binom {2n-2s+1}{ j}}-{\binom {n-s}{ j}}
-\binom{n-s+1}{j}\right),
\end{multline*}
where the last equality follows from~\eqref{eq:BinomialSum}. This,
combined with~\eqref{eq:comb-Sd} and ~\eqref{eq:comb-Se}, leads
to~\eqref{eq:reduction_comblineaire-s_2_N/j<=2n}. This completes the
proof of~\eqref{eq:comblineaire-s_2_N-bis}.
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%         Step 3                   %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsection{ $\prod_{s=1}^{n-1}(x+s+3/2)^{s}$ divides $\Pf
N(x)$}\label{sec:PfN-4} 
Let $s$ be an integer with $1\leq s\leq n-1$. We claim that, for
$a=1,2,\ldots,2s$, we have
\begin{multline}
\label{eq:comblineaire-s-3/2/N}
(-1)^{a}{\binom {n+s-\frac{1}{2}}{ a+1}}\cdot\left( \textrm{row
$(2n-1-a)$ of $N(-s-3/2)$ } \right)\\ 
-\left(n+s-\frac{1}{2}\right) {\binom {2n-1}{ a}}\cdot
\left(\textrm{row $2n-1$ of $N(-s-3/2)$ } \right)\\ 
-a {\binom {2n}{ a+1}}\cdot\left( \textrm{row $(2n)$ of
$N(-s-3/2)$ } \right)=0. 
\end{multline}
Since these are $2s$ linear combinations of the rows of the matrix
$N(x)$ that are linearly independent and vanish at $x=-s-3/2$, this
implies divisibility of $\Pf N(x)$ by~$(x+s+\frac{3}{2})^{s}$.


\begin{proof}[Proof of~\eqref{eq:comblineaire-s-3/2/N}]
Suppose
$1\leq a\leq 2s$. We have to show that, for $1\leq j\leq 2n+2$,
\begin{multline}
\label{eq:Comb-s-3/2-bis}
(-1)^{a}{\binom {n+s-\frac{1}{2}}{ a+1}}N_{2n-1-a,j}(-s-3/2)\\
-\left(n+s-\frac{1}{2}\right) {\binom {2n-1}{ a}}N_{2n-1,j}(-s-3/2)
-a {\binom {2n}{ a+1}}N_{2n,j}(-s-3/2)=0.
\end{multline}


(a) Suppose $2n+1\leq j \leq 2n+2$. It follows
from~\eqref{eq:Qij_B_lastcolumn} that 
$$N_{i,2n+1}(-s-3/2)=-\binom{n-s-3/2}{i}-\binom{n-s-1/2}{i}\quad\text{if
$2n-2s-1\leq i\leq 2n$.}$$ 
Furthermore, by~\eqref{eq:def-Nij-Lr}, we have
$N_{i,2n+2}(-s-3/2)=\binom{n-s-3/2}{i-1}$. 
Therefore, equation~\eqref{eq:Comb-s-3/2-bis} when 
specialized to $j=2n+1$ and $j=2n+2$ reduces to the identities 
\begin{align}
\begin{split}\label{eq:comblineaire-s-3/2/N/j=2n+1}
&(-1)^{a+1}{\binom {n+s-\frac{1}{2}}{ a+1}}
\left(\binom{n-s-3/2}{2n-1-a}+\binom{n-s-1/2}{2n-1-a}\right)\\
&\qquad+\left(n+s-\frac{1}{2}\right) {\binom {2n-1}{ a}}
\left(\binom{n-s-3/2}{2n-1}+\binom{n-s-1/2}{2n-1}\right)\\
&\qquad\qquad+a{\binom {2n}{ a+1}}
\left(\binom{n-s-3/2}{2n}+\binom{n-s-1/2}{2n}\right)=0,
\end{split}\\[0.5cm]
\begin{split}\label{eq:comblineaire-s-3/2/N/j=2n+2}
 &(-1)^{a}{\binom {n+s-\frac{1}{2}}{ a+1}}
{\binom {n-s-\frac{3}{2}}{ 2n-a-2}}\\
 &\qquad-\left(n+s-\frac{1}{2}\right) {\binom {2n-1}{ a}}
{\binom {n-s-\frac{3}{2}}{ 2n-2}}-a{\binom {2n}{ a+1}}
{\binom {n-s-\frac{3}{2}}{ 2n-1}} =0.
\end{split}
\end{align}
The proof of these relations amounts to a routine verification, 
and is consequently left to the reader. 

\medskip
(b) Suppose $2n-2s-1\leq j\leq 2n$. Combining Remark~\ref{rmk:Mn vs
Nn} with~\eqref{eq:Qij-s-1/2_annulation} with the skew-symmetry of 
$N(-s-3/2)$,
we see that the $j$-th column of $N(-s-3/2)$ is zero, and
thus~\eqref{eq:Comb-s-3/2-bis} is clearly true. 

\medskip
(c) Suppose $1\leq j\leq 2n-2s-2$. It follows from Remark~\ref{rmk:Mn
vs Nn} and~\eqref{eq:Qij-s-1/2_simplification_i>} that 
$$
N_{i,j}(-s-3/2)=-{\binom {2n-2s-2}{ j}}\left({\binom
{n-s-\frac{3}{2}}{ i}} 
+{\binom {n-s-\frac{1}{2}} i} \right)
$$
for $2n-2s-1\leq i \leq 2n$. Inserting~this
in~\eqref{eq:Comb-s-3/2-bis} and then 
dividing both sides of the resulted equality by ${\binom {2n-2s-2}{
j}}$ gives 
the relation~\eqref{eq:comblineaire-s-3/2/N/j=2n+1}, as
a routine verification shows. 
This completes the proof of~\eqref{eq:Comb-s-3/2-bis}.
\end{proof}






%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%         Step 4                   %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\newpage
\subsection{ $T(x)=L(x)$ at $x=-1,\ldots,-n$}\label{sec:PfN-5}

Let $\s$ be a given integer with $1\leq \s \leq n$. It is easy to
evaluate the polynomial $L(x)$, defined by~\eqref{eq:def-Lx}, at~$x=-\s$. After a routine
calculation, we obtain
\begin{equation}\label{eq:Ln at -s}
L(-\s)=\frac{2^{5n-1}}{n!\,(4n)!}\,
\prod_{s=2}^{n}\frac{1}{(2s)_{4n-4s+3}}\left(\frac{3}{2}-\s\right)_{2n-1}
\big( (-1)^{\s-1}(2n-1)!!+(2n)!! \big).
\end{equation}

The evaluation of $T(x)$ at $x=-\s$ is much more delicate. For the
same reason as the one in Section~\ref{sec:PfM-5}, we should
write~\eqref{eq:def-Tx} in the form
\begin{multline*}
T(x)=2^{2n-\sigma}\frac{1}{(x+\sigma)^{\sigma}} \Pf N(x)
\;\underset { s\neq \sigma}
{\prod_{s=1}^n}(2x+2s)^{-s}\;\prod_{s=1}^n(2x+4n+2-2s)^{-s}\\
\times\prod_{s=1}^{n-1}\left(2x+3+2s\right)^{-s}\;
\prod_{s=1}^{n-2}\left(2x+4n-1-2s\right)^{-s},
\end{multline*}
and subsequently specialize $x=-\sigma$. After some manipulation,
this leads to
\begin{multline}
\label{eq:Tn at -s}
T(-\sigma)=\left(\frac{1}{(x+\sigma)^{\sigma}} \Pf
N(x)\right)\bigg\vert_{x=-\sigma} 
\;(-1)^{\s-1}2^{3-\s}\frac{(2\s-3)!\,(4n-2\s-1)!}{(\s-2)!\,(2n-\s-1)!}\\
\times\frac{\prod_{s=1}^{n-\s}(2s-1)!}
{\prod_{s=1}^{\s-1}(2s)!\;\prod_{s=n-\s+1}^{2n-\s}(2s)!}.
\end{multline}

Let $B(x)=\big(B_{i,j}(x)\big)_{1\leq i,j\leq 2n+2}$ denote the
matrix obtained from $N(x)$ by first adding 
$$\sum_{i=1}^{2n-2\s+1}(-1)^{i}\left(2^{i-1}
-\binom{i-1}{n-\s}\right)\cdot\left( \textrm{row $i$ of $N(x)$} \right)$$
to row $2n+1$, and then adding
$$\sum_{i=1}^{2n-2\s+1}(-1)^{i}\left(2^{i-1}
-\binom{i-1}{n-\s}\right)\cdot\left( \textrm{column $i$ of $N(x)$}
\right)$$ 
to column $2n+1$.
Of course, we have $\Pf B(x)=\Pf N(x)$, and it follows
from~\eqref{eq:comblineaire-s_N} and~\eqref{eq:comblineaire-s_2_N}
that 
the $i$-th row of $B(-\sigma)$, $2n-2 \sigma+2 \leq i \leq 2n+1$, is
zero, or equivalently, $(x+\s)$ is a factor
of each entry in the $i$-th row in matrix $B(x)$. Applying
Lemma~\ref{lem:FactorizationPfaffian} to the matrix $B(x)$, we obtain 
\begin{equation}\label{eq:factorization_Pfaffian_N}
\left(\frac {1} {(x+\s)^{\s}}\Pf N(x)\right)\bigg\vert_{x=-\s}
=\left(\frac {1} {(x+\s)^{\s}}\Pf B(x)\right)\bigg\vert_{x=-\s}
=\Pf \widetilde B\,\cdot\,\Pf D,
\end{equation}
where $\widetilde B$ is the matrix which arises from $B(x)$ by
deleting rows and columns $2n-2\s+2,2n-2\s+3,\ldots,2n+1$ and 
subsequently specializing $x=-\s$, and where
\begin{equation}\label{eq:def-D}
D=\left( \left(\frac {1}
{x+\s}B_{i,j}(x)\right)\bigg\vert_{x=-\s}\right) _{2n-2\s+2\le
i,j\le 2n+1}.
\end{equation}
By Remark~\ref{rmk:Mn vs Nn} and the definition of the matrix
$B(x)$, we have $\widetilde B=\widetilde M$ where $\widetilde M$ is 
defined by~\eqref{eq:def-Mtil}. We have shown in
Section~\ref{sec:PfM-5} that $\Pf \widetilde M=1$, and thus $\Pf
\widetilde B =1$. 
The evaluation of $\Pf D$ is much more complicated and is therefore 
postponed to the next section for the sake of better readability of the
paper. 

\begin{lem}\label{lem:Pf(S)_N}
We have
\begin{multline*}
\Pf D= (-1)^{\s-1}
2^{5n-2}\frac{(\s-1)!\,(n-\s)!\,(2n-\s)!}{n!^2}\,
\big((2n)!!+(-1)^{\s+1}(2n-1)!!\big) \\
\times\left(\frac{3}{2}-\s\right)_{2n-1}\,
\frac{\prod_{s=1}^{\s-2}(2s)! \big(\prod _{s=n-\s+1}
^{n}(2s)!\big)^2}{\prod_{s=2n-\s}^{2n}(2s)!}.
\end{multline*}
\end{lem}

If we substitute the values obtained for $\Pf \widetilde B$ and $\Pf D$
in~\eqref{eq:factorization_Pfaffian_N}, and then insert the 
obtained result in~\eqref{eq:Tn at -s}, it is easy to check that
$T(-\s)$ is equal to the right-hand side of~\eqref{eq:Ln at -s}. 
Consequently, we have $T(-\s)=L(-\s)$, as desired.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%             Evaluation of Pf D
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\newpage
\section{Evaluation of $\Pf D$: Proof of Lemma~\ref{lem:Pf(S)_N}}
\label{sec:5}

This section is dedicated to the evaluation of the Pfaffian of the
matrix $D$ defined by~\eqref{eq:def-D}. We begin by describing more
explicitly the entries of the matrix $D$ in the next subsection. 
The subsequent subsection provides
an efficient way to compute the Pfaffian of a skew-symmetric matrix
which differs from a Mehta--Wang matrix of even size only in its last
row and column. A more precise statement is given in
Proposition~\ref{prop:PerturbedMehtaWang}. This allows us to write
$\Pf D$ in the form of a multisum, which is evaluated in the last
subsection.

