\magnification=\magstep1 
\input xy
\xyoption{all}

\vcorrection{-0.8cm}
\def\({\left(}
\def\){\right)}
\def\U{{\tilde U}}
\def\be{\beta}
\NoBlackBoxes
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%Input-Datei zum Erzeugen von Gitterpunktwegen.%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%entering a path: 
%  \Pfad(x-coordinate of starting point,y-coordinate of starting point),path
%  as 1-2-3-4-word\endPfad
%(1=step in x-direction, 2=step in y-direction, 3=upward diagonal step, 4=downward diagonal step)
%
%entering a dotted path: 
%  \SPfad(x-coordinate of starting point,y-coordinate of starting point),path
%  as 1-2-3-4-word\endSPfad
%(1=step in x-direction, 2=step in y-direction, 3=upward diagonal step, 4=downward diagonal step)
%
%coordinate-axes:
%  \Koordinatenachsen(length of positive x-axes, length of positive y-axes)(length of negative x-axes, length of negative y-axes)
%The length of negative axes are entered as negative numbers.
%
%lattice:
%  \Gitter(number of points in positive x-direction, number of points in positive y-direction)(number of points in negative x-direction, number of points in negative y-direction)
%
%diagonal lines:
%  \Diagonale(x-coordinate of SW-most point,y-coordinate of SW-most point)length of the projection on the x-axes
%
%antidiagonal lines:
%  \AntiDiagonale(x-coordinate of NW-most point,y-coordinate of NW-most point)length of the projection on the x-axes
%
%vectors:
%  \Vektor(x-coordinate of incline, y-coordinate of incline)length(x-coordinate of starting point, y-coordinate of starting point)
%
%labelling of points:
%  \Label[location?]{[label]}(x-coordinate,y-coordinate)
%where:
%  [location?]=\l,\lo,\lu,\r,\ro,\ru,\o,\u
%and l=left, r=right, u=bottom, o=top.
%In addition, if by \Einheit?cm the basic unit is changed, there exist
%\llo,\loo,\llu,\luu,\rro,\roo,\rru,\ruu.
%
%The basic unit can be changed by entering
%  \Einheit=?cm
%The default is \Einheit=0.5cm.
%
%The thickness of the paths can be changed by entering
%     \PfadDicke{?cm}
%The default is \PfadDicke=1pt.
%
%The following point sizes are available:
%\DuennPunkt, \NormalPunkt, \DickPunkt. Syntax:
%  \DickPunkt(x-coordinate,y-coordinate), etc.
%
%Besides, a circle is available by \Kreis. Syntax:
%  \Kreis(x-coordinate,y-coordinate)
%
\catcode`\@=11
\font\tenln    = line10
\font\tenlnw   = linew10

\newskip\Einheit \Einheit=0.5cm
\newcount\xcoord \newcount\ycoord
\newdimen\xdim \newdimen\ydim \newdimen\PfadD@cke \newdimen\Pfadd@cke

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%LaTeX counters, dimensions, variables for lines%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\newcount\@tempcnta
\newcount\@tempcntb

\newdimen\@tempdima
\newdimen\@tempdimb


\newdimen\@wholewidth
\newdimen\@halfwidth

\newcount\@xarg
\newcount\@yarg
\newcount\@yyarg
\newbox\@linechar
\newbox\@tempboxa
\newdimen\@linelen
\newdimen\@clnwd
\newdimen\@clnht

\newif\if@negarg

\def\@whilenoop#1{}
\def\@whiledim#1\do #2{\ifdim #1\relax#2\@iwhiledim{#1\relax#2}\fi}
\def\@iwhiledim#1{\ifdim #1\let\@nextwhile=\@iwhiledim
        \else\let\@nextwhile=\@whilenoop\fi\@nextwhile{#1}}

\def\@whileswnoop#1\fi{}
\def\@whilesw#1\fi#2{#1#2\@iwhilesw{#1#2}\fi\fi}
\def\@iwhilesw#1\fi{#1\let\@nextwhile=\@iwhilesw
         \else\let\@nextwhile=\@whileswnoop\fi\@nextwhile{#1}\fi}


\def\thinlines{\let\@linefnt\tenln \let\@circlefnt\tencirc
  \@wholewidth\fontdimen8\tenln \@halfwidth .5\@wholewidth}
\def\thicklines{\let\@linefnt\tenlnw \let\@circlefnt\tencircw
  \@wholewidth\fontdimen8\tenlnw \@halfwidth .5\@wholewidth}
\thinlines
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\PfadD@cke1pt \Pfadd@cke0.5pt
\def\PfadDicke#1{\PfadD@cke#1 \divide\PfadD@cke by2 \Pfadd@cke\PfadD@cke \multiply\PfadD@cke by2}
\long\def\LOOP#1\REPEAT{\def\BODY{#1}\ITERATE}
\def\ITERATE{\BODY \let\next\ITERATE \else\let\next\relax\fi \next}
\let\REPEAT=\fi
\def\Punkt{\hbox{\raise-2pt\hbox to0pt{\hss$\ssize\bullet$\hss}}}
\def\DuennPunkt(#1,#2){\unskip
  \raise#2 \Einheit\hbox to0pt{\hskip#1 \Einheit
          \raise-2.5pt\hbox to0pt{\hss$\bullet$\hss}\hss}}
\def\NormalPunkt(#1,#2){\unskip
  \raise#2 \Einheit\hbox to0pt{\hskip#1 \Einheit
          \raise-3pt\hbox to0pt{\hss\twelvepoint$\bullet$\hss}\hss}}
\def\DickPunkt(#1,#2){\unskip
  \raise#2 \Einheit\hbox to0pt{\hskip#1 \Einheit
          \raise-4pt\hbox to0pt{\hss\fourteenpoint$\bullet$\hss}\hss}}
\def\Kreis(#1,#2){\unskip
  \raise#2 \Einheit\hbox to0pt{\hskip#1 \Einheit
          \raise-4pt\hbox to0pt{\hss\fourteenpoint$\circ$\hss}\hss}}

%%%%%%%%%%%%%%%%%%%%%
%LaTeX line macros%
%%%%%%%%%%%%%%%%%%%%%
\def\Line@(#1,#2)#3{\@xarg #1\relax \@yarg #2\relax
\@linelen=#3\Einheit
\ifnum\@xarg =0 \@vline
  \else \ifnum\@yarg =0 \@hline \else \@sline\fi
\fi}

\def\@sline{\ifnum\@xarg< 0 \@negargtrue \@xarg -\@xarg \@yyarg -\@yarg
  \else \@negargfalse \@yyarg \@yarg \fi
\ifnum \@yyarg >0 \@tempcnta\@yyarg \else \@tempcnta -\@yyarg \fi
\ifnum\@tempcnta>6 \@badlinearg\@tempcnta0 \fi
\ifnum\@xarg>6 \@badlinearg\@xarg 1 \fi
\setbox\@linechar\hbox{\@linefnt\@getlinechar(\@xarg,\@yyarg)}%
\ifnum \@yarg >0 \let\@upordown\raise \@clnht\z@
   \else\let\@upordown\lower \@clnht \ht\@linechar\fi
\@clnwd=\wd\@linechar
\if@negarg \hskip -\wd\@linechar \def\@tempa{\hskip -2\wd\@linechar}\else
     \let\@tempa\relax \fi
\@whiledim \@clnwd <\@linelen \do
  {\@upordown\@clnht\copy\@linechar
   \@tempa
   \advance\@clnht \ht\@linechar
   \advance\@clnwd \wd\@linechar}%
\advance\@clnht -\ht\@linechar
\advance\@clnwd -\wd\@linechar
\@tempdima\@linelen\advance\@tempdima -\@clnwd
\@tempdimb\@tempdima\advance\@tempdimb -\wd\@linechar
\if@negarg \hskip -\@tempdimb \else \hskip \@tempdimb \fi
\multiply\@tempdima \@m
\@tempcnta \@tempdima \@tempdima \wd\@linechar \divide\@tempcnta \@tempdima
\@tempdima \ht\@linechar \multiply\@tempdima \@tempcnta
\divide\@tempdima \@m
\advance\@clnht \@tempdima
\ifdim \@linelen <\wd\@linechar
   \hskip \wd\@linechar
  \else\@upordown\@clnht\copy\@linechar\fi}

\def\@hline{\ifnum \@xarg <0 \hskip -\@linelen \fi
\vrule height\Pfadd@cke width \@linelen depth\Pfadd@cke
\ifnum \@xarg <0 \hskip -\@linelen \fi}

\def\@getlinechar(#1,#2){\@tempcnta#1\relax\multiply\@tempcnta 8
\advance\@tempcnta -9 \ifnum #2>0 \advance\@tempcnta #2\relax\else
\advance\@tempcnta -#2\relax\advance\@tempcnta 64 \fi
\char\@tempcnta}

\def\Vektor(#1,#2)#3(#4,#5){\unskip\leavevmode
  \xcoord#4\relax \ycoord#5\relax
      \raise\ycoord \Einheit\hbox to0pt{\hskip\xcoord \Einheit
         \Vector@(#1,#2){#3}\hss}}

\def\Vector@(#1,#2)#3{\@xarg #1\relax \@yarg #2\relax
\@tempcnta \ifnum\@xarg<0 -\@xarg\else\@xarg\fi
\ifnum\@tempcnta<5\relax
\@linelen=#3\Einheit
\ifnum\@xarg =0 \@vvector
  \else \ifnum\@yarg =0 \@hvector \else \@svector\fi
\fi
\else\@badlinearg\fi}

\def\@hvector{\@hline\hbox to 0pt{\@linefnt
\ifnum \@xarg <0 \@getlarrow(1,0)\hss\else
    \hss\@getrarrow(1,0)\fi}}

\def\@vvector{\ifnum \@yarg <0 \@downvector \else \@upvector \fi}

\def\@svector{\@sline
\@tempcnta\@yarg \ifnum\@tempcnta <0 \@tempcnta=-\@tempcnta\fi
\ifnum\@tempcnta <5
  \hskip -\wd\@linechar
  \@upordown\@clnht \hbox{\@linefnt  \if@negarg
  \@getlarrow(\@xarg,\@yyarg) \else \@getrarrow(\@xarg,\@yyarg) \fi}%
\else\@badlinearg\fi}

\def\@upline{\hbox to \z@{\hskip -.5\Pfadd@cke \vrule width \Pfadd@cke
   height \@linelen depth \z@\hss}}

\def\@downline{\hbox to \z@{\hskip -.5\Pfadd@cke \vrule width \Pfadd@cke
   height \z@ depth \@linelen \hss}}

\def\@upvector{\@upline\setbox\@tempboxa\hbox{\@linefnt\char'66}\raise
     \@linelen \hbox to\z@{\lower \ht\@tempboxa\box\@tempboxa\hss}}

\def\@downvector{\@downline\lower \@linelen
      \hbox to \z@{\@linefnt\char'77\hss}}

\def\@getlarrow(#1,#2){\ifnum #2 =\z@ \@tempcnta='33\else
\@tempcnta=#1\relax\multiply\@tempcnta \sixt@@n \advance\@tempcnta
-9 \@tempcntb=#2\relax\multiply\@tempcntb \tw@
\ifnum \@tempcntb >0 \advance\@tempcnta \@tempcntb\relax
\else\advance\@tempcnta -\@tempcntb\advance\@tempcnta 64
\fi\fi\char\@tempcnta}

\def\@getrarrow(#1,#2){\@tempcntb=#2\relax
\ifnum\@tempcntb < 0 \@tempcntb=-\@tempcntb\relax\fi
\ifcase \@tempcntb\relax \@tempcnta='55 \or
\ifnum #1<3 \@tempcnta=#1\relax\multiply\@tempcnta
24 \advance\@tempcnta -6 \else \ifnum #1=3 \@tempcnta=49
\else\@tempcnta=58 \fi\fi\or
\ifnum #1<3 \@tempcnta=#1\relax\multiply\@tempcnta
24 \advance\@tempcnta -3 \else \@tempcnta=51\fi\or
\@tempcnta=#1\relax\multiply\@tempcnta
\sixt@@n \advance\@tempcnta -\tw@ \else
\@tempcnta=#1\relax\multiply\@tempcnta
\sixt@@n \advance\@tempcnta 7 \fi\ifnum #2<0 \advance\@tempcnta 64 \fi
\char\@tempcnta}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\def\Diagonale(#1,#2)#3{\unskip\leavevmode
  \xcoord#1\relax \ycoord#2\relax
      \raise\ycoord \Einheit\hbox to0pt{\hskip\xcoord \Einheit
         \Line@(1,1){#3}\hss}}
\def\AntiDiagonale(#1,#2)#3{\unskip\leavevmode
  \xcoord#1\relax \ycoord#2\relax %\advance\xcoord by -0.05\relax
      \raise\ycoord \Einheit\hbox to0pt{\hskip\xcoord \Einheit
         \Line@(1,-1){#3}\hss}}
\def\Pfad(#1,#2),#3\endPfad{\unskip\leavevmode
  \xcoord#1 \ycoord#2 \thicklines\ZeichnePfad#3\endPfad\thinlines}
\def\ZeichnePfad#1{\ifx#1\endPfad\let\next\relax
  \else\let\next\ZeichnePfad
    \ifnum#1=1
      \raise\ycoord \Einheit\hbox to0pt{\hskip\xcoord \Einheit
         \vrule height\Pfadd@cke width1 \Einheit depth\Pfadd@cke\hss}%
      \advance\xcoord by 1
    \else\ifnum#1=2
      \raise\ycoord \Einheit\hbox to0pt{\hskip\xcoord \Einheit
        \hbox{\hskip-\PfadD@cke\vrule height1 \Einheit width\PfadD@cke depth0pt}\hss}%
      \advance\ycoord by 1
    \else\ifnum#1=3
      \raise\ycoord \Einheit\hbox to0pt{\hskip\xcoord \Einheit
         \Line@(1,1){1}\hss}
      \advance\xcoord by 1
      \advance\ycoord by 1
    \else\ifnum#1=4
      \raise\ycoord \Einheit\hbox to0pt{\hskip\xcoord \Einheit
         \Line@(1,-1){1}\hss}
      \advance\xcoord by 1
      \advance\ycoord by -1
    \fi\fi\fi\fi
  \fi\next}
\def\hSSchritt{\leavevmode\raise-.4pt\hbox to0pt{\hss.\hss}\hskip.2\Einheit
  \raise-.4pt\hbox to0pt{\hss.\hss}\hskip.2\Einheit
  \raise-.4pt\hbox to0pt{\hss.\hss}\hskip.2\Einheit
  \raise-.4pt\hbox to0pt{\hss.\hss}\hskip.2\Einheit
  \raise-.4pt\hbox to0pt{\hss.\hss}\hskip.2\Einheit}
\def\vSSchritt{\vbox{\baselineskip.2\Einheit\lineskiplimit0pt
\hbox{.}\hbox{.}\hbox{.}\hbox{.}\hbox{.}}}
\def\DSSchritt{\leavevmode\raise-.4pt\hbox to0pt{%
  \hbox to0pt{\hss.\hss}\hskip.2\Einheit
  \raise.2\Einheit\hbox to0pt{\hss.\hss}\hskip.2\Einheit
  \raise.4\Einheit\hbox to0pt{\hss.\hss}\hskip.2\Einheit
  \raise.6\Einheit\hbox to0pt{\hss.\hss}\hskip.2\Einheit
  \raise.8\Einheit\hbox to0pt{\hss.\hss}\hss}}
\def\dSSchritt{\leavevmode\raise-.4pt\hbox to0pt{%
  \hbox to0pt{\hss.\hss}\hskip.2\Einheit
  \raise-.2\Einheit\hbox to0pt{\hss.\hss}\hskip.2\Einheit
  \raise-.4\Einheit\hbox to0pt{\hss.\hss}\hskip.2\Einheit
  \raise-.6\Einheit\hbox to0pt{\hss.\hss}\hskip.2\Einheit
  \raise-.8\Einheit\hbox to0pt{\hss.\hss}\hss}}
\def\SPfad(#1,#2),#3\endSPfad{\unskip\leavevmode
  \xcoord#1 \ycoord#2 \ZeichneSPfad#3\endSPfad}
\def\ZeichneSPfad#1{\ifx#1\endSPfad\let\next\relax
  \else\let\next\ZeichneSPfad
    \ifnum#1=1
      \raise\ycoord \Einheit\hbox to0pt{\hskip\xcoord \Einheit
         \hSSchritt\hss}%
      \advance\xcoord by 1
    \else\ifnum#1=2
      \raise\ycoord \Einheit\hbox to0pt{\hskip\xcoord \Einheit
        \hbox{\hskip-2pt \vSSchritt}\hss}%
      \advance\ycoord by 1
    \else\ifnum#1=3
      \raise\ycoord \Einheit\hbox to0pt{\hskip\xcoord \Einheit
         \DSSchritt\hss}
      \advance\xcoord by 1
      \advance\ycoord by 1
    \else\ifnum#1=4
      \raise\ycoord \Einheit\hbox to0pt{\hskip\xcoord \Einheit
         \dSSchritt\hss}
      \advance\xcoord by 1
      \advance\ycoord by -1
    \fi\fi\fi\fi
  \fi\next}
\def\Koordinatenachsen(#1,#2){\unskip
 \hbox to0pt{\hskip-.5pt\vrule height#2 \Einheit width.5pt depth1 \Einheit}%
 \hbox to0pt{\hskip-1 \Einheit \xcoord#1 \advance\xcoord by1
    \vrule height0.25pt width\xcoord \Einheit depth0.25pt\hss}}
\def\Koordinatenachsen(#1,#2)(#3,#4){\unskip
 \hbox to0pt{\hskip-.5pt \ycoord-#4 \advance\ycoord by1
    \vrule height#2 \Einheit width.5pt depth\ycoord \Einheit}%
 \hbox to0pt{\hskip-1 \Einheit \hskip#3\Einheit 
    \xcoord#1 \advance\xcoord by1 \advance\xcoord by-#3 
    \vrule height0.25pt width\xcoord \Einheit depth0.25pt\hss}}
\def\Gitter(#1,#2){\unskip \xcoord0 \ycoord0 \leavevmode
  \LOOP\ifnum\ycoord<#2
    \loop\ifnum\xcoord<#1
      \raise\ycoord \Einheit\hbox to0pt{\hskip\xcoord \Einheit\Punkt\hss}%
      \advance\xcoord by1
    \repeat
    \xcoord0
    \advance\ycoord by1
  \REPEAT}
\def\Gitter(#1,#2)(#3,#4){\unskip \xcoord#3 \ycoord#4 \leavevmode
  \LOOP\ifnum\ycoord<#2
    \loop\ifnum\xcoord<#1
      \raise\ycoord \Einheit\hbox to0pt{\hskip\xcoord \Einheit\Punkt\hss}%
      \advance\xcoord by1
    \repeat
    \xcoord#3
    \advance\ycoord by1
  \REPEAT}
\def\Label#1#2(#3,#4){\unskip \xdim#3 \Einheit \ydim#4 \Einheit
  \def\lo{\advance\xdim by-.5 \Einheit \advance\ydim by.5 \Einheit}%
  \def\llo{\advance\xdim by-.25cm \advance\ydim by.5 \Einheit}%
  \def\loo{\advance\xdim by-.5 \Einheit \advance\ydim by.25cm}%
  \def\o{\advance\ydim by.25cm}%
  \def\ro{\advance\xdim by.5 \Einheit \advance\ydim by.5 \Einheit}%
  \def\rro{\advance\xdim by.25cm \advance\ydim by.5 \Einheit}%
  \def\roo{\advance\xdim by.5 \Einheit \advance\ydim by.25cm}%
  \def\l{\advance\xdim by-.30cm}%
  \def\r{\advance\xdim by.30cm}%
  \def\lu{\advance\xdim by-.5 \Einheit \advance\ydim by-.6 \Einheit}%
  \def\llu{\advance\xdim by-.25cm \advance\ydim by-.6 \Einheit}%
  \def\luu{\advance\xdim by-.5 \Einheit \advance\ydim by-.30cm}%
  \def\u{\advance\ydim by-.30cm}%
  \def\ru{\advance\xdim by.5 \Einheit \advance\ydim by-.6 \Einheit}%
  \def\rru{\advance\xdim by.25cm \advance\ydim by-.6 \Einheit}%
  \def\ruu{\advance\xdim by.5 \Einheit \advance\ydim by-.30cm}%
  #1\raise\ydim\hbox to0pt{\hskip\xdim
     \vbox to0pt{\vss\hbox to0pt{\hss$#2$\hss}\vss}\hss}%
}
\catcode`\@=13