Throughout this section, $\s$ and $n$ are positive integers with
$1\leq \s \leq n$. For the sake of simplicity, 
we also set $N=n-\s$.

\subsection{The entries of the matrix $D$}
\begin{lem}\label{lem:matrixD}
The matrix $D=(D_{i,j})_{1\leq i,j\leq 2\s}$ defined
by~\eqref{eq:def-D} is a skew-symmetric matrix of size $2\s$. 
For $1\leq i,j\leq 2\s-1$, we have
\begin{equation}\label{eq:Dij}
D_{i,j} ={{\left( -1 \right) }^{i + j }}\,{\frac {2\, ( j-i )
\,(2N+i)!\,(2N+j)! }
      {(4N+i+j+2)!}}.
\end{equation}
For $1\leq j\leq 2\s-1$, we have
\begin{multline}\label{eq:Dij-lastrow}
D_{2\s,j}=
\frac {(-1)^{j}2} {\binom {2N+j+1}{j-1}}
\sum _{i=0} ^{2N}(-1)^i\left(2^i-\binom i{N}\right)\frac {2N+j-i}
{i+1}\\ 
\cdot\sum _{\ell=0} ^{i}(-1)^\ell \binom {2N+j}{i-\ell}\binom
{\ell+j-1}\ell\frac {1} {2N+\ell+j+2}\\ 
+\frac {(-1)^{N+j}} {(N+1)\binom
{2N+j+1}{j-1}}+\frac{(-1)^{j}(j-1)N!\,(N+j-1)!}{(2N+j+1)!}.
\end{multline}
\end{lem}

\begin{proof}
Combining the definitions of the matrices $D$
and $B(x)$ (see~\eqref{eq:def-D} and the paragraph above) with 
Remark~\ref{rmk:Mn vs Nn}, we see that, for $1\leq i,j\leq 2\s-1$, the
$(i,j)$-entry of $D$ is equal
to the $(i,j)$-entry of the matrix $S$ defined in
\eqref{eq:def-MatS}. Equation~\eqref{eq:Dij} then follows
from~\eqref{eq:ValSij-simplification}.

Recall that $N=n-\s$ and suppose $1\leq j\leq 2\s-1$. Then,
by~\eqref{eq:def-D} and the definition of the matrix $B(x)$, 
we have
\begin{equation}\label{eq:Dij-lastrow-pf}
D_{2\s,j}=\left(\frac {N_{2n+1,2N+1+j}(x)} {x+\s}
     +\sum_{i=1}^{2N+1}
        (-1)^{i}\left(2^{i-1}-\binom{i-1}{N}\right)
         \frac {N_{i,2N+1+j}(x)} {x+\s}\right)\bigg\vert_{x=-\s}.
\end{equation}

It is a routine matter to derive from~\eqref{eq:Qij_B_lastcolumn} that
\begin{align*}
\frac{N_{2n+1,2N+1+j}(x)}{x+\s}\bigg\vert_{x=-\s}
&=-\frac{\binom{2x+2N+2\s+1}{2N+1+j}}{x+\s}
+\frac{\binom{x+N+\s}{2N+1+j}}{x+\s}
+\frac{\binom{x+N+\s+1}{2N+1+j}}{x+\s}\bigg\vert_{x=-\s}\\
&=(-1)^{j}2\frac{(2N+1)!\,(j-1)!}{(2N+j+1)!}+(-1)^{N+j}
\frac{(N+j)!\,N!}{(2N+j+1)!}\\
&\quad +(-1)^{N+j-1}\frac{(N+1)!\,(N+j-1)!}{(2N+j+1)!}.
\end{align*}
Similarly, combining~\eqref{eq:def-Nij} with~\eqref{eq:def-Tij}
and~\eqref{eq:def-Rij-bis}, for $1\leq i \leq 2N+1$ we obtain 
\begin{align*}
&\frac{N_{i,2N+1+j}(x)}{x+\s}\bigg\vert_{x=-\s}\\
&=\sum_{\ell=0}^{i-1}\frac{2N+1+j-i}{i}{\binom {2N+j}{
i-1-\ell}}
{\binom {\ell+2N+1+j}{ \ell}}(-1)^{\ell+j-1}
2\frac{(2N+2)!\,(\ell+j-1)!}{(2N+j+2+\ell)!}\\
&\quad+{\binom {2N+1}{ i}}
\bigg((-1)^{N+j}\frac{(N+j)!\,N!}{(2N+j+1)!}
+(-1)^{N+j-1}\frac{(N+1)!\,(N+j-1)!}{(2N+j+1)!}
\bigg)\nonumber\\
&\quad-(-1)^{j-1}2\frac{(2N+1)!\,(j-1)!}{(2N+j+1)!}
\left({\binom {N}{ i}}+{\binom {N+1} i}\right)\nonumber
\end{align*}
after a straightforward calculation.
Substituting the last two identities in~\eqref{eq:Dij-lastrow-pf} and
then using 
~\eqref{eq:comb-Sa}--\eqref{eq:comb-Sc}, we obtain 
\eqref{eq:Dij-lastrow} after some simplification.
\end{proof}


\subsection{The Pfaffian of a perturbed Mehta--Wang matrix}

\begin{prop}\label{prop:PerturbedMehtaWang}
Let $s,R$ be positive integers and $A=\left(a_{i,j}\right)_{1\leq
i,j\leq 2s}$ be a skew-symmetric matrix such that, for $1\leq
i,j\leq 2 s-1$,
\begin{equation}\label{eq:defAij}
a_{i,j} =\frac {( j-i )} {(R+i+j)!}.
\end{equation}
Then the Pfaffian of $A$ satisfies the relation
\begin{equation}
\Pf A= -2^{s-1}(s-1)!\left(\sum_{j=1}^{2s-1} a_{2s,j}\cdot
\lambda_{j}\right) \prod_{i=0}^{s-2}
\frac{(2i+1)!}{(R+2s+1+2i)!},
\end{equation}
where, for $1\leq j\leq 2s-1$, the term $\lambda_{j}$ is defined by
\begin{equation}\label{eq:defLambda}
\l_{j}=(-1)^{j+1}\sum_{k=0}^{s-1}
\frac{2^{-k}}{k!}\binom{2k}{j+k-s}(R+2s)_{j+k-s}.
\end{equation}
\end{prop}

\begin{proof} Let $B=\left(B_{i,j}\right)$ be the upper triangular
matrix of size $2s$ defined by 
$$
B=\left(
\begin{array}{c|c}
 \l_1 & \l_2 \cdots \l_{2s-1}\,0 \\ \hline
 0 & \raisebox{-15pt}{{\large\mbox{{$I_{2s-1}$}}}} \\[-4ex]
 \vdots & \\[-0.5ex]
 0 &
\end{array}
\right),
$$
where $I_{2s-1}$ stands for the identity matrix of size $2s-1$.
Moreover, set $\widetilde A=B^t A B$. 
Clearly, $\widetilde A=(\widetilde a_{i,j})_{1\le i,j\le 2s}$ 
is a skew-matrix which differs from
$A$ only in its first column and its first row, and 
we have the relation
\begin{equation}\label{eq:FirstColumn}
\widetilde a_{i,1}=\sum_{j=1}^{2s-1} a_{i,j}\cdot \l_j.
\end{equation}
We claim that $\widetilde a_{i,1}=0$ for $1\leq i\leq 2s-1$, so that
$$
\widetilde{A}= \left(\begin{array}{c|ccccc} 0&0&\cdots &0&-\widetilde
a_{2s,1}\\\hline
0&a_{2,2}&\cdots &a_{2,2s-1}&a_{2,2s}\\
\vdots &\vdots & \ddots &\vdots &\vdots\\
0&a_{2s-1,2}&\cdots &a_{2s-1,2s-1}&a_{2s-1,2s}\\
\widetilde a_{2s,1}&a_{2s,2}&\cdots &a_{2s,2s-1}&a_{2s,2s}
\end{array}\right).
$$
This, combined with the relation $\widetilde A=B^t A B$, would imply
that 
$$
\Pf \widetilde A=-\widetilde a_{2s,1}\,\underset{2\leq i,j\leq
2s-1}\Pf\left(a_{i,j}\right) 
\quad\text{and}\quad
\Pf A=\Pf \widetilde A/\det B=2^{s-1}(s-1)!\Pf \widetilde A.$$
Consequently, we would have
$$\Pf A=-2^{s-1}(s-1)!\,\widetilde a_{2s,1}\,
\underset{2\leq i,j\leq 2s-1}\Pf\left(\frac {j-i}
{(R+i+j)!}\right).
$$
Proposition~\ref{prop:PerturbedMehtaWang} then would immediately
follow 
from~\eqref{eq:FirstColumn} and the Pfaffian evaluation
\eqref{eq:Mehta-Wang}. 

So, to complete the proof,
it remains to verify our claim that $\widetilde a_{i,1}=0$ for $1\leq i\leq
2s-1$. By~\eqref{eq:FirstColumn} and
\eqref{eq:defAij}--\eqref{eq:defLambda}, for $1\leq i\leq 
2s-1$ we have
\begin{align}
\widetilde a_{i,1}&=\sum_{j=1}^{2s-1}\frac {(j-i)} {(R+i+j)!}(-1)^{j+1}
\sum_{k=0}^{s-1} \frac{2^{-k}}{k!}\binom{2k}{j+k-s}(R+2s)_{j+k-s}
\label{eq:comblineaireA_firstRows}\\
      &=\sum_{k=0}^{s-1}\sum_{j=0}^{2k} \frac{2^{-k}}{k!}
\frac {(-1)^{j+s-k} (j+s-k-i)} {(R+i+j+s-k)!}
\binom{2k}{j}(R+2s)_{j},\label{eq:comblineaireA_firstRows-bis}
\end{align}
where~\eqref{eq:comblineaireA_firstRows-bis} follows
from~\eqref{eq:comblineaireA_firstRows} by first 
interchanging the sums over $j$ and $k$ and then shifting the (now)
inner sum over $j$ by $s-k$. The inner sum 
in~\eqref{eq:comblineaireA_firstRows-bis} can be rewritten, by
splitting the term $(j+s-k-i)$, as 
a sum of two ${}_2F_1$-series. After some manipulation, we obtain
\begin{align*}
\widetilde a_{i,1}
&=\sum_{k=0}^{s-1}\frac{2^{-k}}{k!}\frac {(-1)^{s-k+1} (2k)(R+2s)}
{(R+i+s-k+1)!} \, {}_2F_1\!\left[\begin{matrix} {R+2s+1,-2k+1}\\
{R+i+s-k+2}\end{matrix};1\right]\\
&\qquad+\sum_{k=0}^{s-1}\frac{2^{-k}}{k!}\frac
{(-1)^{s-k} (s-k-i)} {(R+i+s-k)!} \,{}_2F_1
\!\left[\begin{matrix}{R+2s,-2k}\\
{R+i+s-k+1}\end{matrix};1\right]\\
&=\sum_{k=1}^{s-1}\frac{2^{-k+1}}{k!}
\frac {(-1)^{s-k+1} k(R+2s)(i-s-k+1)_{2k-1}} {(R+i+s+k)!}\\
&\qquad+\sum_{k=0}^{s-1}\frac{2^{-k}}{k!}\frac {(-1)^{s-k}
(s-k-i)(i-s-k+1)_{2k}} {(R+i+s+k)!},
\end{align*}
where the last equality follows from the Chu--Vandermonde
summation formula. It is easily checked that each summand in the two
above sums 
vanishes when $i=s$, whence~$\widetilde a_{s,1}=0$. 

Now suppose $i\neq s$.
Splitting the term $(s-k-i)$ in the second sum in the above
expression, and rewriting the (now) three sums in hypergeometric
notation, we arrive at
\begin{multline*}
\widetilde a_{i,1}=\frac {(-1)^{s}(R+2s)(i-s)} {(R+i+s+1)!}\,
{}_2F_1\!\left[\begin{matrix}{1+i-s,1+s-i}\\
{R+i+s+2}\end{matrix};\frac{1}{2}\right]\\
+\frac{(-1)^{s}2^{-1}(i-s)_2} {(R+i+s+1)!}\,
{}_2F_1\!\left[\begin{matrix}{2+i-s,1+s-i}\\
{R+i+s+2}\end{matrix};\frac{1}{2}\right]\\
+\frac {(-1)^{s} (s-i)} {(R+i+s)!}\, 
{}_2F_1\!\left[\begin{matrix}{1+i-s,s-i}\\
{R+i+s+1}\end{matrix};\frac{1}{2}\right].
\end{multline*}
To see that the above expression vanishes, it suffices (after
multiplying the above expression by $(-1)^{s}(i-s)^{-1} (R+i+s+1)!$)
to prove that
\begin{equation*}
(c-a)\;
{}_2F_1\!\left[\begin{matrix}{a,b+1}\\{c+1}\end{matrix};x\right]+(1-x)a\;
{}_2F_1\!\left[\begin{matrix}{a+1,b+1}\\{c+1}\end{matrix};x\right] 
-c \; {}_2F_1\!\left[\begin{matrix}{a,b}\\{c}\end{matrix};x\right]=0,
\end{equation*}
where $a=1+i-s$, $b=s-i$, $c=R+i+s+1$ and $x=1/2$. This identity can
be easily derived from Gau{\ss}' contiguous relations for the 
${}_2F_1$-series, or, more directly, 
by extracting the coefficient of $x^n$ in
each side. In our case, we have to check that
\begin{equation*}
(c-a)\frac{(a)_n (b+1)_n}{(c+1)_n(1)_n}+a\left(\frac{(a+1)_n
(b+1)_n}{(c+1)_n(1)_n} 
-\frac{(a+1)_{n-1} (b+1)_{n-1}}{(c+1)_{n-1}(1)_{n-1}}\right)
-c \frac{(a)_n (b)_n}{(c)_n(1)_n}=0,
\end{equation*}
which amounts to a routine computation. To summarize, we have proved
that $\widetilde a_{i,1}=0$ for $1\leq i\leq 2s-1$. This completes the
proof of Proposition~\ref{prop:PerturbedMehtaWang}.
\end{proof}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\newpage

\subsection{Proof of Lemma~\ref{lem:Pf(S)_N}}
It follows from~\eqref{eq:Dij} that we may apply
Proposition~\ref{prop:PerturbedMehtaWang} 
to the matrix
$$\left(\frac{(-1)^{i+j}D_{i,j}}{2(2N+i)!\,(2N+j)!}\right)_{1\leq
i,j\leq 2\s}.$$ 
After some simplification, we obtain
\begin{align*}
\Pf D&=
(-1)^{\s}2^{\s}\left(\prod_{i=1}^{2\s}(2N+i)!\right)\underset{1\leq
i,j\leq
2\s}\Pf\left(\frac{(-1)^{i+j}D_{i,j}}{2(2N+i)!\,(2N+j)!}\right)\\ 
&=(-1)^{\s}2^{2\s-2}(\s-1)!\left(\prod_{i=1}^{2\s-1}(2N+i)!\right)
\left(\prod_{i=0}^{\s-2} 
\frac{(2i+1)!}{(4N+2\s+3+2i)!}\right)\,V(N,\s),
\end{align*}
where
\begin{equation}\label{eq:def-Vns}
V(N,\s)=\sum_{j=1}^{2\s-1}\frac{D_{2\s,j}}{(2N+j)!}
\sum_{k=0}^{\s-1}\frac{2^{-k}}{k!}\binom{2k}{j+k-\s}(4N+2+2\s)_{j+k-\s}.
\end{equation}
This easily leads to the following reformulation of
Lemma~\ref{lem:Pf(S)_N}. 
\begin{prop}\label{prop:reformulation-lemma} 
Let $V(N,\s)$ be defined by~\eqref{eq:def-Vns}. Then
\begin{equation*}
V(N,\s)=\frac{2^{N+2}(2N+\s+1)!}{(N+\s)!\,(4N+2\s+2)!}
\left((-1)^{\s}(2N+2\s)!!-(2N+2\s-1)!!\right).
\end{equation*}
\end{prop}

The first step towards a proof of the preceding proposition is to simplify
expression~\eqref{eq:Dij-lastrow} for $D_{2\s,j}$. 
We shall prove at the end of this section that
\begin{multline}
\label{eq:simplification-Dij}
\frac{(-1)^{j+1}D_{2\s,j}}{(2N+j)!}
=\frac{(j-1)(2N+1)!}{(4N+j+3)!}
\left(\frac{2(2N+1)!}{N!\,(N+1)!}-2^{2N+2}\right)\\
-\frac{(2N+2)!}{(4N+j+3)!}
\left(\frac{4(2N+1)!}{N!\,(N+1)!}+2^{2N+3}\right)
-\frac{(j-1)(N+j-1)!\,N!}{(2N+j)!\,(2N+j+1)!}\\
+\frac{2(j-1)(2N+2)!}{(2N+j+1)!\,(N+1)!}
\sum_{h=N}^{2N+1}\frac{(h+j-1)!\,(h+1)!}{(h-N)!\,(2N+j+h+2)!}.
\end{multline}
Therefore, by~\eqref{eq:def-Vns}, we have
\begin{multline}
\label{eq:decompo-Vns}
V(N,\s)= (2N+1)!\left(\frac{2(2N+1)!}{N!\,(N+1)!}-2^{2N+2}\right)
\sum_{j=1}^{2\s-1}\frac{(-1)^{j+1}(j-1)}{(4N+j+3)!}\l_j(N,\s)\\
-(2N+2)!\left(\frac{4(2N+1)!}{N!\,(N+1)!}+2^{2N+3}\right)
\sum_{j=1}^{2\s-1}\frac{(-1)^{j+1}}{(4N+j+3)!}\l_j(N,\s)\\
-N!\sum_{j=1}^{2\s-1}\frac{(-1)^{j+1}
(j-1)(N+j-1)!}{(2N+j)!\,(2N+j+1)!}\l_j(N,\s)\\
 +\frac{2(2N+2)!}{(N+1)!}\sum_{j=1}^{2\s-1}
\frac{(-1)^{j+1}(j-1)}{(2N+j+1)!}
\sum_{h=N}^{2N+1}\frac{(h+j-1)!\,(h+1)!}{(h-N)!\,(2N+j+h+2)!}\l_j(N,\s),
\end{multline}
where we have set
\begin{equation}\label{eq:def-lamdaNs}
\l_j(N,\s)=\sum_{k=0}^{\s-1}\frac{2^{-k}}{k!}
\binom{2k}{j+k-\s}(4N+2+2\s)_{j+k-\s}.
\end{equation}
All the sums in~\eqref{eq:decompo-Vns} can be evaluated in
closed-form expressions. 

\begin{lem}\label{lem:multisum}
For all positive integers $N$ and $\s$, we have
\begin{align}
\begin{split}\label{eq:multisum1}
& \sum_{j=1}^{2\s-1}\frac{(-1)^{j+1}(j-1) }{(4N+j+3)!}\l_j(N,\s)=0,
\end{split}\\
\begin{split}\label{eq:multisum2}
& \sum_{j=1}^{2\s-1}\frac{ (-1)^{j+1}}{(4N+j+3)!}\l_j(N,\s)
=(-2)^{\s-1}\frac{(2N+\s+1)!}{(2N+2)!\,(4N+2\s+2)!},
\end{split}\\
\begin{split}\label{eq:multisum3}
&\sum_{j=1}^{2\s-1}\frac{(-1)^{j+1}(j-1)(N+j-1)!}
{(2N+j)!\,(2N+j+1)!}\l_j(N,\s)=0,
\end{split}\\
\begin{split}\label{eq:multisum4}
& \sum _{j=1}
^{2\s-1}\sum_{h=N}^{2N+1}\frac{(-1)^{j+1}(j-1)\,(h+j-1)!\,(h+1)!}
   {(h-N)!\,(2N+j+h+2)!\,(2N+j+1)!}\l_j(N,\s)\\
&\qquad=\frac {(2N+\s+1)!} {(N+1)!\,(4N+2\s+2)!}
\left((-2)^{\s-1}-\frac {(N+1)!^2\,(2N+2\s)!}
{2^{\s-1}\,(N+\s)!^2\,(2N+2)!}\right).
\end{split}
\end{align}
\end{lem}

At this point, we should notice that, after substituting these summations
in~\eqref{eq:decompo-Vns}, it becomes a routine matter
to verify Proposition~\ref{prop:reformulation-lemma}. So, to complete
the proof of Lemma~\ref{lem:Pf(S)_N}, 
it remains to prove the preceding lemma and~\eqref{eq:simplification-Dij}.

\begin{proof}[Proof of Lemma~\ref{lem:multisum}] 
(1) The double
sum on the left-hand side of~\eqref{eq:multisum1} is the specialization
$i=1$, $s=\sigma$, and $R=4N+2$ 
of~\eqref{eq:comblineaireA_firstRows}, which was shown to be zero
in the proof of Proposition~\ref{prop:PerturbedMehtaWang},
whence~\eqref{eq:multisum1}. 