\catcode`\@=11
\font@\twelverm=cmr10 scaled\magstep1
\font@\twelveit=cmti10 scaled\magstep1
\font@\twelvebf=cmbx10 scaled\magstep1
\font@\twelvei=cmmi10 scaled\magstep1
\font@\twelvesy=cmsy10 scaled\magstep1
\font@\twelveex=cmex10 scaled\magstep1

\newtoks\twelvepoint@
\def\twelvepoint{\normalbaselineskip15\p@
 \abovedisplayskip15\p@ plus3.6\p@ minus10.8\p@
 \belowdisplayskip\abovedisplayskip
 \abovedisplayshortskip\z@ plus3.6\p@
 \belowdisplayshortskip8.4\p@ plus3.6\p@ minus4.8\p@
 \textonlyfont@\rm\twelverm \textonlyfont@\it\twelveit
 \textonlyfont@\sl\twelvesl \textonlyfont@\bf\twelvebf
 \textonlyfont@\smc\twelvesmc \textonlyfont@\tt\twelvett
%Erg\"anzung des fetten Small-Capitals-Fonts:
%
 \ifsyntax@ \def\big##1{{\hbox{$\left##1\right.$}}}%
  \let\Big\big \let\bigg\big \let\Bigg\big
 \else
  \textfont\z@=\twelverm  \scriptfont\z@=\tenrm  \scriptscriptfont\z@=\sevenrm
  \textfont\@ne=\twelvei  \scriptfont\@ne=\teni  \scriptscriptfont\@ne=\seveni
  \textfont\tw@=\twelvesy \scriptfont\tw@=\tensy \scriptscriptfont\tw@=\sevensy
  \textfont\thr@@=\twelveex \scriptfont\thr@@=\tenex
        \scriptscriptfont\thr@@=\tenex
  \textfont\itfam=\twelveit \scriptfont\itfam=\tenit
        \scriptscriptfont\itfam=\tenit
  \textfont\bffam=\twelvebf \scriptfont\bffam=\tenbf
        \scriptscriptfont\bffam=\sevenbf
  \setbox\strutbox\hbox{\vrule height10.2\p@ depth4.2\p@ width\z@}%
  \setbox\strutbox@\hbox{\lower.6\normallineskiplimit\vbox{%
        \kern-\normallineskiplimit\copy\strutbox}}%
 \setbox\z@\vbox{\hbox{$($}\kern\z@}\bigsize@=1.4\ht\z@
 \fi
 \normalbaselines\rm\ex@.2326ex\jot3.6\ex@\the\twelvepoint@}

\font@\fourteenrm=cmr10 scaled\magstep2
\font@\fourteenit=cmti10 scaled\magstep2
\font@\fourteensl=cmsl10 scaled\magstep2
\font@\fourteensmc=cmcsc10 scaled\magstep2
\font@\fourteentt=cmtt10 scaled\magstep2
\font@\fourteenbf=cmbx10 scaled\magstep2
\font@\fourteeni=cmmi10 scaled\magstep2
\font@\fourteensy=cmsy10 scaled\magstep2
\font@\fourteenex=cmex10 scaled\magstep2
\font@\fourteenmsa=msam10 scaled\magstep2
\font@\fourteeneufm=eufm10 scaled\magstep2
\font@\fourteenmsb=msbm10 scaled\magstep2
\newtoks\fourteenpoint@
\def\fourteenpoint{\normalbaselineskip15\p@
 \abovedisplayskip18\p@ plus4.3\p@ minus12.9\p@
 \belowdisplayskip\abovedisplayskip
 \abovedisplayshortskip\z@ plus4.3\p@
 \belowdisplayshortskip10.1\p@ plus4.3\p@ minus5.8\p@
 \textonlyfont@\rm\fourteenrm \textonlyfont@\it\fourteenit
 \textonlyfont@\sl\fourteensl \textonlyfont@\bf\fourteenbf
 \textonlyfont@\smc\fourteensmc \textonlyfont@\tt\fourteentt
%Erg\"anzung des fetten Small-Capitals-Fonts:
%
 \ifsyntax@ \def\big##1{{\hbox{$\left##1\right.$}}}%
  \let\Big\big \let\bigg\big \let\Bigg\big
 \else
  \textfont\z@=\fourteenrm  \scriptfont\z@=\twelverm  \scriptscriptfont\z@=\tenrm
  \textfont\@ne=\fourteeni  \scriptfont\@ne=\twelvei  \scriptscriptfont\@ne=\teni
  \textfont\tw@=\fourteensy \scriptfont\tw@=\twelvesy \scriptscriptfont\tw@=\tensy
  \textfont\thr@@=\fourteenex \scriptfont\thr@@=\twelveex
        \scriptscriptfont\thr@@=\twelveex
  \textfont\itfam=\fourteenit \scriptfont\itfam=\twelveit
        \scriptscriptfont\itfam=\twelveit
  \textfont\bffam=\fourteenbf \scriptfont\bffam=\twelvebf
        \scriptscriptfont\bffam=\tenbf
  \setbox\strutbox\hbox{\vrule height12.2\p@ depth5\p@ width\z@}%
  \setbox\strutbox@\hbox{\lower.72\normallineskiplimit\vbox{%
        \kern-\normallineskiplimit\copy\strutbox}}%
 \setbox\z@\vbox{\hbox{$($}\kern\z@}\bigsize@=1.7\ht\z@
 \fi
 \normalbaselines\rm\ex@.2326ex\jot4.3\ex@\the\fourteenpoint@}

\font@\seventeenrm=cmr10 scaled\magstep3
\font@\seventeenit=cmti10 scaled\magstep3
\font@\seventeensl=cmsl10 scaled\magstep3
\font@\seventeensmc=cmcsc10 scaled\magstep3
\font@\seventeentt=cmtt10 scaled\magstep3
\font@\seventeenbf=cmbx10 scaled\magstep3
\font@\seventeeni=cmmi10 scaled\magstep3
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\topmatter
\author  
Christian Krattenthaler\footnote{\hbox{Partially supported by the Austrian
Science Foundation FWF, grant P13190-MAT.}}\\
Luigi Orsina\\ Paolo Papi\endauthor
\bigskip
\address
\vskip 0.pt Institut f\"ur Mathematik der Universit\"at Wien,
\vskip 0.pt Strudlhofgasse 4, A-1090 Wien, Austria.
\vskip 0.pt e-mail:{KRATT\@Ap.Univie.Ac.At}
\vskip 3pt
\vskip 5.pt Dipartimento di Matematica, Istituto G. Castelnuovo
\vskip 0.pt Universit\`a di Roma ``La Sapienza"  
\vskip 0.pt Piazzale Aldo Moro 5
\vskip 0.pt 00185 Rome --- ITALY 
\vskip 0.pt e-mail:{\rm \ orsina\@mat.uniroma1.it,\hskip3 pt papi\@mat.uniroma1.it}
 \endaddress

\leftheadtext {C. Krattenthaler, L. Orsina and P. Papi}
\rightheadtext { Enumeration of $ad$-nilpotent
$\frak b$-ideals for simple Lie algebras}
\abstract
We  provide  formulas for the number of $ad$-nilpotent ideals of a Borel subalgebra of a complex simple Lie
algebra 
 having 
fixed class of nilpotence.\endabstract
\title  Enumeration of $ad$-nilpotent
$\frak b$-ideals for simple Lie algebras
\endtitle
%\keywords ad-nilpotent ideal, Lie algebra, order ideal, Dyck path,
%Catalan number, Chebyshev polynomial\endkeywords
%\subjclass Primary: 17B20; Secondary: 05A15 05A19 05E15 17B30\endsubjclass
\endtopmatter
\nologo
\def\endemo{\qed\enddemo}
\def\ua{\underline{a}}
\def\w{\bullet}
\def\g{\frak g}
\def\h{\frak h}
\def\n{\frak n}
\def\bb{\frak b}
\def\D{\Delta}
\def\l{\lambda}
\def\Dp{\Delta^+}
\def\Dm{\Delta^-}
\def\Da{\widehat\Delta}
\def\Dap{\widehat\Delta^+}
\def\d{\delta}
\def\r{R(\h)}
\def\t{\otimes}
\def\a{\alpha}
\def\b{\beta}
\def\f{F(\g)}
\def\I{\Cal I^n}
\def\l{\lambda}
\def\d{\delta}
\def\m{\prec}
\def\th{\theta}
\def\Q{\check Q}
\def\i{{\frak i}}
\def\j{{\bold j}}
\def\N{N(w)}
\def\nd{\noindent}
\def\gg{\gamma}
\def\DD{D_\Phi}
\def\p{\Phi}
\def\nat{\Bbb N}
\def\ganz{\Bbb Z}
\def\compl{\Bbb C} 
\def\real{\Bbb R} 

\TagsOnRight


\def\lM{3.3}
\def\lm{3.4}
\def\summe{4.1}
\def\decomp{4.2}
\def\Summe{4.3}
\def\formela{4.4}
\def\formelb{4.5}
\def\formelc{4.6}
\def\qtCat{6.1}
\def\eqdmin{6.2}
\def\eqdmax{6.3}
\def\eqDmin{6.4}
\def\eqDmax{6.5}
\def\ineqm{6.6}
\def\eqintm{6.7}
\def\ineqM{6.8}
\def\eqstufen{6.9}
\def\eqintM{6.10}
\def\eqintmin{6.11}
\def\ineqA{6.12}
\def\ineqa{6.13}
\def\ineqb{6.14}



\def\l{\lambda}
\def\v#1{\left\vert#1\right\vert}
\def\fl#1{\left\lfloor#1\right\rfloor}
\def\cl#1{\left\lceil#1\right\rceil}
\def\flkl#1{\lfloor#1\rfloor}
\def\clkl#1{\lceil#1\rceil}
\def\dmin#1#2{{\theta^{\text{min}}_{#1}(#2)}}
\def\dmax#1#2{{\theta^{\text{max}}_{#1}(#2)}}
\def\Dmin#1#2{{\Theta^{\text{min}}_{#1}(#2)}}
\def\Dmax#1#2{{\Theta^{\text{max}}_{#1}(#2)}}
\def\({\left(}
\def\){\right)}
\def\T{\bold T}
\def\ll{\tilde\lambda}
\heading\S1 Introduction\endheading\smallskip
In this paper we  provide  formulas for the number of $ad$-nilpotent ideals of a Borel subalgebra of a complex simple Lie
algebra
$\g$ having 
fixed class of nilpotence. Our results complete the analysis made in \cite{OP}, \cite{AKOP}, where
type $A$ has been treated.
The basic framework is the following.
Let $\g$ be a  complex simple Lie algebra of rank $n$. Let
$\h\subset\g$ be a fixed Cartan subalgebra,  
$\D$ the corresponding  root system of $\g$. Fix a positive system
$\Dp$ in $\D$, and let  
$\Pi=\{\a_1,\ldots,\a_n\}$ be the corresponding basis of simple roots.
For each $\a\in \Dp$ let ${\frak g}_\a$ be the root space of $\g$ relative to $\a$, 
$\n = \bigoplus \limits_{\a\in \Dp}{\frak g}_\a$,  and $\bb$ be the Borel subalgebra
$\bb=\h\oplus \n$.\par
Let $\I$ denote the set of $ad$-nilpotent ideals (i.e., consisting of
$ad$-nilpotent elements) of $\bb$. Then $\i\in \I$ if and only if $\i$ decomposes as $\i=\bigoplus \limits_{\a\in \p}{\frak
g}_\a$, $\p\subseteq \Dp$ being a dual order ideal of the root poset $(\Dp,\leq)$. Clearly, an ideal
$\i\in
\I$ is nilpotent; let us denote by $n(\i)$ its nilpotency class.

$ad$-nilpotent ideals were studied by Kostant in \cite{K1}, \cite{K2} in connection with representation theory 
of compact semisimple Lie groups. In particular, in \cite{K2}, using ideas of Peterson, Kostant found an encoding of the
abelian  ideals by means of elements of the affine
Weyl group of $\g$; moreover, he proved that the abelian ideals are $2^n$ in number.
These results lead to ask similar questions for $ad$-nilpotent ideals: find an appropriate encoding and 
enumerate $\I$ by class of nilpotence. The first problem was settled in \cite{CP}, where an encoding in the same spirit 
of Kostant-Peterson's one was found. The following  answer to the second question for $\g$ of type $A$ was given 
in \cite{OP, Th. 1}.
\proclaim{Theorem A.1} Suppose $\g$ of type $A_n$.  Let $\a^*_n(k)$ denote the number of ideals
$\i\in\I $ with
$n(\i)=k$. Then 
$$
\a^*_n(k)=\sum_{0=i_0<i_1<\dots<i_k<i_{k+1}=n+1}\prod_{j=0}^{k-1}
\binom{i_{j+2}-i_j-1}{i_{j+1}-i_j}.\tag A.1$$\endproclaim
The combinatorial meaning of the previous formula has been  analyzed in detail in \cite{AKOP}. In type $A$ the elements of $\i$
can be naturally encoded by partitions contained in the staircase diagram (see \S2). We showed  that  formula (A.1)
reflects a decomposition, having remarkable properties w.r.t. our statistics,  of the interval
$[(0,\dots,0),(n,n-1,\ldots,1)]$ in the Young lattice. This technical fact will be the cornerstone of our  analysis in the
classical cases. Moreover, summing over $k$ in (A.1) from $0$ to $h$, one gets a formula which is known to count  the number
of Dyck paths of length
$2n+2$ and height at most $h+1$.  Generating function results for Dyck paths translate into the
following result for ideals with a given class of nilpotence. Let $U_n(x)$ denote the $n$th Chebyshev
polynomial of the second 
kind, $U_n(\cos t)=\sin((n+1)t)/\sin t$, or, explicitly,
$$U_n(x)=\sum_{j\geq 0}(-1)^j\binom{n-j}{j}(2x)^{n-2j}.$$
\proclaim{Theorem A.2}\cite{AKOP, Th. 4.6} Let $\a_n(h)$ denote the number of ideals in $\I$ with class of
nilpotence at most $h$. Then
$$\align
1+\sum_{n=0}^\infty\a_n(h) x^{n+1}&=
\frac{U_{h+1}\(1/2\sqrt x\)}{\sqrt x\,U_{h+2}\(1/2\sqrt x\)}.
\tag A.2 \endalign$$\endproclaim
Denote by 
 $\a_{n}(h,k)$ the number of such ideals with
dimension $h$ and class
of nilpotence $k$ and set
$$C_n(q,t)=\sum_{h,k\geq 0}\a_n(h,k)t^hq^k.$$
This notation reflects the fact that $C_n(1,1)$ is the $n$-th Catalan number $\frac{1}{n+1}\binom{2n}{n}$. Define the {\it
area}
$A(P)$ of a Dyck path
$P$ as the area of the region between $P$ and the $x$-axis. In \cite{AKOP, Th. 6.1} we proved that 
\proclaim{Theorem A.3}
$$C_n(q,t)=\sum_{k=0}^n
\left(\sum_{0=i_0<i_1<\dots<i_k<i_{k+1}=n+1} 
\prod_{j=0}^{k-1}t^{i_{j+1}(i_{j+3}-i_{j+2})}
\bmatrix {i_{j+2}-i_j-1}\\{i_{j+1}-i_j}\endbmatrix_t
\right)q^k,\tag A.3$$
with $i_{k+2}=n+2$. $C_n(q,1)$ is the generating function for
Dyck paths from $(0,0)$ to $(2n+2,0)$ counted with respect to height.
$C_n(1,t)$ is the
generating function for the same set of Dyck paths with respect to the
weight function ${(n+1)^2}/{2}-A(\cdot)$.\endproclaim
\smallskip
In this paper we obtain analogues of theorems (A.1), (A.2), for $\g$ of type $B,C,D$. In type $C$ we have also an analogue of
(A.3): hence (see Prop. 3.1)  we obtain a $(q,t)$-analog of the binomial coefficient $\binom{2n}{n}$. Now we state these 
results.
\proclaim{Theorem C.1} Suppose $\g$ of type $C_n$. Let $\gamma^*_n(h)$ be the number of $ad$-nilpotent ideals $\i\in\I$ with 
$n(\i)=h$. Set $i_{k+2}=2n+1$. Then 
$$\gamma^*_n(h)=\cases
\sum_{n<i_1<\dots<i_k<i_{k+1}=2n} 
\prod_{j=1}^{k-1}
\binom{i_{j+2}-i_j-1}{i_{j+1}-i_j}
\cdot
\sum _{\ell=0} ^{i_2-i_1-1}\binom{i_1+i_2-2n-1}{\ell}\\
\sum_{2n-i_2<i_1\le n<i_2<\dots<i_k<i_{k+1}=2n} 
2^{i_1+i_2-2n-1}
\prod_{j=1}^{k-1}
\binom{i_{j+2}-i_j-1}{i_{j+1}-i_j}\endcases\tag C.1.1-2$$
according to whether $h=2k$ (upper formula) or $h=2k+1$ (lower formula).\endproclaim
\par\newpage \noindent Let us write $\U_k$ as short-hand for
$U_k\(1/2\sqrt x\)$. Then
\proclaim{Theorem C.2} Let $\gamma_n(h)$ be the number of ideals $\i\in\I$  in type $C_n$ having class of nilpotence at
most $h$. Then 
$$\gamma_n(h)=\cases
\frac {\U_{2k+1}+\U_{2k-1}+\dots+\U_1} 
{\sqrt x\,\U_{2k+2}}\quad&\text{if $h=2k$}\\
\frac {\U_k} {\(\U_k-2\sqrt
x\,\U_{k-1}\)\(\U_k^2-\U_{k-1}^2\)}\quad&\text{if $h=2k+1$}\endcases\tag C.2.1-2$$
\endproclaim
We recover the connections with the theory of Dyck paths by noting that the previous generating functions count the 
number of paths with step vectors $(1,\pm1)$ of length $n$ and height at most $h+1$, starting from the origin, never falling
below the
$x$-axis and with arbitrary endpoint. Differently from type $A$ (see \cite{AKOP,\S5}), we have not been able yet to build up a
bijection between ideals of nilpotency class $h$ and paths of the previous kind and height $h+1$.\par
Finally, we state the result regarding the $(q,t)$-analog of  $\binom{2n}{n}$.
\proclaim{Theorem C.3}
Let $\gamma_n(h,k)$ be the number of ideals in type $C_n$ with
dimension $h$ and class of nilpotence $k$. Then 

$$\multline
\sum _{h,K\ge0} ^{}\gamma_n(h,K)t^hq^{K}=
\sum_{-i_2<i_1<i_2<\dots<i_k<i_{k+1}=n} q^{2k-\chi(i_1\le0)}\\
\cdot \bigg(
\prod_{j=1}^{k-1}t^{(i_{j+1}+n)(i_{j+3}-i_{j+2})}
\bmatrix {i_{j+2}-i_j-1}\\{i_{j+1}-i_j}\endbmatrix_t\\
\cdot
\sum _{\ell=0} ^{i_2-i_1-1}\bmatrix i_1+i_2-1\\\ell\endbmatrix
t^{(i_1+n)(i_3-i_2)-\binom{n-i_2+1}2+\binom {l+1}2}\bigg),
\endmultline$$
where $i_{k+2}=n+1$.\endproclaim

Let us examine the situation in type $B$.
Let $\be^*_n(K)$ denote the number of ideals $\i\in\I$ in type $B_n$ with $n(\i)=K$.  Then 
\proclaim{Theorem B.1}
The generating function $\sum _{n \ge0} ^{}\be^*_n(K)x^n$ is given by
$$\cases \dsize \frac {\U_{2k}+\U_k\U_{k+1}\U_{2k-1}} {\sqrt
x\,\U_{2k} \U_{2k+1}\U_{2k+2}}&\text{if }K=2k,\\
\dsize \frac {\U_{2k}+\U_{k+1}^2\U_{2k-2}+\U_{k-1}^2\U_{2k-2}+\U_2} {
\U_{2k-1} \U_{2k}\U_{2k+1}}&\text{if }K=2k-1.
\endcases\tag B.1.1-2$$
\endproclaim
Analogues of formula (A.1) for type $B$ will be established in the proof of the previous theorem. 
An easy consequence of the previous theorem is the following
\proclaim{Corollary B.2}
Let $\be_n(h)$ denote the number of ideals $\i\in\I$ in type $B_n$ with class of
nilpotence at most $h$. Then the
generating function $\sum _{n \ge0} ^{}\be_n(h)x^n$ is given by 
$$\cases
  {\dfrac{\sum_{i = 1}^{{h /2}}
        \left( 2\,i + 1 \right) \,\U_{2i - 1} + 
      ( h +1) \U_{h+1} + \sum_{i = 1}^{{h /2}}
        2i\,\U_{2h + 3 - 2i}  
      } {\U_{h+1}\U_{h+2}}}&\text{if $h$ is even,}\\
  {\dfrac{\sum_{i = 1}^{{(h - 1)/2}}
        \left( 2\,i + 1 \right) \,\U_{2i - 1} + 
      ( h +1) \U_h + \sum_{i = 1}^{{(h + 1)/2}}
        2i\,\U_{2h + 3 - 2i} 
      } {\U_{h+1}\U_{h+2}}}&\text{if $h$ is odd.}
\endcases\tag B.2.1-2$$\endproclaim
Our findings in type $D$ follow by combining techniques and results used in types $B,C$:
 we refer the reader to Section 6. Finally, in Section 7, we report the data relative to the exceptional types.
\smallskip
The basic ingredients needed in the proof of theorems B, C are an encoding of the dual order ideals by means of {\it shifted}
subdiagrams of the shifted staircase shape (which has been originally introduced by Shi \cite{S}) and a
careful analysis of the main construction of \cite{AKOP}. This analysis  enables us to use a kind of ``type $A$ reduction"
in the calculation of the index of nilpotence. Then the results follow quite easily in type $C$, whereas types $B, D$ need a
more refined (though elementary) analysis.  The computations relative to the exceptional types have been done
in a very straightforward way, by writing down a  computer program which  calculates explicitly the descending
central series  of each ideal.
\medskip
\heading\S2 $ad$-nilpotent ideals in type $A$: encoding and enumeration \endheading
\medskip
Let $\i\in \I$, i.e., $\i$ is 
an $ad$-nilpotent ideal of our fixed Borel subalgebra $\bb$.
Clearly, we can write $\i$ as
$\i=\bigoplus\limits_{\a\in \p}{\frak g}_\a$, for some collection $\p$ of
positive roots. The
collection $\p$ encodes an ideal $\i\in\I$ if and only if
for all $\a\in\p$ and $\b\in\Dp$ such that $\a+\b$ is a root, we have
$\a+\b\in\p.$ If one endows $\Dp$ with the (restriction of the) usual
partial order on the root lattice, that is, for $\a,\b\in\Dp$ we let 
$\a\leq\b$ if and only if $ \b-\a=\sum_{\gg\in\Dp}c_\gg\gg$, for some
nonnegative integers $c_\gg$, then this can be phrased differently as
follows: 
$\p$ encodes an ideal if and only
if it is a dual order ideal in $\Dp$.

In the rest of this section we will exclusively 
deal with $\g$ of type $A_n$, i.e., the Lie
algebra $sl(n+1,\Bbb C)$ of $(n+1)\times (n+1)$ traceless matrices. The last
observation of the previous paragraph allows us to represent
$ad$-nilpotent ideals conveniently in a geometric fashion, which will
be crucial in all subsequent considerations. Clearly, any positive
root in $A_n$ can be written as a sum of simple roots. Explicitly,  
let us write
$\tau_{ij}=\a_i+\dots+\a_{n-j+1},$  $1\leq i\le n$, $1\le j\leq
n-i+1$. If we place the roots $\tau_{ij}$, $j=1,2,\dots,n-i+1$, in the
$i$th row of a diagram, then this defines an arrangement of
the positive roots in a staircase fashion. For example, for
$A_3$ we obtain the arrangement
$$\matrix\format\l\quad &\l\quad &\l\\
\a_1+\a_2+\a_3 &\a_1+\a_2&\a_1\\
\a_2+\a_3 &\a_2 & \\
\a_3 \endmatrix$$
Obviously, the above defines an identification of positive
roots with cells of the staircase diagram $(n,n-1,\dots,1)$, 
in which the root $\tau_{ij}$
is identified with the cell $(i,j)$.\par
Let now $t_{i,j}$ be the maximal
number $m$ such that the root space ${\frak g}_{\tau_{ij}}$ occurs in 
$$\i^{m}:=\underbrace {[\cdots[\i,\i],\dots]}_{m\text{ occurrences of
}\i}.$$
By definition $n(\i)=\max_{i,j} t_{i,j}$ and indeed $n(\i)=t_{1,1}$; it is not difficult to show (see \cite{AKOP, Prop. 3.1})
that the numbers $t_{i,j}$ (hence, in particular, the class of nilpotence $n(\i)$) can be obtained as follows.
Let $\l$ be the subdiagram of $(n,n-1,\dots,1)$ that corresponds to
$\i$ according to the identification explained above. Define a
filling $(t_{i,j})_{1\le i\le n,\,1\le j\le n-i+1}$ 
of the cells of $(n,n-1,\dots,1)$ by recursively setting
$$t_{i,j}=\cases 0\quad &\text{if\ } (i,j)\notin \lambda,\\1\quad &\text{if\
} (i,j)\text{ is a corner cell of }\lambda,\\
\max\limits_{j<k\leq n-i+1}\ \{t_{i,k}+t_{n-k+2,j}\}\quad
&\text{otherwise.}\endcases
\tag 2.1 $$ 
It is easy to see that the above rule uniquely defines a tableau  of
shape $(n,n-1,\dots,1)$, whose nonzero entries are precisely those
corresponding to the cells of $\l$. E.g., when $n=4$, the fillings 
corresponding to 
$(2,1,0,0)$, $(3,3,2,1)$, $(4,3,2,1)$ are respectively
$$
\matrix
1&1&0&0\\1&0&0\\0&0\\0\endmatrix\qquad
\matrix
3&2&1&0\\3&2&1\\2&1\\1\endmatrix\qquad
\matrix
4&3&2&1\\3&2&1\\2&1\\1\endmatrix$$

From the previous algorithm it is possible to  deduce a quick way to obtain the class of nilpotence of a single ideal $\i$. For
a convenient statement of the result, we write, in abuse of notation, $n(\l_1,\l_2,\ldots,\l_n)$ for $n(\i)$, given that the
partition corresponding to $\i$ according to the construction in
Section~2 is $(\l_1,\l_2,\ldots,\l_n)$.

\proclaim{Proposition 2.1}\cite{AKOP, Prop. 3.2}
Let $\i\in\I$ and let 
$\l=(\l_1,\ldots,\l_n)$ be the corresponding partition.
If $\l\ne(0,0,\ldots,0)$ then
$$n(\l_1,\l_2,\ldots,\l_n)=n(\l_{n+2-\l_1},\ldots,\l_n)+1.\tag 2.2$$
\endproclaim

It should be noted that on the left-hand side of (2.2) appears
the class of nilpotence of an ideal in $\I$, whereas on the right-hand
side there appears the class of nilpotence of an ideal in $\Cal
I^{\l_1-1}$ (with corresponding partition
$(\l_{n+2-\l_1},\ldots,\l_n)$). 
The computation, however, can be carried out completely formally,
without reference to ideals, which we now demonstrate by an example.