\medskip
(2) Let $S_1(N,\s)$ denote the double sum on the left-hand side
of~\eqref{eq:multisum2}. 
Interchanging the sums over $j$ and $k$ in $S_1(N,\s)$, we see that the
(now) inner sum over $j$ can be written as a ${}_2F_1$-series which
is summable by the Chu--Vandermonde formula. To be precise, we obtain
\begin{align*}
S_1(N,\s)&=\sum_{j=1}^{2\s-1}\frac{
(-1)^{j+1}}{(4N+j+3)!}\sum_{k=0}^{\s-1}
\frac{2^{-k}}{k!}\binom{2k}{j+k-\s}(4N+2+2\s)_{j+k-\s}\\
&=\sum_{k=0}^{\s-1}\frac{2^{-k}}{k!}\frac
{(-1)^{\s-k+1}}{(4N+\s-k+3)!}
\, {}_2F_1\!\left[\begin{matrix}{2N+2\s+2,-2k}\\
{4N+4+\s-k}\end{matrix};1\right]\\
&=\sum_{k=0}^{\s-1}\frac{2^{-k}}{k!}\frac {(-1)^{\s-k+1}}{(4N+\s-k+3)!}
\frac{(2-\s-k)_{2k}}{(4N+4+\s-k)_{2k}}.
\end{align*}
Writing the above sum in hypergeometric notation, we arrive at
\begin{equation*}
S_1(N,\s)=\frac{(-1)^{\s+1}}{(4N+\s+3)!}
\,{}_2F_1\!\left[\begin{matrix}{\s-1,2-\s}\\{4N+4+\s}\end{matrix};
\frac{1}{2}\right]
=(-1)^{\s+1}2^{\s-1}\frac{(2N+\s+1)!}{(2N+2)!\,(4N+2\s+2)!},
\end{equation*}
where the last equality follows from Bailey's ${}_2F_1$ summation
formula (see \cite[(1.7.1.8); Appendix (III.7)]{SlatAC})
$$
{} _{2} F _{1} \!\left [ \begin{matrix} { a, 1 - a}\\ {
b}\end{matrix} ; {\displaystyle
   {\frac 1 2}}\right ]  =\frac { \Gamma \left( {\frac b 2}\right)\, 
\Gamma \left( {\frac 1 2} +
   {\frac b 2}\right)} { \Gamma \left( {\frac a 2} + {\frac b 2}\right)\, 
\Gamma \left( {\frac 1 2} - {\frac a 2} + {\frac b
   2}\right)}.
$$
This finishes the proof of~\eqref{eq:multisum2}.

\medskip
(3) Let $S_2(N,\s)$ denote the double sum on the left-hand side of
\eqref{eq:multisum3}. We shall establish the recurrence
\begin{equation} \label{eq:rek1}
S_2(N,\s)+(4N+2\s+3)S_2(N,\s+1)=0,\quad \quad \s\ge1.
\end{equation}
Since it is easy to verify directly that $S_2(N,1)=0$, this would
immediately imply the claim.

In order to prove \eqref{eq:rek1}, we use the Gosper--Zeilberger
algorithm (cf.\ \cite{GospAB} and \cite[\S~II.5]{PeWZAA} --- the
particular implementation that we used is the {\sl Mathematica}
implementation by Paule and Schorn \cite{PaScAA}) to find that
\begin{equation} \label{eq:rek1f}
F_2(N,\s,j,k) +(4N+2\s+3)F_2(N,\s+1,j,k)= G_2(N,\s,j,k+1)-G_2(N,\s,j,k),
\end{equation}
where $F_2(N,\s,j,k)$ is the summand of the sum on the left-hand side
of \eqref{eq:multisum3}, that is,
$$
F_2(N,\s,j,k)=(-1)^{j+1}\frac {(j-1)\,(N+j-1)!} {(2N+j)!\,(2N+j+1)!}
\frac {1} {2^k\,k!}\binom {2k}{j+k-\s} (4N+2\s+2)_{j+k-\s},
$$
and
$$
G_2(n,\s,j,k)=\frac {j+k-\s} {4N+2\s+2}F_2(N,\s,j,k).
$$
We now sum both sides of \eqref{eq:rek1f} over $k$ between $0$ and
$\s-1$ and subsequently over $j$ between $1$ and $2\s-1$. Taking into
account the telescoping effect on the right-hand side when we
perform summation over $k$, we arrive at
\begin{multline*} %\label{}
S_2(N,\s)+(4N+2\s+3)S_2(N,\s+1)
-(4N+2\s+3)\sum _{j=1} ^{2\s-1}F_2(N,\s+1,j,\s)\\
-(4N+2\s+3)\big(F_2(N,\s+1,2\s,\s)+F_2(N,\s+1,2\s,\s-1)
+F_2(N,\s+1,2\s+1,\s)\big)\\
=\sum _{j=1} ^{2\s-1}(G_2(N,\s,j,\s)-G_2(N,\s,j,0)).
\end{multline*}
After some simplification, this becomes
\begin{align*} \label{}
S_2(N,\s)&{}+(4N+2\s+3)S_2(N,\s+1)\\
&=
 \left( 4 N + 4 \s + 3 \right)
  \sum_{j = 1}^{2 \s-1}
   {\frac {{{\left( -1 \right) }^{j}}\,\left( j-1 \right) \,
     \left( N+j-1 \right) !\,\left( 2 \s \right) !\,
     ({ \textstyle 4 N + 2 \s+3}) _{j-1} }
    {{2^\s}\,\left( j-1 \right) !\,\left( 2 N+j \right) !\,
     \left( 2 N +j+1 \right) !\,\s!\,\left( 2 \s-j+1 \right)!}}\\
&\kern2cm + {\frac {{2^{1 - \s}} \left( 4 N + 4 \s + 3 \right) \,
   \left( N + 2 \s -1\right) !\,
   ({ \textstyle 4 N + 2 \s+3}) _{2 \s-1} }
  {\left( N + \s+1 \right) \,\left( \s-2 \right) !\,
   {{\left( 2 N + 2 \s \right) !}^2}}} \\
&= \left( 4 N + 4 \s + 3 \right)
  \sum_{j = 1}^{2 \s+1}
   {\frac {{{\left( -1 \right) }^{j}}\,\left( j-1 \right) \,
     \left( N+j-1 \right) !\,\left( 2 \s \right) !\,
     ({ \textstyle 4 N + 2 \s+3}) _{j-1} }
    {{2^\s}\,\left( j-1 \right) !\,\left( 2 N+j \right) !\,
     \left( 2 N +j+1 \right) !\,\s!\,\left( 2 \s-j+1 \right) !}}
.
\end{align*}
It remains to show that the right-hand side of this equation
vanishes. In order to see this, we write the sum over $j$ in
hypergeometric notation. Thereby we obtain
\begin{multline*} %\label{}
S_2(N,\s)+(4N+2\s+3)S_2(N,\s+1)\\= {\frac {2 \s\left( 4 N + 4 \s
+3\right) \,\left( 4 N + 2 \s +3\right) \,
   \left( N+1 \right) !}
  {{2^\s}\,\s!\,\left( 2 N+2 \right) !\,\left( 2 N+3 \right) !}}
   {} _{3} F _{2} \!\left [ \begin{matrix} 
 { - 2 \s+1, 4 N + 2 \s+4,N+2}\\ {
   2 N+3, 2 N+4}\end{matrix} ; {\displaystyle 1}\right ].
\end{multline*}
This $_3F_2$-series can be evaluated by means of Watson's
$_3F_2$-summation (see \cite[(2.3.3.13); Appendix~(III.23)]{SlatAC})
\begin{equation*} %\label{eq:S3233}
{} _{3} F _{2} \!\left [ \begin{matrix} { a, b, c}\\ { {\frac {1 + a
+ b} 2}, 2
  c}\end{matrix} ; {\displaystyle 1}\right ]
=
 \frac {\Gamma\left( {\frac 1 2}\right)\, 
 \Gamma\left({\frac 1 2} + c\right)\,
  \Gamma\left( {\frac 1 2} + {\frac a 2} +
  {\frac b 2}\right)\,
  \Gamma\left( {\frac 1 2} -{\frac a 2}-{\frac b 2}+c\right)}
 {\Gamma\left({\frac 1 2} + {\frac a
  2}\right)\, \Gamma\left( {\frac 1 2} + {\frac b 2}\right)\,
  \Gamma\left( {\frac 1 2} - {\frac a 2} + c\right)\,
  \Gamma\left( {\frac 1 2} -
  {\frac b 2} + c\right)}.
\end{equation*}
In fact, because of the term $\Gamma(\frac {1} {2}+\frac {a} {2})$
in the denominator on the right-hand side, the series vanishes
whenever $a$ is an odd negative integer. The recurrence
\eqref{eq:rek1} follows immediately. This completes the proof
of~\eqref{eq:multisum3}. 

\medskip
(4) Again, we prove the claim by induction on $\s$. To start the
induction, we have to show that \eqref{eq:multisum4} holds for $\s=1$.
This is trivially the case, since in this case the sum over $j$
reduces to just one term, the term for $j=1$, so that both sides of
\eqref{eq:multisum4} vanish.

Let $S_3(N,\s)$ denote the triple sum on the left-hand side of
\eqref{eq:multisum4}. We now claim that
\begin{multline} \label{eq:rek2}
S_3(N,\s)+(4N+2\s+3)S_3(N,\s+1)\\= -\frac{ (4 N+4
  \s+3)\,(N+1)!\, (2 N+\s+1)!\, (2 N+2\s)!}
{2^\s\,(2 N+2)!\, (N+\s)!\,
  (N+\s+1)!\, (4 N+2 \s+2)!},
\quad \quad \s\ge1.
\end{multline}
Since the right-hand side of \eqref{eq:multisum4} satisfies the above
recurrence --- as is not difficult to check --- this would prove the
lemma.