\example{Example}
Let $\i\in\Cal I^{13}$ be the ideal which corresponds to the partition 
$(10,10,9,6,\mathbreak 5,4,4,3,1,1,1,1,0)$. (This is the partition in
Figure~1. At this point, all dotted lines should be ignored.) 
Then, by applying Proposition~2.1
iteratively, we obtain for the class of nilpotence of $\i$
$$\align
n(\i)&=n(10,10,9,6,5,4,4,3,1,1,1,1,0)\\ 
&= n(5,4,4,3,1,1,1,1,0)+1\\
&= n(1,1,1,0)+2\\
&=3.
\endalign$$
\endexample

As is obvious from the example, iterated application of
Proposition~2.1 provides a very efficient algorithm of determining the
class of nilpotence of a given ideal $\i$. 

Since it will be essential in the following, 
we wish to point out that this algorithm has a very nice geometric
rendering. Let, as before, $\l=(\l_1,\l_2,\dots,\l_n)$ 
be the partition corresponding to
$\i$. Consider the Ferrers diagram of $\l$. As it is contained in the
staircase diagram $(n,n-1,\dots,1)$, it must not cross the
antidiagonal line $x+y=n+1$.
We draw a zig-zag line
as follows (see Figure~1, where $n=13$ and
$\l=(10,10,9,6,5,4,4,3,1,1,1,1,0)$): 
we start on the vertical edge on the right of cell
$(1,\l_1)$, and move downward until we touch the antidiagonal
$x+y=n+1$. At the touching point we turn direction from 
vertical-down to horizontal-left, and move on until we touch a vertical
part of the Ferrers diagram. At the touching point we turn direction
from  horizontal-left to vertical-down. Now the procedure is
iterated, until we reach the line $x=0$. The class of nilpotence of
the ideal $\i$ is equal to the number of touching points on
$x+y=n+1$. In Figure~1, the resulting zig-zag line is the dotted line
outside the Ferrers diagram of $(10,10,9,6,5,4,4,3,1,1,1,1,0)$.
There are three touching points on $x+y=n+1=14$. (At this
point, the dotted lines inside the diagram should still be ignored.) 

\midinsert
\vskip10pt
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\thinlines
\Diagonale(0,-1){14}
\Label\r{x+y=n+1}(15,12)
\DuennPunkt(1,0)
\DuennPunkt(5,4)
\DuennPunkt(10,9)
\hskip7cm
$$
\centerline{\eightpoint Figure 1}
}
\vskip10pt
\endinsert

Let us formalize the main technical lemma of \cite{AKOP}. Given 
a $k$-element subset $1\leq i_1<i_2<\ldots<i_k\leq n$ of $[n]$, set 
$$\align
\l^M_{i_1,\dots,i_k}&=\left( i_k^{n-i_k+1},
i_{k-1}^{i_k-i_{k-1}},\ldots,i_1^{i_2-i_1},0^{i_1-1}\right),
\\
\l^m_{i_1,\dots,i_k}&=\left( i_k, i_{k-1}^{n-i_k+1},
i_{k-2}^{i_k-i_{k-1}},\ldots,i_1^{i_3-i_2},0^{i_2-2}\right).\endalign
$$
\proclaim{Proposition 2.2}\cite{AKOP, Prop 4.2}  The interval
$I=[(0,\ldots,0),(n,n-1,\ldots,1)]$ in the Young lattice can be
decomposed into disjoint subintervals as  
$$I=\bigcup_{k=0}^n\,\,\bigcup_{0=i_0<i_1<\dots<i_k<i_{k+1}=n+1}
[\l^m_{i_1,\dots,i_k},\l^M_{i_1,\dots,i_k}].$$
Moreover, let $\i$ be an ideal with corresponding partition
$\l$. If $\l\in[\l^m_{i_1,\dots,i_k},\l^M_{i_1,\dots,i_k}]$,
then the class of nilpotence of $\i$ is equal to $k$.\endproclaim\par\newpage 
In the previous example, the partitions $\l^M_{i_1,\dots,i_k}, \l^m_{i_1,\dots,i_k}$ correspond (see Figure~1) 
to the bottom-most, top-most zig-zag lines respectively. So the previous Proposition states that
any partition in between $\l^M_{i_1,\dots,i_k}, \l^m_{i_1,\dots,i_k}$ (w.r.t. inclusion of diagrams) has the same class of
nilpotence
$k$. Conversely, given a partition $\l$ contained in the staircase shape, one recovers the subinterval
$[\l^m_{i_1,\dots,i_k}, \l^M_{i_1,\dots,i_k}]$ to which $\l$ belongs by performing the geometric construction previously
described, and
more precisely by determining the touching points $i_1,\ldots,i_k$.\par
Note that the previous Proposition affords a fairly easy proof of $(A.1), (A.3)$. 
%\par\newpage
\medskip
\heading\S3 $ad$-nilpotent ideals in types $B, C, D$ \endheading
\medskip Let $\g$ be a  complex simple Lie algebra of  type  $B_n,\,C_n,$ or $D_n$. We describe a  diagramatic encoding of the
positive roots (cf. \cite{S, \S2-3}) similar to the one introduced in the previous Section for type $A$.\par Display the whole
$\Dp$ in a shifted staircase diagram 
$T_n$ of shape
$(2n-1,2n-3,\ldots,1)$ for
$B_n$ and
$C_n$ and of shape $(2n-2,2n-4,\ldots,2)$ for $D_n$ as shown in the following two examples (the precise formula is given
below):\smallskip\noindent
$\g\cong B_3$
$$\alignat5
&\a_1+2\a_2+2\a_3\qquad &&\a_1+\a_2+2\a_3\qquad &&\a_1+\a_2+\a_3\qquad &&\a_1+\a_2\qquad &&\a_1\\
& &&\a_2+2\a_3 &&\a_2+\a_3 &&\a_2 && \\
& && &&\a_3 &&  &&\endalignat$$
$\g\cong C_3$
$$\alignat5
&2\a_1+2\a_2+\a_3\qquad &&\a_1+2\a_2+\a_3\qquad &&\a_1+\a_2+\a_3\qquad &&\a_1+\a_2\qquad &&\a_1\\
& &&2\a_2+\a_3 &&\a_2+\a_3 &&\a_2 && \\
& && &&\a_3 &&  &&\endalignat$$
$\g\cong D_4$
$$\eightpoint\alignat6
&\a_1+2\a_2+\a_3+\a_4\qquad &&\a_1+\a_2+\a_3+\a_4\qquad&&\a_1+\a_2+\a_4\qquad &&\a_1+\a_2+\a_3\qquad &&\a_1+\a_2\qquad&&\a_1\\
 & &&\a_2+\a_3+\a_4 &&\a_2+\a_4 &&\a_2+\a_3&&\a_2 && \\
& && &&\a_4&&\a_3 &&  &&\endalignat$$
We need the following {\it ad hoc} definition for type $D$: if $A\subseteq (2n-2,2n-4,\ldots,2)$ is a collection of boxes,
denote by $A^\bullet$ the collection of boxes obtained from $A$ by switching columns $n-1,n$.
\proclaim{Proposition 3.1} The dual order ideals of $(\Dp,\leq)$ correspond bijectively to the
subdiagrams of
$T_n$ when
$\g$ is of type $B_n$ or $C_n$; if $\g$ is of type $D_n$ there is a bijection between the dual order ideals and the collection
of boxes
 $A\subseteq T_n$ such that  either $A$  or $A^\bullet$ is a subdiagram. Moreover 
$$
|\I|=\cases\binom{2n}{n}\qquad &\text{ if $\g$ is of type $B_n$ or $C_n$,}\\
\binom{2n}{n}-\binom{2n-2}{n-1}\qquad &\text{ if $\g$ is of type $D_n$.}\endcases$$\endproclaim
\noindent{\sl Remark.} The previous Proposition is proved in \cite{S, \S3}. In \cite{CP, Th. 3.1}, using a result of Proctor,
a quick non-bijective proof of the enumerative result is given. A simple and bijective proof of the enumerative result in
the relevant cases $B, C$ (case $D$ follows as a consequence) can be obtained as follows.
Consider the right outer border of a given subdiagram (i.e. the split line connecting the point $(2n,0)$ with  the bottom-most
point in the diagram lying on the line $x=y$. Up to a rotation of $\frac{\pi}{4}$ in the positive direction followed by a
suitable rescaling and reflection, this border  may be viewed as a path  with step vectors $(1,\pm 1)$  starting from the
origin (with arbitrary end point), of length
$n$, and never falling below the $x$-axis. In the Appendix  we  build up a bijection between such paths and the set of paths
with step vectors $(1,\pm 1)$  starting from the origin and ending in $(2n,0)$. Since the paths in the latter class 
are clearly $\binom{2n}{n}$ in number, the claim follows.
%\par\newpage
\smallskip
Suppose $\g$ of type $B_n$ or $C_n$ (the $D_n$ case will be treated in remark at the end of this section).
We explain how to calculate the class of nilpotence by means of the   algorithm (2.1) used in type $A$. The general
procedure will be as follows: we  embed the shifted staircase $T_n$ into the ordinary staircase $\T_{2n-1}$, in such a way
that  to any shifted subdiagram $\l\subseteq T_n$ corresponds uniquely an ordinary subdiagram $\ll$ of $\T_{2n-1}$. 
We will eventually  prove that $n(\l)=n(\ll)$, so we will be reduced, as far as the determination of the class of nilpotence
is concerned,  to a calculation in type
$A$. Moreover we shall be able to determine the image of the map $\l\mapsto\ll$.\smallskip
Fill in $\T_{2n-1}$ with elements in the root lattice $Q$ of $B_n,C_n$ as follows (in these examples $n=3$):
$$\alignat5
&2\a_1+2\a_2+\a_3\qquad &&\a_1+2\a_2+\a_3\qquad &&\a_1+\a_2+\a_3\qquad &&\a_1+\a_2\qquad &&\a_1\\
&\a_1+2\a_2+\a_3 &&2\a_2+\a_3 &&\a_2+\a_3 &&\a_2 && \\
&\a_1+\a_2+\a_3 &&\a_2+\a_3 &&\a_3\\
&\a_1+\a_2 &&\a_2\\
&\a_1\endalignat$$
$$\alignat5
&\a_1+2\a_2+2\a_3\qquad &&\a_1+\a_2+2\a_3\qquad &&\a_1+\a_2+\a_3\qquad &&\a_1+\a_2\qquad &&\a_1\\
&2\a_2+2\a_3 &&\a_2+2\a_3 &&\a_2+\a_3 &&\a_2 && \\
&\a_2+2\a_3 &&2\a_3 &&\a_3\\
&\a_2+\a_3 &&\a_2\\
&\a_3\endalignat$$
In general:
$$\alignat2
&\g\cong C_n:\qquad &&t_{ij}=\cases \a_i+\ldots+2(\a_{j}+\ldots\a_{n-1})+\a_n\qquad & 1\leq j\leq n-1,\\
\a_i+\ldots+\a_{2n-j}\qquad &n\leq j\leq 2n-1,\endcases\\
&\g\cong B_n:\qquad &&t_{ij}=\cases \a_i+\ldots+2(\a_{j+1}+\ldots+\a_n)\qquad & 1\leq j\leq n-1,\\
\a_i+\ldots+\a_{2n-j}\qquad &n\leq j\leq 2n-1.\endcases\endalignat$$

Fill in $\T_{2n-2}$ with elements in the root lattice $Q$ of $D_n$ as follows (in this example $n=4$):
$$\eightpoint\alignat6
&\a_1+2\a_2+\a_3+\a_4\qquad &&\a_1+\a_2+\a_3+\a_4\qquad&&\a_1+\a_2+\a_4\qquad &&\a_1+\a_2+\a_3\qquad &&\a_1+\a_2\qquad&&\a_1\\
 &2\a_2+\a_3+\a_4 &&\a_2+\a_3+\a_4 &&\a_2+\a_4 &&\a_2+\a_3&&\a_2 && \\
&\a_2+\a_3+\a_4 &&\a_3+\a_4\a_4 &&\a_4&&\a_3\\
&\a_2+\a_4 &&\a_4&&\times\\
&\a_2+\a_3 &&\a_3\\
&\a_2\endalignat$$
In general: 
$$\g\cong D_n:\quad t_{ij}=\cases \a_i+\ldots+2(\a_{j+1}+\ldots+\a_{n-2})+\a_{n-1}+\a_n\ & 1\leq j\leq n-2,\\
\a_i+\ldots+\a_{n-2}+\a_n\ &j=n-1,\\
\a_i+\ldots+\a_{2n-j}\ &n\leq j\leq 2n-1.\endcases$$
\smallskip
 Now consider a shifted subdiagram  $\l\subseteq T$:  we get in a natural way an ordinary  subdiagram $\ll=\phi(\l)$ of $\T$
 by taking  the union of the North-West corners of the roots enclosed in $\l$
(if a  root appears twice in the filling of $\Cal T$, two North-west
corners should be considered). E.g. for $B_3$ and $\l=(4,2,1)$ we obtain $\ll=(4,3,3,2,0)$.\par
Note that in types $B,D$ the elements in the tableau which are not roots are exactly the ones lying on the diagonal
$x=y-1$. Then, by considering the displacement of the roots 
in $\T$,  it is not difficult to deduce the following
\proclaim{Proposition 3.2} If $\g\cong C_n$, then $Im (\phi)$ consists of the partitions $\mu\subseteq \T_{2n-1}$ which are
self-conjugated. If $\g\cong B_n$, then $Im (\phi)$ consists of the partitions\break  $\mu=(\mu_1,\ldots,\mu_{2n-1})\subseteq
\T_{2n-1}$ such that $(\mu_2,\ldots,\mu_{2n-1})$ is self-conjugated and $\mu_i\ne i-1$ for $i\geq 2$.\endproclaim
\proclaim{Proposition 3.3} If $\g\cong C_n, B_n$ then $n(\l)=n(\ll)$.\endproclaim
\demo{Proof} Suppose that $\l$ encodes the ideal $\i_\p$; then, by definition, 
$$\align
n(\l)&= \max_{\a\in\p}\left\{k_\a\mid \a=\b_1+\cdots+\b_{k_\a},\ \b_j\in\p\right\}\\
&= \max\left\{k\mid \theta=\b_1+\cdots+\b_{k},\ \b_j\in\p\right\},\endalign$$
$\theta$ being the highest root of $\Dp$.
Now remark that performing the  algorithm (2.1) on $\ll$ all the decompositions of any root in $\i_\p$ are
considered, so the claim follows in this case. When considering case $B_n$, the  algorithm (2.1) involves decompositions
of the form 
$\a=\b+\xi,\ \a\,\b\in\Dp,\ \xi\in Q\setminus \Dp$; on the other hand the
hypothesis 
$\mu_i\ne i-1$ for
$i\geq 2$ guarantees that the box of $\T$ relative to  $\xi$ is filled with $h>0$  if and only if
$\xi$ is in turn a sum of $h$ roots in
$\p$ (in other terms, no box of $\T$ filled with elements from $Q\setminus \Dp$ can be a corner cell of an admissible diagram
$\ll$).
Thus, considering these elements does not affect the final result of the algorithm.\endemo
\smallskip\noindent{\sl Remark.} Let us sketch our strategy in type $D$. Recall that the  ideals are parameterized by
subdiagrams of 
$T=(2n-2,2n-4,...,2)$ {\it and} subsets of boxes of $T$ which are not
subdiagrams and which become subdiagrams after switching columns $n-1,n$.  The switching operation  mirrors an automorphism
of $\g$, so  any ideal parameterized by a ``non-diagram" $A$ is isomorphic to the ideal parameterized to the corresponding
diagram $A^\bullet$. Consider the following classes of subdiagrams:
\roster
\item partitions
$(\mu_1,...,\mu_{2n-2})\subseteq \T_{2n-2}$ with $(\mu_2,...,\mu_{2n-2})$ self-conjugate and $\mu_i\not= i-1$ for $i\geq 2$;
\item partitions as above with columns $n-1,n$ of the same length.
\endroster
Then, if $\sum\limits_{n\geq 0}\d^*_n(k)x^n,\,\sum\limits_{n\geq 0}\d^1_n(k)x^n,\,\sum\limits_{n\geq 0}\d^2_n(k)x^n$
denote  the generating functions for all the ideals $\i\in \I$ (resp. the ones relative to partitions of class (1), (2)) of
type
$D_n$ with $n(\i)=k$, one has
$$\d^*_n(k)=2\d^1_n(k)-\d^2_n(k).\tag 3.1$$
It turns out that $\d^2_n(k)=\b^*_{n-1}(k)$, so we are reduced to determine $\d^1_n(k)$, and this will be done mimicking 
the type $B$ case: details are given in Section $6$.
%\medskip
\par\newpage
\heading \S4 Type $C$\endheading
\medskip
The strategy used in  case $C_n$ is based on the following observation:
instead of making the (strict) partition corresponding to an ideal
into a self-conjugate partition and
then applying the $A_{2n-1}$ algorithm to the self-conjugate
partition, we may proceed as follows. Draw the strict partition
$(\l_1,\l_2,\dots)$. Start
as in the $A_{2n-1}$ algorithm. I.e., begin on the vertical edge on
the right of cell $(1,\l_1)$, and move 
downward until we touch the antidiagonal
$x+y=2n$. At the touching point we turn direction from 
vertical-down to horizontal-left, and move on until we touch a vertical
part of the Ferrers diagram. At the touching point we turn direction
from  horizontal-left to vertical-down. We iterate until we meet the
diagonal $x=y$. Let $k$ be the number of touching points on the
antidiagonal $x+y=2n$. If we meet the diagonal $x=y$ while traveling
horizontal-left, then the class of nilpotence is $2k$, if we meet the
diagonal while traveling vertical-downward, then the class of
nilpotence is $2k+1$. For example, if $n=9$, then for the partition
$(16,13,11,8,7,5,3)$ in Figure~2 the class of nilpotence is $2\cdot
3=6$, whereas for the partition 
$(16,13,11,8,5,3,1)$ in Figure~3 the class of nilpotence is $2\cdot
2+1=5$. \par
 
%\midinsert
\vskip10pt
\vbox{
$$
\Pfad(-9,9),1111111111111111\endPfad
\Pfad(-3,2),11121212211212112\endPfad
\Pfad(-9,8),2\endPfad
\Pfad(-9,8),1\endPfad
\Pfad(-8,7),2\endPfad
\Pfad(-8,7),1\endPfad
\Pfad(-7,6),2\endPfad
\Pfad(-7,6),1\endPfad
\Pfad(-6,5),2\endPfad
\Pfad(-6,5),1\endPfad
\Pfad(-5,4),2\endPfad
\Pfad(-5,4),1\endPfad
\Pfad(-4,3),2\endPfad
\Pfad(-4,3),1\endPfad
\Pfad(-3,2),2\endPfad
\SPfad(4,7),1112\endSPfad
\SPfad(2,4),1122\endSPfad
\SPfad(-1,1),1122\endSPfad
\thinlines
\Diagonale(-1,-1){11}
\Label\r{x+y=2n}(10,8)
\AntiDiagonale(-10,10){11}
\Label\o{x=y}(-10,10)
\DuennPunkt(1,1)
\DuennPunkt(4,4)
\DuennPunkt(7,7)
\hskip0cm
$$
\centerline{\eightpoint Figure 2}
}
%\vskip10pt
%\endinsert

%\midinsert
%\vskip10pt
\vbox{
$$
\Pfad(-9,9),1111111111111111\endPfad
\Pfad(-3,2),12121211211212112\endPfad
\Pfad(-9,8),2\endPfad
\Pfad(-9,8),1\endPfad
\Pfad(-8,7),2\endPfad
\Pfad(-8,7),1\endPfad
\Pfad(-7,6),2\endPfad
\Pfad(-7,6),1\endPfad
\Pfad(-6,5),2\endPfad
\Pfad(-6,5),1\endPfad
\Pfad(-5,4),2\endPfad
\Pfad(-5,4),1\endPfad
\Pfad(-4,3),2\endPfad
\Pfad(-4,3),1\endPfad
\Pfad(-3,2),2\endPfad
\SPfad(4,7),1112\endSPfad
\SPfad(-1,4),1111122\endSPfad
\SPfad(-1,1),22\endSPfad
\thinlines
\Diagonale(-1,-1){11}
\Label\r{x+y=2n}(10,8)
\AntiDiagonale(-10,10){11}
\Label\o{x=y}(-10,10)
\DuennPunkt(4,4)
\DuennPunkt(7,7)
\hskip0cm
$$
\centerline{\eightpoint Figure 3}
}
\vskip10pt
%\endinsert
\par\newpage
To prove these claims, first remark that the previous procedure affords a split line which can be regarded as the right border 
of a strict partition $\l^M_{i_1,\ldots,i_k}$. Denote by $\l^m_{i_1,\ldots,i_k}$ the strict partition 
whose corner cells are $(1,i_k),\,(2n-i_k+1,i_{k-1}),\ldots,$ $(2n-i_2+1,i_1)$. We claim  that any strict partition
$\l,\ \l^m_{i_1,\ldots,i_k}\subseteq\l\subseteq \l^M_{i_1,\ldots,i_k}$ has class of nilpotence $2k$ in the first case or
$2k-1$ in the other. Set $\tilde \l_m=\widetilde{\l^m_{i_1,\ldots,i_k}},\,$ $\tilde\l_M=\widetilde{\l^M_{i_1,\ldots,i_k}}$.
Clearly, $\tilde \l_m\subseteq\tilde \l\subseteq \tilde\l^M$; by Proposition 3.3, $n(\l)=n(\tilde\l)$,  so it suffices to show
that $n(\tilde \l_m)=n(\tilde \l_M)=2k$ in the former case and $n(\tilde \l_m)=n(\tilde \l_M)=2k-1$ in the latter.
To determine $n(\tilde \l_M)$ we should find, according to \S2, the relative ``maximal" partition, but since we are dealing
with self-conjugate partitions we can also proceed as follows. Begin as above moving on the split line which is the 
right border of $\l^M_{i_1,\ldots,i_k}$. The final point of this split line lies on the line $x=y$; now we move
horizontal-right until we meet the line $x+y=2n$; at the touching point we turn direction from horizontal-right to
vertical-up. When we reach the right border of our partition we again turn direction from vertical-up to horizontal-right
following the shape of the partition. We iterate the procedure, which finishes when we reach the point $(2n,0)$. The total
number of touching points with $x+y=2n$ is the class of nilpotence. Now the geometric rendering of
$\l^m_{i_1,\ldots,i_k},\l^M_{i_1,\ldots,i_k}$ makes clear that we have the desired result.
We have also implicitly determined  the decomposition of
the interval $[\l^m_{i_1,\ldots,i_k},\l^M_{i_1,\ldots,i_k}]$ (w.r.t. inclusion of strict partitions) as a 
product of Young lattices; for instance, in the first case we have 
$$\multline
\left[\emptyset,\((i_k-i_{k-1})^{n-i_k}\)\right]\times
\left[\emptyset,\((i_{k-1}-i_{k-2})^{i_k-i_{k-1}-1}\)\right]\\
\times
\left[\emptyset,\((i_{k-2}-i_{k-3})^{i_{k-1}-i_{k-2}-1}\)\right]\times\dots
\times\left[\emptyset,\((i_{2}-i_{1})^{i_{3}-i_{2}-1}\)\right]\\ \times
\coprod_{i_1\leq j< i_2}\left[\emptyset,\(i_1-2n+j)^{i_2-j-1}\)\right].
\endmultline$$
Note the contribution in the last line of the previous formula is relative
to the lattice paths form $(2n-j,2n-j)$ to $(i_1, i_2-2n+1), \ i_1\leq j< i_2$. 
Summarizing we
obtain
$$\multline
\sum _{h\ge0} ^{}\gamma_n(h,2k)t^h=
\sum_{n<i_1<\dots<i_k<i_{k+1}=2n} 
\prod_{j=1}^{k-1}t^{i_{j+1}(i_{j+3}-i_{j+2})}
\bmatrix {i_{j+2}-i_j-1}\\{i_{j+1}-i_j}\endbmatrix_t\\
\cdot
\sum _{\ell=0} ^{i_2-i_1-1}\bmatrix i_1+i_2-2n-1\\\ell\endbmatrix
t^{i_1(i_3-i_2)-\binom{2n-i_2+1}2+\binom {l+1}2}
\endmultline\tag C.1.1$$
(where $i_{k+2}=2n+1$) for the generating function for ideals with
nilpotence $2k$, and
$$\multline
\sum _{h\ge0} ^{}\gamma_n(h,2k-1)t^h=
\sum_{2n-i_2<i_1\le n<i_2<\dots<i_k<i_{k+1}=2n} 
t^{i_1(i_3-i_2)-\binom{2n-i_2+1}2}\\
\cdot
(1+t)(1+t^2)\cdots(1+t^{i_1+i_2-2n-1})
\prod_{j=1}^{k-1}t^{i_{j+1}(i_{j+3}-i_{j+2})}
\bmatrix {i_{j+2}-i_j-1}\\{i_{j+1}-i_j}\endbmatrix_t
\endmultline\tag C.1.2$$
(where again $i_{k+2}=2n+1$) for the generating function for ideals with
nilpotence $2k-1$. Now recall 
that the generating function $\sum _{\l} ^{}t^{\v\l}$
summed over all partitions $\l$ which
are contained in an $a\times b$ rectangle
is the $t$-binomial coefficient
$\left[\smallmatrix {a+b}\\ {b} \endsmallmatrix\right]_t$. 
 After having done the substitution
$i_j\to i_j+n$ and combining both formulas, we have the following analog of Theorem A.3.
\proclaim{Theorem C.3}
$$\multline
\sum _{h,K\ge0} ^{}\gamma_n(h,K)t^hq^{K}=
\sum_{-i_2<i_1<i_2<\dots<i_k<i_{k+1}=n} q^{2k-\chi(i_1\le0)}\\
\cdot \bigg(
\prod_{j=1}^{k-1}t^{(i_{j+1}+n)(i_{j+3}-i_{j+2})}
\bmatrix {i_{j+2}-i_j-1}\\{i_{j+1}-i_j}\endbmatrix_t\\
\cdot
\sum _{\ell=0} ^{i_2-i_1-1}\bmatrix i_1+i_2-1\\\ell\endbmatrix
t^{(i_1+n)(i_3-i_2)-\binom{n-i_2+1}2+\binom {l+1}2}\bigg),
\endmultline$$
where $i_{k+2}=n+1$ and $\chi(X)$ is the characteristic function of the set $X$.\endproclaim
\medskip
Note that the specialization $t=1$ affords Theorem C.1. 
Now we seek for a type $C$ analogue of Theorem A.2.
\demo{Proof of (C.2.1)}
We want to compute next  the number of ideals with class of
nilpotence at most $2k$. In principle, we could sum up the above
formulas with $t=1$, but it is more convenient to use an algorithm
which determines whether the class of nilpotence is at most $k$. This
algorithm works as follows. Again we are given a strict partition
$\l=(\l_1,\l_2,\dots)$. This time we start at the diagonal $x=y$ (!),
at the lowest intersection point with the Ferrers diagram of $\l$. We
move right until we touch the antidiagonal $x+y=2n$. At the
touching point we turn direction from horizontal-right to vertical-up,
and move on until we touch a horizontal part of the Ferrers
diagram. At the touching point we turn direction from vertical-up to
horizontal-right. Etc. The class of nilpotence is at most twice the
number of touching points on $x+y=2n$. See Figure~4, where 
this procedure is applied to the partition of Figure~2.

\midinsert
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\thinlines
\Diagonale(-1,-1){11}
\Label\r{x+y=2n}(10,8)
\AntiDiagonale(-10,10){11}
\Label\o{x=y}(-10,10)
\DuennPunkt(2,2)
\DuennPunkt(6,6)
\DuennPunkt(8,8)
\hskip0cm
$$
\centerline{\eightpoint Figure 4}
}
\vskip10pt
\endinsert

If we now use the idea of the proof of Theorem~6.1 to this situation
(in a slightly modified form),
then we get 
$$\underset j_1,j_2,\dots,j_k\ge1,\,j_{k+1}\ge0\to
{\sum _{j_1+j_2+\dots+j_{k+1}=n}} ^{}\binom {j_1+j_2-1}{j_1}
\binom {j_2+j_3-1}{j_2}\cdots \binom {j_{k-1}+j_k-1}{j_{k-1}}
\binom {j_{k}+2j_{k+1}}{j_{k}}.$$
for the number of ideals with class of nilpotence exactly $2k$, and
$$\underset j_1,j_2,\dots,j_k,j_{k+1}\ge0\to
{\sum _{j_1+j_2+\dots+j_{k+1}=n}} ^{}\binom {j_1+j_2-1}{j_1}
\binom {j_2+j_3-1}{j_2}\cdots \binom {j_{k-1}+j_k-1}{j_{k-1}}
\binom {j_{k}+2j_{k+1}}{j_{k}}.$$
for the number of ideals with class of nilpotence at most $2k$.
Let us denote the latter number by $\bar\gamma_n(2k)$. Now we can
easily compute the generating function $\sum _{n\ge0}
^{}\bar\gamma_n(2k)x^n$ for the ideals of nilpotence at most $k$. We
substitute the previous expression in the generating function,
$$\multline
{\sum _{j_1,j_2,\dots,j_k,j_{k+1}\ge0}} ^{}
x^{j_1+\cdots+j_k+j_{k+1}}
\binom {j_1+j_2-1}{j_1}
\binom {j_2+j_3-1}{j_2}\\
\cdots \binom {j_{k-1}+j_k-1}{j_{k-1}}
\binom {j_{k}+2j_{k+1}}{j_{k}}.
\endmultline$$
Now we perform the summation over $j_1$ and obtain $(1/(1-x))^{j_2}$
in the sum. Next the summation over $j_2$ is performed, etc.  Let us write $\U_k$ as short-hand for $U_k\(1/2\sqrt
x\)$. If we use
the fact that 
$$\frac{\U_{h}}{\sqrt x\,\U_{h+1}}
=\cfrac 1\\1-
\cfrac x\\1-
\cfrac x\\
\cfrac \ddots\\1-x\endcfrac\endcfrac\endcfrac\endcfrac\ ,
\tag 4.1$$
where $x$ occurs $h$ times in the continued fraction, then we finally
end up with
$$\sum _{j_{k+1}\ge0} ^{}\frac {1} {\sqrt x}\(\frac {\U_k} {\U_{k+1}}\)^{2j_{k+1+1}}.$$
The sum is easily evaluated as it is just a geometric series. This
gives
$$\frac {\U_k\U_{k+1}} 
{\sqrt x\,\(\U_{k+1}^2-\U_{k}^2\)}.$$
Now, it is easy to verify that
$$U_k(x)U_{k+1}(x)=U_{2k+1}(x)+U_{2k-1}(x)+\dots+U_1(x)$$
and
$$U_{k+1}^2(x)-U_k^2(x)=U_{2k+2}(x).$$
Hence, our generating function equals
$$\frac {\U_{2k+1}+\U_{2k-1}+\dots+\U_1} 
{\sqrt x\,\U_{2k+2}}.$$
This proves the first formula of Theorem (C.2).\endemo
\noindent{\sl Remark.}
As it is also known that the generating function $\sum _{P}
^{}x^{\ell(P)/2}$ for Dyck paths starting at the origin, never
exceeding height $2k+1$, and ending
at height $h$ ($\ell(P)$ denoting the length of $P$) is given by
$$\frac {\U_{2k+1-h}} {\sqrt x\,\U_{2k+2}},$$
(see for example [Kr, Theorem~A2])
it follows that the number of ideals in type $C_n$ with class of
nilpotence at most $2k$ is equal to the number of Dyck paths of length
$2n$ which start at the origin and never exceed height $2k+1$.
\demo{Proof of (C.2.2)}
We compute  $\sum _{n\ge0}
\gamma^*(2k-1)x^n$. The case $k=1$ has to be done
separately, but this is trivial. (This is already included in
Peterson's result.) So let us assume $k\ge2$. If we substitute the
formula (C.1.2) from the first page in the generating function, we obtain
$$\multline
\sum _{n\ge0} ^{}
\sum_{2n-i_2<i_1\le n<i_2<\dots<i_k<2n} x^n
\pmatrix {2n-i_{k-1}-1}\\{i_{k}-i_{k-1}}\endpmatrix
\pmatrix {i_{k}-i_{k-2}-1}\\{i_{k-1}-i_{k-2}}\endpmatrix\\
\cdots
\pmatrix {i_{3}-i_1-1}\\{i_{2}-i_1}\endpmatrix
2^{i_1+i_2-2n-1}
\endmultline$$
for our generating function. As before, the sums over
$n,i_k,i_{k-1},\dots,i_3$ are easily computed. On using again that the
resulting continued fractions can be expressed in terms of
Chebyshev polynomials, and after having replaced $i_1$ by $i_1+n$ and
$i_2$ by $i_2+n$, we obtain 
$$\sum _{-i_2<i_1\le 0<i_2} ^{}x^{i_2+k-1}\(\frac {\U_1}
{x^{k/2-1}\,\U_{k-1}}\)^2 \(\frac {\U_{k-1}} 
{\sqrt x\,\U_k}\)^{i_2-i_1+1}2^{i_1+i_2-1}.$$
Next we sum over $i_1$, and subsequently over $i_2$. In both cases, it
is just geometric series that have to be summed, 
a terminating series when summing over $i_1$
and two nonterminating series when summing over $i_2$. The result,
after cancellation of factors, is
$$\frac {\U_k} {\(\U_k-2\sqrt
x\,\U_{k-1}\)\(\U_k^2-\U_{k-1}^2\)}.$$
It is now routine to verify that this is equal to
$$
\frac {\U_{2k}+\U_{2k-2}+\dots+\U_0} 
{\sqrt x\,\U_{2k+1}}
-\frac {\U_{2k-1}+\U_{2k-3}+\dots+\U_1} 
{\sqrt x\,\U_{2k}}.
$$
At this point Theorem (C.2) is completely proved.\endemo\par\newpage
\noindent{\sl Remark.}
By the result on Dyck paths  quoted before,
the previous formula appears as the generating function for Dyck paths, starting at the
origin, ending arbitrarily, with height {\it exactly} $2k$. (The first
part counts the paths with height at most $2k$, whereas the part which
is subtracted counts the paths with height at most $2k-1$.)
Therefore, if we combine, we obtain the following result.
\proclaim{Proposition 4.1}  The $ad$-nilpotent  ideals of a Borel subalgebra of a Lie algebra of type $C_n$ 
having nilpotency class $K$ are as many as the   Dyck paths of length $2n$, starting at the
origin, ending arbitrarily, with height  $K+1$. Moreover the number of ideals having nilpotency class at most $h$ is given by
$$\sum _{s=0} ^{\fl{h/2}}\sum _{k\in\Bbb Z} ^{}\frac {1+2s+2k(h+2)}
{2n+1} \binom {2n+1}{n-s-k(h+2)}.$$
\endproclaim
\demo{Proof} We have only to prove the second statement. This statement follows from 
 the iterated reflection formula  of Mohanty's book \cite{M} 
 (cf. \cite{AKOP, Th. 4.5} for an analog result in type $A$).\endemo
\smallskip
Using Theorem C.2, it is easy to deduce the following result (cf. \cite{AKOP, Corollary 4.7}). Let $F_n$ denote the n-th
Fibonacci number.
\proclaim{Corollary 4.2}
For $n\ge1$
the number of ideals $\i\in\I$ in type $C_n$ with $n(\i)\leq 2$ is $F_{2n}$, and 
the number of ideals $n(\i)\leq 3$ is $2\cdot 3^{n-1}$.
\endproclaim
%\par\newpage
\medskip
\heading\S5 Type $B$\endheading
\medskip