In order to prove \eqref{eq:rek2}, we multiply both sides of
\eqref{eq:rek1} by
$$
\frac {(h+j-1)!\,(h+1)!} {(h-N)!\,(2N+j+h+2)!} \bigg/ \frac
{(N+j-1)!} {(2N+j)!}
$$
to find that
\begin{multline} \label{eq:rek2f}
F_3(N,\s,j,h,k) +(4N+2\s+3)F_3(N,\s+1,j,h,k)\\=
G_3(N,\s,j,h,k+1)-G_3(N,\s,j,h,k),
\end{multline}
where $F_3(N,\s,j,h,k)$ is the summand of the sum on the left-hand
side of \eqref{eq:multisum4}, that is,
\begin{multline*}
F_3(N,\s,j,h,k)=(-1)^{j+1}\frac {(j-1)\,(h+j-1)!\,(h+1)!}
{(h-N)!\,(2N+j+h+2)!\,(2N+j+1)!}\\
\times \frac {1} {2^k\,k!}\binom {2k}{j+k-\s} (4N+2\s+2)_{j+k-\s},
\end{multline*}
and
$$
G_3(n,\s,j,h,k)=\frac {j+k-\s} {4N+2\s+2}F_3(N,\s,j,h,k).
$$
We now sum both sides of \eqref{eq:rek2f} over $k$ between $0$ and
$\s-1$, and subsequently over $j$ between $1$ and $2\s-1$, and over
$h$ between $N$ and $2N+1$. Taking into account the telescoping
effect on the right-hand side when we perform summation over $k$, we
arrive at
\begin{multline*} %\label{}
S_3(N,\s)+(4N+2\s+3)S_3(N,\s+1)
-(4N+2\s+3)\sum _{j=1} ^{2\s-1}\sum_{h=N}^{2N+1}F_3(N,\s+1,j,h,\s)\\
-(4N+2\s+3)\sum_{h=N}^{2N+1}
\big(F_3(N,\s+1,2\s,h,\s)\kern6cm\\
\kern5cm
+F_3(N,\s+1,2\s,h,\s-1)+F_3(N,\s+1,2\s+1,h,\s)\big)\\
=\sum _{j=1}
^{2\s-1}\sum_{h=N}^{2N+1}\big(G_3(N,\s,j,h,\s)-G_3(N,\s,j,h,0)\big).
\end{multline*}
After some simplification, this becomes
\begin{align*} \label{}
S_3&(N,\s)+(4N+2\s+3)S_3(N,\s+1)\\
&= -
 \sum _{h=N} ^{2N+1}\sum_{j = 2}^{2 \s-1}
  {\frac {{{\left( -1 \right) }^{j}}\,\left( 4 N + 4 \s + 3 \right) \,
    \left( h+j-1 \right) !\,\left( h+1 \right) !\,\left( 2 \s \right) !\,
    ({ \textstyle 4 N + 2 \s+3}) _{j-1} }
   {{2^\s}\,\left( j-2 \right) !\,
    \left( h-N\right) !\,\left( 2 N+j +h+2\right) !\,
    \left( 2 N +j+1 \right) !\,\s!\,\left( 2 \s-j+1 \right)!}}
\\
&\kern.5cm + \sum _{h=N} ^{2N+1}
 4\,\left( 4 N + 4 \s+3 \right)
  \big( 3 + 2 h + 5 N + 3 h N + 2 {N^2} + \s + h \s - 2 N \s \\
&\kern5cm -
    2 h N \s - 4 {N^2} \s - 4 {\s^2} - 2 h {\s^2} - 8 N {\s^2} -
    4 {\s^3} \big)\\
&\kern4cm \cdot {\frac { \left( h+1 \right) !\,
   \left( h + 2 \s -1\right) !\,
   ({ \textstyle 4 N + 2 \s+3}) _{2 \s-1} }
  {{2^\s}\,\left( h - N \right) !\,\left( \s-1 \right) !\,
   \left( 2 N + 2 \s+2 \right) !\,\left( 2 N + 2 \s+h+3 \right) !}}\\
&= -
 \sum _{h=N} ^{2N+1}\sum_{j = 2}^{2 \s+1}
  {\frac {{{\left( -1 \right) }^{j}}\,\left( 4 N + 4 \s + 3 \right) \,
    \left( h+j-1 \right) !\,\left( h+1 \right) !\,\left( 2 \s \right) !\,
    ({ \textstyle 4 N + 2 \s+3}) _{j-1} }
   {{2^\s}\,\left( j-2 \right) !\,
    \left( h-N\right) !\,\left( 2 N+j +h+2\right) !\,
    \left( 2 N +j+1 \right) !\,\s!\,\left( 2 \s-j+1 \right)!}}
.
\end{align*}
By writing the sum over $j$ in hypergeometric notation, this turns
into
\begin{multline*} \label{}
S_3(N,\s)+(4N+2\s+3)S_3(N,\s+1)\\
=-\sum_{h=N}^{2N+1} {\frac {\left( 4 N + 2 \s+3 \right) \,
   \left( 4 N + 4 \s +3\right) \,
   {{\left( h+1 \right) !}^2}}
  {2^{\s-1}\,\left( h - N \right) !\,\left( 2 N+3 \right) !\,
   \left( 2 N+h+4 \right) !\,\left( \s-1 \right) !}}\\
\cdot
   {} _{3} F _{2} \!\left [ \begin{matrix} 
     { h+2, 4 N + 2 \s+4,- 2 \s+1}\\ {
    2 N+4, 2 N+h+5}\end{matrix} ; {\displaystyle 1}\right ]
.
\end{multline*}
Next we apply the contiguous relation
$$%C26
{} _{3} F _{2} \!\left [ \begin{matrix} { a, b, c}\\ {
d,e}\end{matrix} ;
  {\displaystyle z}\right ] =
 {\frac{b } {b - a}}
 {} _{3} F _{2} \!\left [ \begin{matrix} 
   { a, b + 1, c}\\ {d,e}\end{matrix} ;
    {\displaystyle z}\right ] +
  {\frac {a } {a - b}}
 {} _{3} F _{2} \!\left [ \begin{matrix} 
   { a + 1, b, c}\\ { d,e}\end{matrix} ;
    {\displaystyle z}\right ].
$$
We obtain
\begin{multline*} \label{}
S_3(N,\s)+(4N+2\s+3)S_3(N,\s+1)\\
= -\sum_{h=N}^{2N+1} {\frac {\left( 4 N + 2 \s+3 \right) \,
    \left( 4 N + 2 \s+4 \right) \,\left( 4 N + 4 \s+3 \right) \,
    {{\left( h +1\right) !}^2}}
   {{2^{\s-1}}\,\left( 4 N + 2 \s-h+2 \right) \,\left( h - N \right) !\,
    \left( 2 N +3\right) !\,\left( 2 N+h+4 \right) !\,
    \left( \s -1\right) !}}\\
\cdot
    {} _{3} F _{2} \!\left [ \begin{matrix} 
  { h+2, 4 N + 2 \s+5, - 2 \s+1}\\
    { 2 N+4, 2 N+h+5}\end{matrix} ; {\displaystyle 1}\right ] \\
+\sum_{h=N}^{2N+1}
 {\frac {\left( 4 N + 2 \s+3 \right) \,
   \left( 4 N + 4 \s+3 \right) \,
   \left( h+1 \right) !\,\left( h+2 \right) !}
  {{2^{\s-1}}\,\left( 4 N + 2 \s-h+2 \right) \,\left( h - N \right) !\,
   \left( 2 N+3 \right) !\,\left( 2 N+h+4 \right) !\,
   \left( \s-1 \right) !}}\\
\cdot
   {} _{3} F _{2} \!\left [ \begin{matrix} 
  { h+3, 4 N + 2 \s+4, 1 - 2 \s}\\ {
    2 N+4, 2 N+h+5}\end{matrix} ; {\displaystyle 1}\right ]
.
\end{multline*}
Both $_3F_2$-series can be evaluated by means of the
Pfaff--Saalsch\"utz summation formula \eqref{eq:PfSa}.
If we apply the formula, then,
after some simplification, the above recurrence reduces to
\begin{multline*} 
S_3(N,\s)+(4N+2\s+3)S_3(N,\s+1)\\
= 4\,\left( 4 N + 2 \s+3 \right) \,\left( 4 N + 4 \s+3 \right)
\sum_{h=N}^{2N+1}
\left( 1 + h - N + h N - 2 {N^2} - 3 \s - 4 N \s - 2 {\s^2} \right)\\
\cdot {\frac {
   {{\left( h+1 \right) !}^2}\,({ \textstyle 2 N+3}) _{2 \s-2} \,
   ({ \textstyle 2 - h + 2 N}) _{2 \s-2} }
  {{2^\s}\,\left( h - N \right) !\,\left( \s-1 \right) !\,
   \left( 2 N + 2 \s+2 \right) !\,\left( 2 N + 2 \s+h+3 \right) !}}
.
\end{multline*}
Let $S_4(N,\s)$ denote the right-hand sum. The Gosper--Zeilberger
algorithm then yields the recurrence
\begin{multline} \label{eq:rek3}
(2 N + 2 \s+1) (4 N + 4 \s+7)S_4(N,\s)\\-
 4 (2 N+3) (N + \s+2) (4 N + 2 \s+3) (4 N + 2 \s+5) (4 N +
   4 \s+3)S_4(N+1,\s)=0,
\end{multline}
with a half-page certificate, which we omit here for the sake of
brevity. Since it is straightforward to check that $S_4(1,\s)$ is
equal to the right-hand side of \eqref{eq:rek2} with $N=1$, and that
the right-hand side of \eqref{eq:rek2} satisfies the recurrence in
\eqref{eq:rek3}, the claimed recurrence \eqref{eq:rek2} follows.

This completes the proof of~\eqref{eq:multisum4}, and thus of the lemma.
\end{proof}

We now turn our attention to the proof of~\eqref{eq:simplification-Dij}.

\begin{proof}[Proof of~\eqref{eq:simplification-Dij}]
By~\eqref{eq:Dij-lastrow}, we have
\begin{equation}\label{eq:decompo-Dij}
D_{2\s,j}=\frac {(-1)^{j}2} {\binom {2N+j+1}{j-1}}\cdot S+\frac
{(-1)^{N+j}} {(N+1)\binom 
{2N+j+1}{j-1}}+\frac{(-1)^{j}(j-1)\,N!\,(N+j-1)!}{(2N+j+1)!},
\end{equation}
where $S$ stands for the double sum in~\eqref{eq:Dij-lastrow},
that is,
\begin{equation*}
S=\sum _{i=0}
^{2N}(-1)^i\left(2^i-\binom i{N}\right)
\frac {2N+j-i} {i+1}\sum _{\ell=0} ^{i}(-1)^\ell \binom {2N+j}
{i-\ell}\binom 
{\ell+j-1}\ell\frac {1} {2N+\ell+j+2}.
\end{equation*}

By extending the sum over $i$, we may rewrite $S$ as
\begin{multline*}
\sum _{i=0} ^{2N+1}(-1)^i\left(2^i-\binom i{N}\right)
\frac {2N+j-i} {i+1}\\
\cdot \sum _{\ell=0} ^{i}(-1)^\ell \binom {2N+j}
{i-\ell}\binom {\ell+j-1}\ell\frac {1} {2N+\ell+j+2}\\
+\left(2^{2N+1}-\binom
{2N+1}{N}\right)
\frac {j-1} {2N+2}\kern5cm\\
\cdot \sum _{\ell=0} ^{2N+1}(-1)^\ell \binom {2N+j}
{2N+1-\ell}\binom {\ell+j-1}\ell\frac {1} {2N+\ell+j+2}.
\end{multline*}