Let $\be^*_n(K)$ denote the number of ideals in $B_n$ with class of
nilpotence $K$. As above, $\U_k\equiv U_k\(1/2\sqrt
x\)$. We prove the following theorem.

\proclaim{Theorem B.1}
The generating function $\sum _{n \ge0} ^{}\be^*_n(K)x^n$ is given by
$$\cases \dsize \frac {\U_{2k}+\U_k\U_{k+1}\U_{2k-1}} {\sqrt
x\,\U_{2k} \U_{2k+1}\U_{2k+2}}&\text{if }K=2k,\\
\dsize \frac {\U_{2k}+\U_{k+1}^2\U_{2k-2}+\U_{k-1}^2\U_{2k-2}+\U_2} {
\U_{2k-1} \U_{2k}\U_{2k+1}}&\text{if }K=2k-1.
\endcases$$
\endproclaim
\demo{Proof}
For the proof, we split the set of (strict) partitions corresponding to
ideals into 7 subclasses. To determine, to which subclass 
a particular partition belongs, one proceeds as follows: We draw two
broken rays. Ray~1 starts on the vertical edge on the right of cell
$(1,\l_1)$ (that is, on the vertical right border of
the first row), whereas ray~2 starts on the vertical edge on the
right of cell $(2,\l_2+1)$ (that is, on the vertical right border of
the second row). Both rays are determined in an algorithmic manner:
starting on that edge, we move downward until we touch the antidiagonal
$x+y=2n$. At the touching point we turn direction from 
vertical-down to horizontal-left, and move on until we touch a vertical
part of the Ferrers diagram. At the touching point we turn direction
from  horizontal-left to vertical-down. Now the procedure is
iterated, until we reach the diagonal $x=y-1$. To which subclass the
partition belongs depends on whether the rays reach $x=y-1$ while
traveling vertical-down or horizontally-left, and whether one ray
is above the other or not, as is detailed below. In particular, see
Figures~8--12 for typical examples, with ray~1 (starting at the
right border of the first row) marked as a thin line, and ray~2
(starting at the right border of the second row) marked as a dotted line.
It is easy to check that all the possibilities are displayed in the following synopsis.
%\par\newpage
\smallskip
\noindent{\it Case 1}.  Ray~2 reaches
$x=y-1$ while traveling horizontal-left, and  either ray~1 reaches
$x=y-1$ while traveling horizontal-left weakly below ray~2 or ray~1 reaches
$x=y-1$ while traveling vertical-down strictly to the right of ray~2.
$$\text{ k touching points for ray 2}\quad\xymatrix{
 \ar@{=}[dddrrr] & & & & \\
& \ar@{.}[rr]& & & \\
 & & \ar@{-}[rr]& & \\
 & & & & \\
}\text{class = 2k+1}$$

$$\text{ k touching points for ray 2}\quad\xymatrix{
 \ar@{=}[dddrrr] & & \ar@{-}[dd]& & \\
& \ar@{.}[rr]& & & \\
 & & & & \\
 & & & & \\
}\text{class = 2k+1}$$
{\it Case 2}. Ray~2 reaches
$x=y-1$ while traveling horizontal-left, and ray~1 reaches
$x=y-1$ while traveling vertical-down weakly to the left of ray~2.
$$\text{ k touching points for ray 2}\quad\xymatrix{
 \ar@{=}[dddrrr] & \ar@{-}[d]& & & \\
& & & & \\
 & & \ar@{.}[rr]& & \\
 & & & & \\
}\text{class = 2k}$$
{\it Case 3}. Ray~2 reaches
$x=y-1$ while traveling vertical-down, and  ray~1 reaches
$x=y-1$ while traveling horizontal-left strictly above ray~2.
$$\text{ k touching points for ray 1}\quad\xymatrix{
 \ar@{=}[dddrrr] & & \ar@{.}[dd]& & \\
& \ar@{-}[rr]& & & \\
 & & & & \\
 & & & & \\
}\text{class = 2k}$$
{\it Case 4}. Ray~2 reaches
$x=y-1$ while traveling vertical-down, and  ray~1 reaches
$x=y-1$ while traveling vertical-down weakly to the left of ray~2.
$$\text{ k touching points for ray 1}\quad\xymatrix{
 \ar@{=}[dddrrr] & \ar@{-}[d]& \ar@{.}[dd]& & \\
& & & & \\
 & & & & \\
 & & & & \\
}\text{class = 2k}$$
{\it Case 5}. Ray~2 reaches
$x=y-1$ while traveling vertical-down, and ray~1 reaches
$x=y-1$ while traveling horizontal-left weakly below ray~2.
$$\text{ k touching points for ray 1}\quad\xymatrix{
 \ar@{=}[dddrrr] & \ar@{.}[d]& & & \\
& & & & \\
 & & \ar@{-}[rr]& & \\
 & & & & \\
}\text{class = 2k-1}$$
{\it Case 6}. Ray~2 reaches
$x=y-1$ while traveling vertical-down, and  ray~1 reaches
$x=y-1$ while traveling vertical-down weakly to the right of ray~2.
$$\text{ k-1 touching points for ray 1}\quad\xymatrix{
 \ar@{=}[dddrrr] & \ar@{.}[d]& \ar@{-}[dd]& & \\
& & & & \\
 & & & & \\
 & & & & \\
}\text{class = 2k-1}$$
{\it Case 7}. Ray~2 reaches
$x=y-1$ while traveling horizontal-left, and  ray~1 reaches
$x=y-1$ while traveling horizontal-left weakly above ray~2.
$$\text{ k touching points for ray 2}\quad\xymatrix{
 \ar@{=}[dddrrr] & & & & \\
& \ar@{-}[rr]& & & \\
 & & \ar@{.}[rr]& & \\
 & & & & \\
}\text{class = 2k}$$

\medskip
Finally, suppose that 
both rays reach $x=y-1$ horizontally (resp. vertically) in the same point. We regard this possibility as 
case 1 or 7 (resp. case 4 or 6) according to whether the number of touching points of ray 1 is one more the number of touching
points of ray 2 or it is the same.\smallskip
%\par\newpage
We now examine each case to get formulas in the same spirit of (A.1), (A.2).\smallskip

{\it Case 1}. A partition belongs to subclass 1, if ray~2 reaches
$x=y-1$ while traveling horizontal-left, and if either ray~1 reaches
$x=y-1$ while traveling horizontal-left weakly below ray~2 (see
Figure~5) or ray~1 reaches
$x=y-1$ while traveling vertical-down strictly to the right of ray~2 (see
Figure~6; that some edges are thick is irrelevant at the moment).

\midinsert
\vskip10pt
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$$
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\Pfad(-11,11),1111111111111111111\endPfad
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\Pfad(-10,9),2\endPfad
\Pfad(-10,9),1\endPfad
\Pfad(-9,8),2\endPfad
\Pfad(-9,8),1\endPfad
\Pfad(-8,7),2\endPfad
\Pfad(-8,7),1\endPfad
\Pfad(-7,6),2\endPfad
\Pfad(-7,6),1\endPfad
\Pfad(-6,5),2\endPfad
\Pfad(-6,5),1\endPfad
\Pfad(-5,4),2\endPfad
\Pfad(-5,4),1\endPfad
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\Pfad(1,5),2\endPfad
\Pfad(4,7),2\endPfad
\Pfad(6,9),2\endPfad
\Pfad(8,10),2\endPfad
\thinlines
\Diagonale(-1,-1){12}
\Label\r{x+y=2n}(11,9)
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\DuennPunkt(0,0)
\DuennPunkt(1,1)
\DuennPunkt(4,4)
\DuennPunkt(6,6)
\DuennPunkt(8,8)
\Kreis(0,11)
$$
\centerline{\eightpoint Figure 5}
}
\vskip10pt
\endinsert


\midinsert
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$$
\Einheit.4cm
\Pfad(-3,2),11221212111221212112\endPfad
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\Pfad(1,5),2\endPfad
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\Pfad(6,9),2\endPfad
\Pfad(8,10),2\endPfad
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\DuennPunkt(4,4)
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\DuennPunkt(8,8)
\Kreis(0,11)
$$
\centerline{\eightpoint Figure 6}
}
\vskip10pt
\endinsert

Let $k$ be the number of touching points of ray~2 on the antidiagonal
$x+y=2n$. Then
the class of nilpotence of all ideals in this subclass is $2k+1$. 
For given $n$, the number of ideals in this case is given by the
multiple summation
$$\sum _{-i_2\le i_1\le i_2\le \dots\le i_{2k+1}\le n-1} ^{}
\bigg(\prod _{j=4} ^{2k+2}\binom {i_j-i_{j-1}+i_{j-2}-i_{j-3}-1}
{i_{j-2}-i_{j-3}}\bigg)\kern-4pt \sum _{\ell=0} ^{i_3-i_2-1}\binom {i_3+i_1-1}\ell,$$
where, as always in the sequel, $i_{2k+2}:=n-1$. This expression is
obtained from the geometrical presentation as in Figures~5 or 6, by
denoting the deviation of the first vertical edges on each downward
travel along a ray (in Figures~5 and 6 these are the thick vertical
edges; the thick horizontal edge marks the downmost and leftmost edge
of the Ferrers diagram; it must touch the diagonal $x=y-1$) from the
``reference point" $(n,0)$ (in Figures~5 and 6 it is marked by a circle) by $i_{2k+1},i_{2k},\dots,i_1$, from right to left.
Indeed, we can use the same procedure of case $C$, having replaced  this time the line
$x=y$ with the line $x=y-1$ (just to take into account the condition
$\l_i\ne i-1,\,i\geq2$ of Prop. 3.1). Then,  arguing as before, if we set   $\mu\equiv(\l_2,\ldots,\l_n)$, we have to examine 
the set $[\l^m_{i_1,i_3,\ldots,i_{2k+1}},\l^M_{i_1,i_3,\ldots,i_{2k+1}}]\cap[\mu^m_{i_2,i_4,\ldots,i_{2k}},
\mu^M_{i_2,i_4,\ldots,i_{2k}}]$. By expliciting the previous formula as a product of Young lattices, it is easy
to see, for fixed $i_{2k+1},i_{2k},\dots,i_1$, that the number of
partitions is equal to the summand in the above sum.

Let us denote the above number of ideals by $\be_n^{(1)}(2k+1)$.
We now compute the generating function $\sum _{n\ge0}
^{}\be_n^{(1)}(2k+1)x^n$. By definition, this is
$$\multline \sum _{n\ge0} ^{}x^n
\sum _{-i_2\le i_1\le i_2\le \dots\le i_{2k+1}\le n-1} ^{}
\bigg(\prod _{j=4} ^{2k+2}\binom {i_j-i_{j-1}+i_{j-2}-i_{j-3}-1}
{i_{j-2}-i_{j-3}}\bigg)\\
\cdot \sum _{\ell=0} ^{i_3-i_2-1}\binom {i_3+i_1-1}\ell.
\endmultline$$
Now we would like to perform the sum over $n$, then the sum over
$i_{2k+1}$, then over $i_2k$, etc., one after the other. In order to
do this conveniently, we have to split the range of the sum into the
possibilities 
$$-i_2\le i_1\le i_2,\quad i_2+\ell+1\le i_3< \dots< i_{2k+1}< n-1$$
and
$$\multline
0\le i_1=i_2,\quad i_2+\ell+1\le i_3=i_4<i_5=i_6<
\cdots\\
<i_{2j-1}=i_{2j}<i_{2j+1}< \dots< i_{2k+1}< n-1,
\endmultline$$
for some $j$ between $2$ and $k+1$. These two do indeed cover all
possibilities which contribute to the sum. For, if we have
$i_s=i_{s-1}$ for some $s\ge4$, then, in order that the summand does
not vanish, we must also have $i_{s-2}=i_{s-3}$, etc. The case of $s$
being even is covered by the second possibility above. If $s$ should
be odd, then we would also have $i_3=i_2$. Since then the sum over
$\ell$ is empty, this does not contribute to the sum.

Now, in each of the two cases the sums over $n$, $i_{2k+1}$, \dots,
$i_4$ are easily carried out by the binomial theorem. Thereby,
continued fractions as in the l.h.s. of (4.1)
are slowly built up.