We concentrate now on the evaluation of the second sum over $\ell$.
It can be written as a $_2F_1$-series which is summable by means of
the Chu--Vandermonde summation formula
(in hypergeometric form; see \cite[(1.7.7), Appendix~(III.4)]{SlatAC}), 
so that we obtain
\begin{align*}
 & \left(2^{2N+1}-\binom{2N+1}{N}\right)\frac {j-1} {2N+2}\sum
 _{\ell=0} ^{2N+1}(-1)^\ell \binom {2N+j} 
{2N+1-\ell}\binom {\ell+j-1}\ell\frac {1} {2N+\ell+j+2}\\
 &\quad =\left(2^{2N+1}-\binom{2N+1}{N}\right)\frac {j-1} {2N+2}\binom
 {2N+j} {2N+1}\frac {1} {2N+j+2} 
   {}_2 F_1\!\left[\begin{matrix} 2N+j+2,-2N-1\\2N+j+3\end{matrix};
   1\right]\\ 
 &\quad = \left(2^{2N+1}-\binom{2N+1}{N}\right)\frac {j-1} {2N+2}\binom
 {2N+j} {2N+1}\frac {1} {2N+j+2}\frac {(1)_{2N+1}} {(2N+j+3)_{2N+1}}\\ 
 &\quad = \frac {(j-1)\,(2N+j)!\,(2N+j+1)!}
 {(2N+2)\,(j-1)!\,(4N+j+3)!}\left(2^{2N+1}-\binom {2N+1}{N}\right). 
\end{align*}
Consequently, our sum $S$ is equal to
\begin{multline*}
\sum _{i=0}
^{2N+1}(-1)^i\left(2^i-\binom i{N}\right)
\frac {2N+j-i} {i+1}\sum _{\ell=0} ^{i}(-1)^\ell \binom {2N+j}
{i-\ell}\binom {\ell+j-1}\ell\frac {1} {2N+\ell+j+2}\\
+\frac {(j-1)\,(2N+j)!\,(2N+j+1)!}
{(2N+2)\,(j-1)!\,(4N+j+3)!}\left(2^{2N+1}-\binom {2N+1}{N}\right). 
\end{multline*}

Next we apply the partial fraction expansion
$$
\frac {1} {(i+1)(2N+\ell+j+2)}=\frac {1} {2N+\ell+j-i+1}\left( \frac
{1} {i+1}-\frac {1} {2N+\ell+j+2}\right).
$$
Thus, we have
\begin{equation} \label{eq:1}
S=S_1-S_2 + \frac {(j-1)\,(2N+j)!\,(2N+j+1)!}
{(2N+2)\,(j-1)!\,(4N+j+3)!}\left(2^{2N+1}-\binom {2N+1}{N}\right), 
\end{equation}
where
\begin{multline*}
S_1=\frac {1} {(2N+j+1)} \sum _{i=0}
^{2N+1}(-1)^i\left(2^i-\binom i{N}\right)
\frac {2N+j-i} {i+1}\\
\cdot \sum _{\ell=0} ^{i}(-1)^\ell \binom {2N+j+1} {i-\ell}\binom
{\ell+j-1}\ell
\end{multline*}
and
\begin{multline*}
S_2=\frac {1} {(2N+j+1)} \sum _{i=0}
^{2N+1}(-1)^i\left(2^i-\binom i{N}\right)
(2N+j-i)\\
\cdot \sum _{\ell=0} ^{i}(-1)^\ell \binom {2N+j+1} {i-\ell}\binom
{\ell+j-1}\ell\frac {1} {2N+\ell+j+2}.
\end{multline*}

We start with the evaluation of $S_1$. We have
\begin{multline*}
\sum _{\ell=0} ^{i}(-1)^\ell \binom {2N+j+1} {i-\ell}\binom
{\ell+j-1}\ell =\binom {2N+j+1}i
{}_2 F_1\!\left[\begin{matrix} j,-i\\
2N+j-i+2\end{matrix}; 1\right]\\
=\binom {2N+j+1}i\frac {(2N-i+2)_i} {(2N+j-i+2)_i} =\binom {2N+1}i,
\end{multline*}
and therefore
\begin{align*} %\label{}
S_1&=\frac {1} {(2N+j+1)} \sum _{i=0}
^{2N+1}(-1)^i\left(2^i-\binom i{N}\right)
\frac {2N+j-i} {2N+2}\binom {2N+2}{i+1}\\
&=\frac {1} {(2N+j+1)}
\sum _{i=0} ^{2N+1}(-1)^i\left(2^i-\binom i{N}\right)\\
&\kern3cm \cdot \frac {1} {2N+2}\left( (2N+2)\binom
{2N+1}{i+1}+(j-1)\binom {2N+2}{i+1}
\right)\\
&=\frac {1} {(2N+j+1)} \bigg( -\frac {1}
{2}(-1)^{2N+1}+\frac {1} {2} +\frac {j-1} {2N+2}\left(-\frac {1}
{2}(-1)^{2N+2}+\frac {1} {2}\right)
\bigg)\\
&\kern1cm -\frac {1} {(2N+j+1)} \bigg(
(-1)^N\binom {2N+1}{N+1}
\,{}_2 F_1\!\left[\begin{matrix} N+1,-N\\
N+2\end{matrix}; 1\right]\\
&\kern3cm + \frac {j-1} {2N+2}(-1)^N\binom {2N+2}{N+1}
\,{}_2 F_1\!\left[\begin{matrix} N+1,-N-1\\
N+2\end{matrix}; 1\right]
\bigg)\\
&=\frac {1} {(2N+j+1)} \bigg( 1-
(-1)^N\binom {2N+1}{N+1}\frac {N!\,(N+1)!} {(2N+1)!}\\
&\kern5cm -(-1)^N \frac {j-1} {2N+2}\binom {2N+2}{N+1}\frac
{(N+1)!^2} {(2N+2)!}
\bigg)\\
&=\frac {1} {2N+j+1}-\frac{(-1)^N} {2N+2}.
\end{align*}

Next we consider the evaluation of $S_2$. We have
\begin{multline*}
S_2=\frac {1} {(2N+j+1)} \sum _{i=0}
^{2N+1}(-1)^i\left(2^i-\binom i{N}\right)
(2N+j-i)\\
\cdot \sum _{\ell=0} ^{i}(-1)^\ell \binom {\ell+j-1}\ell \frac {1}
{2\pi \sqrt{-1}} \bigg(\int _{C} ^{}\frac {(1+z)^{2N+j+1}}
{z^{i-\ell+1}}\,dz\bigg) \bigg(\int _{0}
^{1}x^{2N+\ell+j+1}\,dx\bigg),
\end{multline*}
where $C$ is a small contour in the complex plane encircling the
origin in positive orientation. The sum over $\ell$ can be extended
to a sum from $0$ to $\infty$ since the terms corresponding to
$\ell$'s which are larger than $i$ vanish. Hence, by evaluating the
(geometric) sum over $\ell$, we may rewrite this as
\begin{align*} %\label{}
S_2&= \frac {1} {(2N+j+1)2\pi \sqrt{-1}} \int
_{C} ^{}\int _{0} ^{1}
\sum _{i=0} ^{2N+1}(-1)^i\left(2^i-\binom i{N}\right)\\
&\kern7cm \cdot \frac {d} {dz}\big(z^{2N+j-i}\big) \frac
{(1+z)^{2N+j+1}} {z^{2N+j}}\frac {x^{2N+j+1}} {(1+xz)^j}\,dz
\,dx\\
&= -\frac {1} {(2N+j+1)2\pi \sqrt{-1}} \int _{C}
^{}\int _{0} ^{1}
\sum _{i=0} ^{2N+1}(-1)^i\left(2^{2N+1-i}-\binom {2N+1-i}{N}\right)\\
&\kern7cm \cdot \frac {d} {dz}\big(z^{i+j-1}\big) \frac
{(1+z)^{2N+j+1}} {z^{2N+j}}\frac {x^{2N+j+1}} {(1+xz)^j}\,dz \,dx.
\end{align*}
Consequently,
\begin{equation} \label{eq:2}
S_2=\frac {1} {(2N+j+1)}(-S_3+S_4),
\end{equation}
where
$$
S_3=\frac {1} {2\pi \sqrt{-1}} \int _{C} ^{}\int _{0} ^{1} \sum
_{i=0} ^{2N+1}(-1)^i2^{2N+1-i} \frac {d} {dz}\big(z^{i+j-1}\big)
\frac {(1+z)^{2N+j+1}} {z^{2N+j}}\frac {x^{2N+j+1}} {(1+xz)^j}\,dz
\,dx
$$
and
\begin{multline*}
S_4=\frac {1} {2\pi \sqrt{-1}} \int _{C} ^{}\int _{0} ^{1}
\sum _{i=0} ^{2N+1}(-1)^i\binom {2N+1-i}{N}\\
\cdot \frac {d} {dz}\big(z^{i+j-1}\big) \frac {(1+z)^{2N+j+1}}
{z^{2N+j}}\frac {x^{2N+j+1}} {(1+xz)^j}\,dz \,dx.
\end{multline*}

Now we evaluate $S_3$. Similarly to before, we may extend the sum
over $i$ to a sum from $0$ to $\infty$. The resulting sum is again a
geometric series, so that we obtain
\begin{align*}
S_3&=\frac {1} {2\pi \sqrt{-1}} \int _{C} ^{}\int _{0} ^{1} 2^{2N+1}
\frac {d} {dz}\left(\frac {z^{j-1}} {1+\frac {z} {2}}\right) \frac
{(1+z)^{2N+j+1}} {z^{2N+j}}\frac {x^{2N+j+1}} {(1+xz)^j}\,dz
\,dx\\
&=\frac {2^{2N+1}} {2\pi \sqrt{-1}} \int _{C} ^{}\int _{0} ^{1}
\frac {\left(j-1+(j-2)\frac {z} {2}\right)} {\left(1+\frac {z}
{2}\right)^2} \frac {(1+z)^{2N+j+1}} {z^{2N+2}}\frac {x^{2N+j+1}}
{(1+xz)^j}\,dz \,dx.
\end{align*}
Now we do the substitution $z\to z/(1-z)$. Thereby, we obtain
\begin{align*}
S_3&=\frac {2^{2N+1}} {2\pi \sqrt{-1}} \int _{C} ^{}\int _{0} ^{1}
\frac {\left(j-1-j\frac {z} {2}\right)} {\left(1-\frac {z}
{2}\right)^2} \frac {x^{2N+j+1}} {z^{2N+2}\,(1-z(1-x))^j}\,dz
\,dx\\
&=\frac {2^{2N+1}} {2\pi \sqrt{-1}} \int _{C} ^{}\int _{0} ^{1}
\frac {\left(j-1-j\frac {z} {2}\right)} {\left(1-\frac {z}
{2}\right)^2} \frac {x^{2N+j+1}} {z^{2N+2}} \sum_{h=0}^{2N+1}\binom
{h+j-1}h z^h(1-x)^h\,dz
\,dx\\
&=\frac {2^{2N+1}} {2\pi \sqrt{-1}} \int _{C} ^{} \frac
{\left(j-1-j\frac {z} {2}\right)} {\left(1-\frac {z} {2}\right)^2}
\sum_{h=0}^{2N+1}\binom {h+j-1}h \frac {(2N+j+1)!\,h!} {(2N+j+h+2)!}
\frac {1} {z^{2N+2-h}}
\,dz\\
&=2^{2N+1}
\sum_{h=0}^{2N+1}\binom {h+j-1}h \frac {(2N+j+1)!\,h!} {(2N+j+h+2)!}\\
&\kern5cm \cdot \left((j-1)(2N+2-h)-j(2N+1-h)
\right)2^{-2N-1+h}\\
&= \sum_{h=0}^{2N+1}\binom {h+j-1}h \frac {(2N+j+1)!\,h!}
{(2N+j+h+2)!}
\left(-2N-2+h+j\right)2^{h}\\
&= \sum_{h=0}^{2N+1} \bigg(\binom {h+j}{h+1} \frac
{(2N+j+1)!\,(h+1)!} {(2N+j+h+2)!}2^{h+1} - \binom {h+j-1}h \frac
{(2N+j+1)!\,h!} {(2N+j+h+1)!}2^h
\bigg)\\
&=\binom {2N+j+1}{2N+2} \frac {(2N+j+1)!\,(2N+2)!}
{(4N+j+3)!}2^{2N+2}
-1\\
&=2^{2N+2} \frac {(2N+j+1)!^2} {(4N+j+3)!\,(j-1)!} -1 .
\end{align*}