To finish the calculation, we use the identity
$$\sum _{\ell=0} ^{i_3-i_2-1}\binom {i_3+i_1-1}\ell=
\sum _{\ell=0} ^{i_3-i_2-1}\binom {i_1+i_2+\ell}\ell\cdot
\cases 2^{i_3-i_2-\ell-2}&\text{if }\ell<i_3-i_2-1,\\
1&\text{if }\ell=i_3-i_2-1.\endcases$$
For, now the sums over $i_3$, $\ell$, $i_1$, and finally $i_2$, in
that order, can easily be carried out, as this amounts again to just
summing binomial or even just geometric series. To take into account  the
continued fractions, we use relation (4.1)
The result is
$${\frac{{\sqrt{x}}\,\U_{k+1}\,
      \left( {{\U_{k}}^2} + {{\U_{k+1}}^2} \right) }{{{\U_{k}}^2}\,
   {{\U_{2k+2}}^2}   \left(   \U_{k}-2\,{\sqrt{x}}\,\U_{k-1} \right) \,
      }}.
$$

The computation for the second possible range is similar. The result
is
$${\frac{{\sqrt{x}}\,{{\U_{k+1}}^2}}
    { \U_{k-j+1}\,\U_{k-j+2}\,
\U_{k}\, \U_{2k+2}\left(   \U_{k}-2\,{\sqrt{x}}\,\U_{k-1} \right) \,
    }}
$$

In order to obtain the overall generating function for the ideals in
this subclass, we have to take the first expression and add to it the
sum of the second expressions over $j$ from $2$ to $k+1$. Using the
identity
$$\frac {1} {U_0(x)U_1(x)}+\frac {1} {U_1(x)U_2(x)}+\dots+\frac {1}
{U_s(x)U_{s+1}(x)}=\frac {U_s(x)} {U_{s+1}(x)},$$
we obtain
$$ {\frac{{\sqrt{x}}\,\U_{k+1}\,
      \left( {{\U_{k}}^2} + {{\U_{k+1}}^2} + 
        \U_{k-1}\,\U_{k+1}\,\U_{2k+2} \right) }{{{\U_{k}}^2}\,
   {{\U_{2k+2}}^2}   \left(   \U_{k}-2\,{\sqrt{x}}\,\U_{k-1} \right) \,
      }}.
$$

\smallskip
{\it Case 2}. A partition belongs to subclass 2, if ray~2 reaches
$x=y-1$ while traveling horizontal-left, and if ray~1 reaches
$x=y-1$ while traveling vertical-down weakly to the left of ray~2 (see
Figure~7).

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\PfadDicke{.2pt}
\Pfad(-2,1),22211111122221111222\endPfad
\SPfad(-2,1),11122222111112222\endSPfad
\PfadDicke{3pt}
\Pfad(-3,2),1\endPfad
\Pfad(-2,3),2\endPfad
\Pfad(1,5),2\endPfad
\Pfad(4,7),2\endPfad
\Pfad(6,9),2\endPfad
\Pfad(8,10),2\endPfad
\thinlines
\Diagonale(-1,-1){12}
\Label\r{x+y=2n}(11,9)
\AntiDiagonale(-12,11){12}
\Label\o{x=y-1}(-14,10)
\DuennPunkt(1,1)
\DuennPunkt(4,4)
\DuennPunkt(6,6)
\DuennPunkt(8,8)
\Kreis(0,11)
$$
\centerline{\eightpoint Figure 7}
}
\vskip10pt
\endinsert

Let $k$ be the number of touching points of ray~2 on the antidiagonal
$x+y=2n$. Then
the class of nilpotence of all ideals in this subclass is $2k$.

For given $n$, the number of ideals in this case is given by the
multiple summation
$$\underset -i_3+1\le i_1\le -i_2-1\to
{\sum _{0\le i_2\le \dots\le i_{2k+1}\le n-1} ^{}}
\bigg(\prod _{j=4} ^{2k+2}\binom {i_j-i_{j-1}+i_{j-2}-i_{j-3}-1}
{i_{j-2}-i_{j-3}}\bigg) 2^{i_3+i_1-1}.$$
Let us denote this number by $\be_n^{(2)}(2k)$.
We now compute the generating function $\sum _{n\ge0}
^{}\be_n^{(2)}(2k)x^n$. 

We proceed as in Case 1. Here, the analysis is in fact easier. First,
there is no sum over $\ell$, so that we do not have to use the
above transformation formula. Second, both $i_2=i_1$ and $i_3=i_2$
produce vanishing summands, so that we can restrict the summation to
$$-i_3+1\le i_1\le -i_2-1,\quad 0\le i_2< \dots< i_{2k+1}< n-1.$$
As a result, we obtain for the generating function the expression
$${\frac{{\sqrt{x}}\,\U_{k-1}\,\U_{k+1}}
    {{{\U_{k}}^2}\,\U_{2k+2}\left(   \U_{k}-2\,{\sqrt{x}}\,\U_{k-1} \right) \,
      \left(   \U_{k+1}-\U_{k-1} \right) }}.
$$

\smallskip
{\it Case 3}. A partition belongs to subclass 3, if ray~2 reaches
$x=y-1$ while traveling vertical-down, and if ray~1 reaches
$x=y-1$ while traveling horizontal-left strictly above ray~2 (see
Figure~8).

\midinsert
\vskip10pt
\vbox{
$$
\Einheit.4cm
\Pfad(-4,3),12112211212112112112\endPfad
\Pfad(-11,11),1111111111111111111\endPfad
\Pfad(-11,10),2\endPfad
\Pfad(-11,10),1\endPfad
\Pfad(-10,9),2\endPfad
\Pfad(-10,9),1\endPfad
\Pfad(-9,8),2\endPfad
\Pfad(-9,8),1\endPfad
\Pfad(-8,7),2\endPfad
\Pfad(-8,7),1\endPfad
\Pfad(-7,6),2\endPfad
\Pfad(-7,6),1\endPfad
\Pfad(-6,5),2\endPfad
\Pfad(-6,5),1\endPfad
\Pfad(-5,4),2\endPfad
\Pfad(-5,4),1\endPfad
\Pfad(-4,3),2\endPfad
%\Pfad(-4,3),1\endPfad
%\Pfad(-3,2),2\endPfad
\PfadDicke{.2pt}
\Pfad(-3,2),11111222222111111222\endPfad
\SPfad(-1,0),22222211111112222\endSPfad
\PfadDicke{3pt}
\Pfad(-4,3),1\endPfad
\Pfad(-1,5),2\endPfad
\Pfad(2,7),2\endPfad
\Pfad(6,9),2\endPfad
\Pfad(8,10),2\endPfad
\thinlines
\Diagonale(-1,-1){12}
\Label\r{x+y=2n}(11,9)
\AntiDiagonale(-12,11){12}
\Label\o{x=y-1}(-14,10)
\DuennPunkt(2,2)
\DuennPunkt(6,6)
\DuennPunkt(8,8)
\Kreis(0,11)
$$
\centerline{\eightpoint Figure 8}
}
\vskip10pt
\endinsert

Let $k$ be the number of touching points of ray~1 on the antidiagonal
$x+y=2n$. Then
the class of nilpotence of all ideals in this subclass is $2k$.

For given $n$, the number of ideals in this case is given by the
multiple summation
$$\sum _{-i_2\le i_1<0\le i_2\le \dots\le i_{2k}\le n-1} ^{}
\kern-2pt
\bigg(\prod _{j=4} ^{2k+1}\binom {i_j-i_{j-1}+i_{j-2}-i_{j-3}-1}
{i_{j-2}-i_{j-3}}\bigg)\kern-4pt \sum _{\ell=0} ^{i_3-i_2-1}\binom
{i_3+i_1-1}\ell .$$
Let us denote this number by $\be_n^{(3)}(2k)$.
We now compute the generating function $\sum _{n\ge0}
^{}\be_n^{(3)}(2k)x^n$. 

We proceed as in Case 1. Since both $i_2=i_1$ and $i_3=i_2$
produce vanishing summands, we can restrict the summation to
$$-i_2\le i_1<0\le i_2< \dots< i_{2k}< n-1.$$
As a result, we obtain for the generating function the expression
$${\frac{{\sqrt{x}}\,\U_{k-1}\,\U_{k+1}}
    {{{\U_{k}}^2}\, \U_{2k}\,\left(   \U_{k+1}-\U_{k-1} \right)
\left(   \U_{k}-2\,{\sqrt{x}}\,\U_{k-1} \right) \,
      }}.
$$

\smallskip
{\it Case 4}. A partition belongs to subclass 4, if ray~2 reaches
$x=y-1$ while traveling vertical-down, and if ray~1 reaches
$x=y-1$ while traveling vertical-down weakly to the left of ray~2 (see
Figure~9).

\midinsert
\vskip10pt
\vbox{
$$
\Einheit.4cm
\Pfad(-4,3),11221211111221212112\endPfad
\Pfad(-11,11),1111111111111111111\endPfad
\Pfad(-11,10),2\endPfad
\Pfad(-11,10),1\endPfad
\Pfad(-10,9),2\endPfad
\Pfad(-10,9),1\endPfad
\Pfad(-9,8),2\endPfad
\Pfad(-9,8),1\endPfad
\Pfad(-8,7),2\endPfad
\Pfad(-8,7),1\endPfad
\Pfad(-7,6),2\endPfad
\Pfad(-7,6),1\endPfad
\Pfad(-6,5),2\endPfad
\Pfad(-6,5),1\endPfad
\Pfad(-5,4),2\endPfad
\Pfad(-5,4),1\endPfad
\Pfad(-4,3),2\endPfad
%\Pfad(-4,3),1\endPfad
%\Pfad(-3,2),2\endPfad
\PfadDicke{.2pt}
\Pfad(-2,1),22211111122221111222\endPfad
\SPfad(-1,0),22222211111112222\endSPfad
\PfadDicke{3pt}
\Pfad(-4,3),1\endPfad
\Pfad(-2,3),2\endPfad
\Pfad(-1,5),2\endPfad
\Pfad(4,7),2\endPfad
\Pfad(6,9),2\endPfad
\Pfad(8,10),2\endPfad
\thinlines
\Diagonale(-1,-1){12}
\Label\r{x+y=2n}(11,9)
\AntiDiagonale(-12,11){12}
\Label\o{x=y-1}(-14,10)
\DuennPunkt(4,4)
\DuennPunkt(6,6)
\DuennPunkt(8,8)
\Kreis(0,11)
$$
\centerline{\eightpoint Figure 9}
}
\vskip10pt
\endinsert

Let $k$ be the number of touching points of ray~1 on the antidiagonal
$x+y=2n$. Then
the class of nilpotence of all ideals in this subclass is $2k$.

For given $n$, the number of ideals in this case is given by the
multiple summation
$$
{\sum _{-i_3+1\le i_1\le i_2<0\le i_3\le \dots\le i_{2k+1}\le n-1} ^{}}
\bigg(\prod _{j=4} ^{2k+2}\binom {i_j-i_{j-1}+i_{j-2}-i_{j-3}-1}
{i_{j-2}-i_{j-3}}\bigg) 2^{i_3+i_1-1}.$$
Let us denote this number by $\be_n^{(4)}(2k)$.
We now compute the generating function $\sum _{n\ge0}
^{}\be_n^{(4)}(2k)x^n$. 

We proceed as in Case 1. We split the range of summation into the
possibilities 
$$-i_3+1\le i_1\le i_2<0\le i_3< \dots< i_{2k+1}< n-1$$
and
$$ -i_3+1\le i_1= i_2<0\le i_3=i_4< \cdots
<i_{2j-1}=i_{2j}<i_{2j+1}<\dots<i_{2k+1}< n-1.
$$
The contribution to the generating function of the first range is
$${\frac{{\sqrt{x}}\,{{\U_{k-1}}^2}}
    {{{\U_{k}}^2}\, \U_{2k}\,\left(   \U_{k+1}-\U_{k-1} \right) 
\left(   \U_{k}-2\,{\sqrt{x}}\,\U_{k-1} \right) \,
     }},
$$
while the contribution of the second range is
$${\frac{{\sqrt{x}}\,{{\U_{k-1}}^2}}
    {\U_{k-j+1}\,\U_{k-j+2}\,\U_{k}\,\U_{2k}
\left(   \U_{k}-2\,{\sqrt{x}}\,\U_{k-1} \right) \,
      }}.
$$
Summing the second expression over $j$ from $2$ to $k+1$, and adding
the result to the first expression gives
$${\frac{{\sqrt{x}}\,{{\U_{k-1}}^2}\,
      }
      {{{\U_{k}}^2}\, \U_{2k}
\left(   \U_{k}-2\,{\sqrt{x}}\,\U_{k-1} \right) \,
     }}
 \left(   {\frac{1}{  \U_{k+1}-\U_{k-1}}}+\U_{k-1} \right).
$$

\smallskip
{\it Case 5}. A partition belongs to subclass 5, if ray~2 reaches
$x=y-1$ while traveling vertical-down, and if ray~1 reaches
$x=y-1$ while traveling horizontal-left weakly below ray~2 (see
Figure~10).

\midinsert
\vskip10pt
\vbox{
$$
\Einheit.4cm
\Pfad(-4,3),12221112121112112112\endPfad
\Pfad(-11,11),1111111111111111111\endPfad
\Pfad(-11,10),2\endPfad
\Pfad(-11,10),1\endPfad
\Pfad(-10,9),2\endPfad
\Pfad(-10,9),1\endPfad
\Pfad(-9,8),2\endPfad
\Pfad(-9,8),1\endPfad
\Pfad(-8,7),2\endPfad
\Pfad(-8,7),1\endPfad
\Pfad(-7,6),2\endPfad
\Pfad(-7,6),1\endPfad
\Pfad(-6,5),2\endPfad
\Pfad(-6,5),1\endPfad
\Pfad(-5,4),2\endPfad
\Pfad(-5,4),1\endPfad
\Pfad(-4,3),2\endPfad
%\Pfad(-4,3),1\endPfad
%\Pfad(-3,2),2\endPfad
\PfadDicke{.2pt}
\Pfad(-2,1),11122222221111111222\endPfad
\SPfad(-3,2),22221111111112222\endSPfad
\PfadDicke{3pt}
\Pfad(-4,3),1\endPfad
\Pfad(-3,5),2\endPfad
\Pfad(1,7),2\endPfad
\Pfad(6,9),2\endPfad
\Pfad(8,10),2\endPfad
\thinlines
\Diagonale(-1,-1){12}
\Label\r{x+y=2n}(11,9)
\AntiDiagonale(-12,11){12}
\Label\o{x=y-1}(-14,10)
\DuennPunkt(1,1)
\DuennPunkt(6,6)
\DuennPunkt(8,8)
\Kreis(0,11)
$$
\centerline{\eightpoint Figure 10}
}
\vskip10pt
\endinsert

Let $k$ be the number of touching points of ray~1 on the antidiagonal
$x+y=2n$. Then
the class of nilpotence of all ideals in this subclass is $2k-1$.

For given $n$, the number of ideals in this case is given by the
multiple summation
$$\underset -i_3+1\le i_1\le -i_2-1\to
{\sum _{0\le i_2\le \dots\le i_{2k}\le n-1} ^{}}
\bigg(\prod _{j=4} ^{2k+1}\binom {i_j-i_{j-1}+i_{j-2}-i_{j-3}-1}
{i_{j-2}-i_{j-3}}\bigg) 2^{i_3+i_1-1}.$$
Let us denote this number by $\be_n^{(5)}(2k-1)$.
We now compute the generating function $\sum _{n\ge0}
^{}\be_n^{(5)}(2k-1)x^n$. 

We proceed as in Case 1. Since both $i_2=i_1$ and $i_3=i_2$
produce vanishing summands, we can restrict the summation to
$$ -i_3+1\le i_1\le -i_2-1,\quad 0\le i_2< \dots< i_{2k}< n-1$$
As a result, we obtain for the generating function the expression
$${\frac{{\sqrt{x}}\,\U_{k-1}}
    {{{\U_{2k}}^2}\left(   \U_{k}-2\,{\sqrt{x}}\,\U_{k-1} \right) }
    }.
$$

\smallskip
{\it Case 6}. A partition belongs to subclass 6, if ray~2 reaches
$x=y-1$ while traveling vertical-down, and if ray~1 reaches
$x=y-1$ while traveling vertical-down weakly to the right of ray~2 (see
Figure~11).