Finally, we compute $S_4$. In the earlier definition of $S_4$, we
may again extend the sum over $i$ to a sum from $0$ to $\infty$.
Using
$$
\binom {2N+1-i}{N}=\frac {1} {2\pi \sqrt{-1}} \int_{\tilde C}\frac
{(1+u)^{2N+1-i}} {u^{N+1}}\,du,
$$
where $\tilde C$ is a small contour in the complex plane encircling
the origin in positive orientation, we then obtain
\begin{align*}
S_4&=\frac {1} {(2\pi \sqrt{-1})^2} \int _{\tilde C} ^{} \int _{C}
^{}\int _{0} ^{1} \sum _{i=0} ^{\infty}(-1)^i \frac {d}
{dz}\big(z^{i+j-1}\big) \frac {(1+u)^{2N+1-i}} {u^{N+1}}
\\
&\kern7cm \cdot \frac {(1+z)^{2N+j+1}} {z^{2N+j}}\frac {x^{2N+j+1}}
{(1+xz)^j}
\,du\,dz\,dx\\
&=\frac {1} {(2\pi \sqrt{-1})^2} \int _{\tilde C} ^{} \int _{C}
^{}\int _{0} ^{1} \frac {d} {dz}\left(\frac {z^{j-1}} {1+\frac {z}
{1+u}}\right) \frac {(1+u)^{2N+1}} {u^{N+1}}
\\
&\kern7cm \cdot \frac {(1+z)^{2N+j+1}} {z^{2N+j}}\frac {x^{2N+j+1}}
{(1+xz)^j}
\,du\,dz\,dx\\
&=\frac {1} {(2\pi \sqrt{-1})^2} \int _{\tilde C} ^{} \int _{C}
^{}\int _{0} ^{1} \frac {(j-1)+(j-2)\frac {z} {1+u}} {\left(1+\frac
{z} {1+u}\right)^2} \frac {(1+u)^{2N+1}} {u^{N+1}}
\\
&\kern7cm \cdot \frac {(1+z)^{2N+j+1}} {z^{2N+2}}\frac {x^{2N+j+1}}
{(1+xz)^j} \,du\,dz\,dx.
\end{align*}
Again we do the substitution $z\to z/(1-z)$. Thereby, we obtain
\begin{align*}
S_4&=\frac {1} {(2\pi \sqrt{-1})^2} \int _{\tilde C} ^{} \int _{C}
^{}\int _{0} ^{1} \frac {(j-1)(1+u)-z(1+(j-1)u)} {\left(1-\frac {uz}
{1+u}\right)^2}
\\
&\kern7cm \cdot \frac {(1+u)^{2N}} {u^{N+1}} \frac {x^{2N+j+1}}
{z^{2N+2}(1-z(1-x))^j}
\,du\,dz\,dx\\
&=\frac {1} {(2\pi \sqrt{-1})^2} \int _{\tilde C} ^{} \int _{C}
^{}\int _{0} ^{1} \frac {(j-1)(1+u)-z(1+(j-1)u)} {\left(1-\frac {uz}
{1+u}\right)^2}
\\
&\kern4.5cm \cdot \frac {(1+u)^{2N}} {u^{N+1}} \frac {x^{2N+j+1}}
{z^{2N+2}} \sum _{h=0}^{2N+1}\binom {h+j-1}h z^h(1-x)^h
\,du\,dz\,dx\\
&=\frac {1} {(2\pi \sqrt{-1})^2} \int _{\tilde C} ^{} \int _{C} ^{}
(j-1)(1+u)-z(1+(j-1)u) \sum _{s=0}^{\infty}(s+1)\left(\frac {uz}
{1+u}\right)^s \frac {(1+u)^{2N}} {u^{N+1}}
\\
&\kern4.5cm \cdot \sum _{h=0}^{2N+1}\binom {h+j-1}h \frac
{(2N+j+1)!\,h!} {(2N+j+h+2)!}\frac {1} {z^{2N+2-h}}
\,du\,dz\\
&=\sum _{h=0}^{2N+1}\binom {h+j-1}h \frac {(2N+j+1)!\,h!}
{(2N+j+h+2)!}
\bigg((j-1)(2N+2-h)\binom {h}{h-N-1}\\
&\kern3cm -(2N+1-h)\binom {h}{h-N} -(j-1)(2N+1-h)\binom {h}{h-N-1}
\bigg)\\
&=\sum _{h=0}^{2N+1}\binom {h+j-1}h \frac {(2N+j+1)!\,h!}
{(2N+j+h+2)!}
\frac {h!} {(h-N)!\,(N+1)!}\\
&\kern3cm \cdot \big((j-1)(h-N) -(2N+1-h)(N+1)
\big)\\
&=\sum _{h=0}^{2N+1} \bigg(
\frac {(h+j-1)!\,h!\,(2N+j+1)!} {(j-1)!\,(h-N-1)!\,(N+1)!\,(2N+j+h+1)!}\\
&\kern3cm -\frac {(h+j)!\,(h+1)!\,(2N+j+1)!}
{(j-1)!\,(h-N)!\,(N+1)!\,(2N+j+h+2)!}
\bigg)\\
&\kern1cm +(j-1)\sum _{h=0}^{2N+1}
\frac {(h+j-1)!\,(h+1)!\,(2N+j+1)!} {(j-1)!\,(h-N)!\,(N+1)!\,(2N+j+h+2)!}\\
&= -\frac {(2N+1+j)!^2\,(2N+2)!} {(j-1)!\,(N+1)!^2\,(4N+j+3)!}
\\
&\kern1cm +(j-1)\sum _{h=0}^{2N+1} \frac
{(h+j-1)!\,(h+1)!\,(2N+j+1)!} {(j-1)!\,(h-N)!\,(N+1)!\,(2N+j+h+2)!}.
\end{align*}

If we substitute all our findings in \eqref{eq:1} and
\eqref{eq:2}, then we have shown that
\begin{align*}
S&=S_1-S_2 + \frac {(j-1)\,(2N+j)!\,(2N+j+1)!}
{(2N+2)\,(j-1)!\,(4N+j+3)!}\left(2^{2N+1}-\binom {2N+1}{N}\right)\\ 
&=S_1+\frac {S_3-S_4} {2N+j+1}+ \frac {(j-1)\,(2N+j)!\,(2N+j+1)!}
{(2N+2)\,(j-1)!\,(4N+j+3)!}\left(2^{2N+1}-\binom {2N+1}{N}\right)\\ 
&= \frac {1} {2N+j+1}-\frac {(-1)^N}{2N+2}+ \frac
{(j-1)\,(2N+j)!\,(2N+j+1)!}
{(2N+2)\,(j-1)!\,(4N+j+3)!}\left(2^{2N+1}-\binom {2N+1}{N}\right)\\ 
&\quad+\frac {1} {2N+j+1} \Bigg( 2^{2N+2}\frac {(2N+j+1)!^2}
{(4N+j+3)!\,(j-1)!}-1
+\frac {(2N+1+j)!^2\,(2N+2)!}{(j-1)!\,(N+1)!^2\,(4N+j+3)!}\\
&\kern2cm -(j-1)\sum _{h=0}^{2N+1} \frac
{(h+j-1)!\,(h+1)!\,(2N+j+1)!}
{(j-1)!\,(h-N)!\,(N+1)!\,(2N+j+h+2)!}\Bigg).
\end{align*}
A combination of the latter identity with~\eqref{eq:decompo-Dij}
leads, after some trivial simplification,
to~\eqref{eq:simplification-Dij}. 
This finishes the proof.
\end{proof}



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%             ASYMPTOTIQUE DU RATIO
%
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%\newpage
\section{Proof of Corollary~\ref{cor:Ncentered(2n,2x+1)}}
\label{sec:6}

Throughout this section, all asymptotics are taken as $n\to\infty$ and
$x\sim an$. By Theorem~\ref{thm:Ncentered(2n,2x+1)}, the probability 
that a random vertically symmetric rhombus tiling of a 
$(2n,2x+1,2n)$-hexagon is
centered is equal to $R(n,x)$. Using expressions~\eqref{eq:def R(x,n)}
and~\eqref{eq:def-Ux}, we see that $R(n,x)$ can be written in 
hypergeometric notation (after reversing the order of summation
in~\eqref{eq:def-Ux} and 
dividing the sum in four parts) as
\begin{align*}
R&(n,x)=2^{3n-2}\frac{(2x+2)!\,(x+2n)!}{n!\,(x+1)!\,(2x+4n)!}\\
    \times&\bigg{(}
    \frac{(2n-1)!!\,(\frac{3}{2}-n)_{2n-1}(x+1)_{n-1}(x+n+1)_{n}}{(n-1)!\,n!} 
{}_{4}F_3\!\left[\begin{matrix}{1,n+\frac{1}{2},1-n,-n-x}\\ 
{n+1,\frac{3}{2}-n, 1-n-x}\end{matrix};-1\right]\\
   &- \frac{(2n-1)!!\,(\frac{3}{2}-n)_{2n-1}(x+1)_{n}
   (x+n+2)_{n-1}}{(n-1)!\,n!}
{}_{4}F_3\!\left[\begin{matrix}{1,x+n+1,n+\frac{1}{2},1-n}\\
 {n+1,\frac{3}{2}-n, x+n+2}\end{matrix};-1\right]\\
   &+\frac{(-1)^{n+1}(2n)!!\,
(\frac{3}{2}-n)_{2n-1}(x+1)_{n-1}(x+n+1)_{n}}{(n-1)!\,n!}
{}_{4}F_3\!\left[\begin{matrix}{1,n+\frac{1}{2},1-n,-n-x}\\ 
{n+1,\frac{3}{2}-n, 1-n-x}\end{matrix};1\right]\\
   &- \frac{(-1)^{n+1}(2n)!!\,
(\frac{3}{2}-n)_{2n-1}(x+1)_{n}(x+n+2)_{n-1}}{(n-1)!\,n!}
{}_{4}F_3\!\left[\begin{matrix}{1,x+n+1,n+\frac{1}{2},1-n}\\ 
{n+1,\frac{3}{2}-n, x+n+2}\end{matrix};1\right]\bigg{)}.
\end{align*}
This simplifies to
\begin{multline}\label{eq:simplification-ratio}
R(n,x)=(-1)^{n+1}2^{2n-2}\frac{ (x+2n)}{n(x+n)}
     \frac{\Gamma(2x+2)\,\Gamma(x+2n)^2
       \,\Gamma(n+\frac{1}{2})}
   {\Gamma(x+1)^2\,\Gamma(2x+4n)\,\Gamma(\frac{3}{2}-n)\,\Gamma(n)^2}\\
    \times\bigg{(} (-1)^{n+1}\frac{(2n)!}{4^n\,n!^2}
{}_{4}F_3\!\left[\begin{matrix}{1,n+\frac{1}{2},1-n,-n-x}\\ 
{n+1,\frac{3}{2}-n, 1-n-x}\end{matrix};-1\right]\\
     - (-1)^{n+1}\frac{(2n)!}{4^n\,n!^2}\frac{x+n}{x+n+1}
{}_{4}F_3\!\left[\begin{matrix}{1,x+n+1,n+\frac{1}{2},1-n}\\ 
{n+1,\frac{3}{2}-n, x+n+2}\end{matrix};-1\right]\\
   + {}_{4}F_3\!\left[\begin{matrix}
  {1,n+\frac{1}{2},1-n,-n-x}\\ 
{n+1,\frac{3}{2}-n, 1-n-x}\end{matrix};1\right]\\
   -\frac{x+n}{x+n+1} 
{}_{4}F_3\!\left[\begin{matrix}{1,x+n+1,n+\frac{1}{2},1-n}\\ 
{n+1,\frac{3}{2}-n, x+n+2}\end{matrix};1\right]\bigg{)}.
\end{multline}

Using Stirling's formula, it is a routine matter to determine the
asymptotic behaviour 
of the term in the first row in~\eqref{eq:simplification-ratio}:
\begin{equation}\label{eq:asympt-facteur-ratio}
(-1)^{n+1}2^{2n-2}\frac{ (x+2n)}{n(x+n)}
     \frac{\Gamma(2x+2)\,\Gamma(x+2n)^2\,\Gamma(n+\frac{1}{2})}
  {\Gamma(x+1)^2\,\Gamma(2x+4n)\,\Gamma(\frac{3}{2}-n)\,\Gamma(n)^2}
     \sim\frac{\sqrt{a(a+2)}}{\pi (a+1)}\frac{1}{n}.
\end{equation}
To deal with the ${}_{4}F_3$-series
in~\eqref{eq:simplification-ratio}, we shall use the next lemma. 

\begin{lem}\label{lem:asympt-hypergeoseries}
Let $b$ be a real number with $|b|>1$. Then, for any non-negative
integer $r$ 
and any sequence $(b_n)_{n\geq 1}$ with $b_n\to b$ as
$n\to\infty$, we have
\begin{equation}\label{eq:asympt-hypergeoseries}
\lim_{n\to\infty} \frac{1}{n} \,
{}_{4}F_3\!\left[\begin{matrix}{1,n+\frac{1}{2},1-n, b_n n+r}\\
{n+1,\frac{3}{2}-n, b_n n+r+1}\end{matrix};1\right]
=\frac{2b}{b+1}\sqrt{\frac{b+1}{b-1}}
\arctan\left(\sqrt{\frac{b-1}{b+1}}\right).
\end{equation}
\end{lem}

Before we prove this result, we show how it leads to
Corollary~\ref{cor:Ncentered(2n,2x+1)}. First,
Lemma~\eqref{eq:asympt-hypergeoseries} 
provides the asymptotic behaviour of the last two ${}_{4}F_3$-series
in~\eqref{eq:simplification-ratio}. Moreover, 
it is easily checked that the absolute value of the first (respectively
second) ${}_{4}F_3$-series is asymptotically
less than the third ${}_{4}F_3$-series 
(respectively fourth) ${}_{4}F_3$-series in~\eqref{eq:simplification-ratio},
which is 
$O(n)$ by Lemma~\ref{lem:asympt-hypergeoseries}. This, combined with
the fact that $\frac{(2n)!}{4^n\,n!^2}\sim (\pi n)^{-1/2}$, 
shows that the contribution of the first two ${}_{4}F_3$-series
in~\eqref{eq:simplification-ratio} is negligible in the limit. 
Altogether, by~\eqref{eq:asympt-facteur-ratio} and
Lemma~\ref{lem:asympt-hypergeoseries}, we see that $R(n,x)$ is 
asymptotically equal to
\begin{multline*}
\frac{\sqrt{a(a+2)}}{\pi (a+1)} \left(
 \frac{2(a+1)}{a}\sqrt{\frac{a}{a+2}}
 \arctan\left(\sqrt{\frac{a+2}{a}}\right)\right. \\
\left.
-\frac{2(a+1)}{a+2}\sqrt{\frac{a+2}{a}}
\arctan\left(\sqrt{\frac{a}{a+2}}\right)
\right)\\
=\frac{2}{\pi}\arctan\left(\frac{1}{\sqrt{a(a+2)}}\right)
=\frac{2}{\pi}\arcsin\left(\frac{1}{a+1}\right),
\end{multline*}
as desired.
To conclude the proof of Corollary~\ref{cor:Ncentered(2n,2x+1)}, it
remains to prove Lemma~\ref{lem:asympt-hypergeoseries}. 


\begin{proof}[Proof of Lemma~\ref{lem:asympt-hypergeoseries}] 
If we write
the ${}_{4}F_3$-series in~\eqref{eq:asympt-hypergeoseries}
explicitly as a sum over $k$, after some simplification, we obtain
\begin{align}
{}_{4}F_3&\!\left[\begin{matrix}{1,n+\frac{1}{2},1-n, b_n n+r,}\\
{n+1,\frac{3}{2}-n, b_n n+r+1}\end{matrix};1\right]\nonumber\\
&\qquad= \sum_{k=0}^{n-1}\frac{(n+\frac{1}{2})_k(1-n)_k (b_n
n+r)_k}{(n+1)_k(\frac{3}{2}-n)_k (b_n n+r+1)_k}\nonumber\\ 
&\qquad= \sum_{k=0}^{n-1}\frac{(b_n n+r)n(n-\frac{1}{2})}{(b_n n+r+k)}
   \frac{ \Gamma(n)^2\,\Gamma(n+k+\frac{1}{2})\,\Gamma(n-k-\frac{1}{2})}
      { \Gamma(n+\frac{1}{2})^2\,\Gamma(n+k+1)\,\Gamma(n-k)}.
\label{eq:hypergeo-explicite_lem}
\end{align}
Let us denote the summand in the latter sum by $F(n,k)$. It is easy to
check that ${F(n,k)>0}$ for $0\leq k\leq n-1$, and that
\begin{equation}\label{eq:asymptoLastTerm}
F(n,0)=1\quad\text{and}\quad F(n,n-1)\sim
\frac{b}{b+1}\sqrt{\frac{\pi }{2}}n^{1/2} \quad\text{as
$n\to\infty$.}
\end{equation}
Moreover, a routine computation shows that
$(\frac{\partial^2}{\partial k^2} F(n,k))/F(n,k)$ is equal to 
\begin{multline*}
\bigg{(}\psi(n+k+1/2)-\psi(n-k-1/2)-\psi(n+k+1)+\psi(n-k)-\frac{1}{b_n+k+r}
\bigg{)}^2\\ 
+\bigg{(}\psi_1(n+k+1/2)+\psi_1(n-k-1/2)-\psi_1(n+k+1)
-\psi_1(n-k)+\frac{1}{(b_n+k+r)^2}
\bigg{)},
\end{multline*}
where $\psi$ and $\psi_1$ are the digamma and trigamma functions
defined by $\psi(x)=\frac{d}{dx}\log(\Gamma(x))$ and
$\psi_1(x)=\frac{d^2}{dx^2}\log(\Gamma(x))=\frac{d}{dx}\psi(x)$.
Since the trigamma function $\psi_1$ is decreasing, the above
expression is positive. 
Consequently, for fixed $n\geq 1$, the summand $F(n,k)$ is convex as a
function in
$k$. Combined with~\eqref{eq:asymptoLastTerm}, this implies that the
sum in~\eqref{eq:hypergeo-explicite_lem} may be approximated by an 
integral and
\begin{equation}\label{eq:serie-integrale}
\sum_{k=0}^{n-1}F(n,k)=\int_0^{n-1}F(n,k)\, dk+O(n^{1/2})
   =\int_{n^{1/3}}^{n-n^{1/3}}F(n,k) \,dk+O(n^{1/2+1/3}).
\end{equation}
Using the expression~\eqref{eq:hypergeo-explicite_lem} for $F(n,k)$
and the asymptotic approximation
\begin{equation*}
\Gamma(z+\tfrac{1}{2})=z^{-1/2}\Gamma(z+1)\,
\left(1+O\left(z^{-1}\right)\right),\qquad z\to\infty,
\end{equation*}
we see after some manipulation that
\begin{equation*}
F(n,k)=\frac{b_n n}{(b_n n+k) \sqrt{1-\frac{k^2}{n^2}}}
 \left(1+O(n^{-1/3})\right),
\qquad\text{for $n^{1/3}\leq k\leq
n-n^{1/3}$}.
\end{equation*}
Combined with~\eqref{eq:serie-integrale}, this leads to
\begin{equation*}
\sum_{k=0}^{n-1}F(n,k) =\left(\int_{n^{1/3}}^{n-n^{1/3}}\frac{b_n
n}{(b_n n+k) \sqrt{1-\frac{k^2}{n^2}}} \,dk\right)
   \left( 1+O(n^{-1/3})   \right) +O(n^{5/6}),
\end{equation*}
which, after the substitution $y=k/n$, gives
\begin{equation*}
\sum_{k=0}^{n-1}F(n,k) 
=n\,\left(\int_{n^{-2/3}}^{1-n^{-2/3}}\frac{b_n}{(b_n+y)
\sqrt{1-y^2}} \,dy\right)
   \left( 1+O(n^{-1/3})   \right) +O(n^{5/6}).
\end{equation*}
Dividing both sides by $n$ and taking the limit, we obtain
\begin{equation*}
\lim_{n\to\infty} \frac{1}{n} \,
{}_{4}F_3\!\left[\begin{matrix}{1,n+\frac{1}{2},1-n, b_n n+r,}\\
{n+1,\frac{3}{2}-n, b_n n+r+1}\end{matrix};1\right]
=\int_0^1\frac{b}{b+y}\frac{1}{\sqrt{1-y^2}} dy.
\end{equation*}
To finish the proof of the lemma, it remains to check that the
above integral is equal to the right-hand side
of~\eqref{eq:asympt-hypergeoseries}, 
which amounts to a routine computation.
\end{proof}




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\end{document}

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