\midinsert
\vskip10pt
\vbox{
$$
\Einheit.4cm
\Pfad(-4,3),12222121111112112112\endPfad
\Pfad(-11,11),1111111111111111111\endPfad
\Pfad(-11,10),2\endPfad
\Pfad(-11,10),1\endPfad
\Pfad(-10,9),2\endPfad
\Pfad(-10,9),1\endPfad
\Pfad(-9,8),2\endPfad
\Pfad(-9,8),1\endPfad
\Pfad(-8,7),2\endPfad
\Pfad(-8,7),1\endPfad
\Pfad(-7,6),2\endPfad
\Pfad(-7,6),1\endPfad
\Pfad(-6,5),2\endPfad
\Pfad(-6,5),1\endPfad
\Pfad(-5,4),2\endPfad
\Pfad(-5,4),1\endPfad
\Pfad(-4,3),2\endPfad
%\Pfad(-4,3),1\endPfad
%\Pfad(-3,2),2\endPfad
\PfadDicke{.2pt}
\Pfad(-2,1),22222221111111111222\endPfad
\SPfad(-3,2),22221111111112222\endSPfad
\PfadDicke{3pt}
\Pfad(-4,3),1\endPfad
\Pfad(-3,5),2\endPfad
\Pfad(-2,7),2\endPfad
\Pfad(6,9),2\endPfad
\Pfad(8,10),2\endPfad
\thinlines
\Diagonale(-1,-1){12}
\Label\r{x+y=2n}(11,9)
\AntiDiagonale(-12,11){12}
\Label\o{x=y-1}(-14,10)
\DuennPunkt(6,6)
\DuennPunkt(8,8)
\Kreis(0,11)
$$
\centerline{\eightpoint Figure 11}
}
\vskip10pt
\endinsert

Let $k-1$ be the number of touching points of ray~1 on the antidiagonal
$x+y=2n$. Then
the class of nilpotence of all ideals in this subclass is $2k-1$.

For given $n$, the number of ideals in this case is given by the
multiple summation
$$
{\sum _{-i_3+1\le i_1\le i_2<0\le i_3\le \dots\le i_{2k}\le n-1} ^{}}
\bigg(\prod _{j=4} ^{2k+1}\binom {i_j-i_{j-1}+i_{j-2}-i_{j-3}-1}
{i_{j-2}-i_{j-3}}\bigg) 2^{i_3+i_1-1}.$$
Let us denote this number by $\be_n^{(6)}(2k-1)$.
We now compute the generating function $\sum _{n\ge0}
^{}\be_n^{(6)}(2k-1)x^n$. 

We proceed as in Case 1. We split the range of summation into the
possibilities 
$$-i_3+1\le i_1\le i_2<0\le i_3< \dots< i_{2k}< n-1$$
and
$$\multline 
-i_3+1\le i_1= i_2<0\le i_3=i_4<\cdots\\
<i_{2j-1}=i_{2j}<i_{2j+1}< \dots< i_{2k}< n-1.
\endmultline$$
The contribution to the generating function of the first range is
$${\frac{{\sqrt{x}}\,\U_{k-1}}
    {{{\U_{2k}}^2}\left(   \U_{k}-2\,{\sqrt{x}}\,\U_{k-1} \right) }
    },
$$
while the contribution of the second range is
$${\frac{{\sqrt{x}}\,{{\U_{k-1}}^2}}
    {\U_{k-j}\,\U_{k-j+1}\,\U_{k}\, \U_{2k}
\left(   \U_{k}-2\,{\sqrt{x}}\,\U_{k-1} \right) \,
     }}.
$$
Summing the second expression over $j$ from $2$ to $k$, and adding
the result to the first expression gives
$${\frac{{\sqrt{x}}\,\U_{k-1}\,
     }{{{\U_{2k}}^2}
      \left(   \U_{k}-2\,{\sqrt{x}}\,\U_{k-1} \right) 
      }}
 \left( 1 + {\frac{\U_{k-2}\,\U_{2k}}{\U_{k}}} \right) .
$$

\smallskip
{\it Case 7}. A partition belongs to subclass 7, if ray~2 reaches
$x=y-1$ while traveling horizontal-left, and if ray~1 reaches
$x=y-1$ while traveling horizontal-left weakly above ray~2 (see
Figure~12). 

\midinsert
\vskip10pt
\vbox{
$$
\Einheit.4cm
\Pfad(-5,4),11121112111221212112\endPfad
\Pfad(-11,11),1111111111111111111\endPfad
\Pfad(-11,10),2\endPfad
\Pfad(-11,10),1\endPfad
\Pfad(-10,9),2\endPfad
\Pfad(-10,9),1\endPfad
\Pfad(-9,8),2\endPfad
\Pfad(-9,8),1\endPfad
\Pfad(-8,7),2\endPfad
\Pfad(-8,7),1\endPfad
\Pfad(-7,6),2\endPfad
\Pfad(-7,6),1\endPfad
\Pfad(-6,5),2\endPfad
\Pfad(-6,5),1\endPfad
\Pfad(-5,4),2\endPfad
\PfadDicke{.2pt}
\Pfad(-5,4),11111111122221111222\endPfad
\SPfad(-2,1),11122222111112222\endSPfad
\PfadDicke{3pt}
\Pfad(-5,4),1\endPfad
\Pfad(1,5),2\endPfad
\Pfad(4,7),2\endPfad
\Pfad(6,9),2\endPfad
\Pfad(8,10),2\endPfad
\thinlines
\Diagonale(-1,-1){12}
\Label\r{x+y=2n}(11,9)
\AntiDiagonale(-12,11){12}
\Label\o{x=y-1}(-14,10)
\DuennPunkt(1,1)
\DuennPunkt(4,4)
\DuennPunkt(6,6)
\DuennPunkt(8,8)
\Kreis(0,11)
$$
\centerline{\eightpoint Figure 12}
}
\vskip10pt
\endinsert

Let $k$ be the number of touching points of ray~2 on the antidiagonal
$x+y=2n$. Then
the class of nilpotence of all ideals in this subclass is $2k$.

For given $n$, the number of ideals in this case is given by the
multiple summation
$$\sum _{0\le i_1\le i_2\le \dots\le i_{2k}\le n-1} ^{}
\bigg(\prod _{j=4} ^{2k+1}\binom {i_j-i_{j-1}+i_{j-2}-i_{j-3}-1}
{i_{j-2}-i_{j-3}}\bigg) \sum _{\ell=0} ^{i_3-i_2-1}\binom {i_3+i_1-1}\ell.$$
Let us denote this number by $\be_n^{(7)}(2k)$.
We now compute the generating function $\sum _{n\ge0}
^{}\be_n^{(7)}(2k)x^n$. 

We proceed as in Case 1. We split the range of summation into the
possibilities 
$$0\le i_1\le i_2< \dots< i_{2k}< n-1$$
and
$$0\le i_1= i_2<i_3=i_4<\dots<i_{2j-1}<i_{2j}<i_{2j+1}< \dots< i_{2k}<
n-1.$$
The contribution to the generating function of the first range is
$${\frac{{\sqrt{x}}\,{{\U_{k+1}}^3}}
    {\U_{k-1}\,{{\U_{k}}^2}\, \U_{2k+2}
 \left(   \U_{k+1}-\U_{k-1} \right)
      \left(   \U_{k}-2\,{\sqrt{x}}\,\U_{k-1} \right) \,
     }},$$
while the contribution of the second range is
$$ {\frac{{\sqrt{x}}\,{{\U_{k+1}}^2}}
    { \U_{k-j}\,\U_{k-j+1}\,\U_{k}\,\U_{2k+2}
\left(   \U_{k}-2\,{\sqrt{x}}\,\U_{k-1} \right) \,
     }}.
$$
Summing the second expression over $j$ from $2$ to $k$, and adding
the result to the first expression gives
$$ {\frac{{\sqrt{x}}\,{{\U_{k+1}}^2}\,
     }{
      \U_{k-1}\,{{\U_{k}}^2}\,\U_{2k+2}
      \left(   \U_{k}-2\,{\sqrt{x}}\,\U_{k-1} \right) }} \left( 
        {\frac{\U_{k+1}}{  \U_{k+1}-\U_{k-1}}}+
 \U_{k-2}\,\U_{k}  \right) .
$$

\medskip
Finally, in order to complete the proof of the theorem, in case that
$K$ is even, the expressions of Cases~2, 3, 4 and 7 have to be summed,
in case that $K$ is odd, the expressions of Cases~1 (with $k$ replaced
by $k-1$), 5 and 6 have to be summed. Some simplification yields the
claimed expressions.\endemo

\proclaim{Corollary 5.1}
Let $\be_n(h)$ denote the number of ideals in $B_n$ with class of
nilpotence at most $h$. Then the
generating function $\sum _{n \ge0} ^{}\be_n^*(h)x^n$ is given by 
$$\cases
  {\dfrac{\sum_{i = 1}^{{h /2}}
        \left( 2\,i + 1 \right) \,\U_{2i - 1} + 
      ( h +1) \U_{h+1} + \sum_{i = 1}^{{h /2}}
        2i\,\U_{2h + 3 - 2i}  
      } {\U_{h+1}\U_{h+2}}}&\text{if $h$ is even,}\\
  {\dfrac{\sum_{i = 1}^{{(h - 1)/2}}
        \left( 2\,i + 1 \right) \,\U_{2i - 1} + 
      ( h +1) \U_h + \sum_{i = 1}^{{(h + 1)/2}}
        2i\,\U_{2h + 3 - 2i} 
      } {\U_{h+1}\U_{h+2}}}&\text{if $h$ is odd.}
\endcases$$
\endproclaim
This is easily verified by induction on $h$. We have also a type $B$ analog of Corollary 4.2

\proclaim{Corollary 5.2}
For $n\ge1$
the number of ideals $\i\in\I$  in type $B_n$ with $n(\i)\leq 2$ is $F_{2n}+F_{2n-2}-2^{n-1}$, and 
the number of ideals $n(\i)\leq 3$ is $\frac {1} {2}(5\cdot 3^{n-1}+1)-F_{2n-2}$.
\endproclaim

\medskip
\heading\S6 Type D\endheading
\medskip
We keep the notation introduce in the final Remark of Section 3. First we prove relation (3.1).
\proclaim{Proposition 6.1} $\d^*_n(k)=2\d^1_n(k)-\d^2_n(k).$\endproclaim
\demo{Proof} Let $\tilde\d_n(k)$ denote the number of ideals $\i\in\I$ parameterized by subset of boxes
$A\subseteq \T_{2n-2}$ such that $A^\bullet$ is a subdiagram. By Proposition 3.1
we have $\d^*_n(k)=\d^1_n(k)+\tilde\d_n(k)$; on the other hand the automorphism of the Dynkin diagram 
switching nodes $n-1,n$ induces an automorphism of $\g$ mapping bijectively ideals parameterized by subsets $A\subseteq \Dp$
as above into  ideals parameterized by the subdiagrams having columns $n-1,n$ of different lengths. Of course 
$n(\i_A)=n(\i_{A^\bullet})$, so that $\tilde\d_n(k)=\d^1_n(k)-\d^2_n(k)$ and the claim follows.
\endemo
With a slight abuse of notation we will write $\p\in \I(X)$ to mean that $\p$ is a dual order ideal 
of a root system of type $X_n$.
\proclaim{Proposition 6.2} $\d^2_n(k)=\b_{n-1}^*(k)$.\endproclaim
\demo{Proof} Let $f:\Dp(D_n)\to\Dp(B_{n-1})$ the map that on the root $\be\in\Dp(D_n)$ sets $\a_n=\a_{n-1}$; it is easy to show
that $f$ is well-defined (i.e. $f(\be)\in\Dp(B_{n-1})$). Moreover $f$ induces a map, still denoted by $f$, from
$\I(D)$ to $\Cal I^{n-1}(B)$. \par If $\p\subseteq\I(D)$, denote by $\l_\p$ the  subdiagram of $\T_{2n-2}$ naturally
associated to
$\p$. We claim that $f$ restricts to a bijection
$$\overline f:\{\p\in\I(D)\mid \l_\p \text{ is of type (2)}\}\to \Cal I^{n-1}(B).$$
(If $\p'\subseteq \Cal I^{n-1}(B)$,  $f^{-1}(\p')$ is obtained by substituting  $\frac{\a_{n-1}+\a_n}{2}$ to $\a_{n-1}$ in the
roots of $\p'$ in which
$\a_{n-1}$ appears with coefficient 2, by adding  the root $\a_i+\ldots+\a_{n-2}+\a_n$ if the root
$\a_i+\ldots+\a_{n-1}, 1\leq i\leq n-1$ belongs to $\p'$ and leaving  fixed the other roots in $\p'$.)
On subdiagrams, $\overline f$ is the map which erases the (n-1)-th column, i.e. $\l_{\overline f(\p)}\subseteq\T_{2n-3}$
is obtained from $\l_\p\subseteq\T_{2n-2}$ by erasing the (n-1)-th column. We claim that
$n(\l_\p)=n(\l_{\overline f(\p)})$, which finishes the proof. To prove the claim, denote by $\theta_D, \theta_B$ the highest
roots in
$\Dp(D_n),\Dp(B_{n-1})$ respectively. Note that $\overline f(\theta_D)=\theta_B$. Given a decomposition
$\theta_D=\d_1+\ldots+\d_k,\,\d_j\in\p$, we have $\theta_B=f(\d_1)+\ldots+f(\d_k)$ so that $n(\l_\p)\leq n(\l_{\overline
f(\p)})$. On the other hand,  we note that if $\theta_B=\b_1+\ldots+\b_k,\,\b_j\in\overline f(\p)$, it is always possible to
choose $\d_j\in f^{-1}(\b_j),\,1\leq j\leq k$ in such a way that $\theta_D=\d_1+\ldots+\d_k$. Therefore we have $n(\l_\p)\geq
n(\l_{\overline f(\p)})$ and equality follows.
\endemo
\smallskip
To calculate $\sum\limits_{n\geq 0}\d_n^1(k)x^n$ (see \S3) we proceed imitating case B. Take as reference point $(n,0)$: this
induces a coordinate system on the line
$x+y=2n-1$; now for touching point we mean a point in which  a path intersects the line $x+y=2n-1$ {\it with non-negative
coordinate}. The subdivision into seven subcases is the same as above; again,  if
both rays reach $x=y-1$ horizontally (resp. vertically) in the same point, we regard this possibility as 
case 1 or 7 (resp. case 4 or 6) according to whether the number of touching points of ray 1 is one more the number of touching
points of ray 2 or it is the same. Finally, set
 $i_{2k+2}:=n-1$ in  cases 1,2,4,7, $i_{2k+1}:=n-1$ in cases 3,5,6.
\bigskip\noindent
{\it Case 1}: $n(\i)=2k+1$ if there are $k$ touching points of ray 2.
$$\sum _{-i_2-1\le i_1\le i_2\le \dots\le i_{2k+1}\le n-2} ^{}
\bigg(\prod _{j=4} ^{2k+2}\binom {i_j-i_{j-1}+i_{j-2}-i_{j-3}-1}
{i_{j-2}-i_{j-3}}\bigg)\kern-4pt \sum _{\ell=0} ^{i_3-i_2-1}\binom {i_3+i_1}\ell.$$
\medskip\noindent
{\it Case 2}: $n(\i)=2k$ if there are $k$ touching points of ray 2.
$$\underset -i_3\le i_1\le -i_2-2\to
{\sum _{0\le i_2\le \dots\le i_{2k+1}\le n-2} ^{}}
\bigg(\prod _{j=4} ^{2k+2}\binom {i_j-i_{j-1}+i_{j-2}-i_{j-3}-1}
{i_{j-2}-i_{j-3}}\bigg) 2^{i_3+i_1}.$$
\medskip\noindent
{\it Case 3}: $n(\i)=2k$ if there are $k$ touching points of ray 1.
$$\sum _{-i_2-1\le i_1<0\le i_2\le \dots\le i_{2k}\le n-2} ^{}
\bigg(\prod _{j=4} ^{2k+1}\binom {i_j-i_{j-1}+i_{j-2}-i_{j-3}-1}
{i_{j-2}-i_{j-3}}\bigg)\kern-4pt \sum _{\ell=0} ^{i_3-i_2-1}\binom {i_3+i_1}\ell.$$
\medskip\noindent
{\it Case 4}: $n(\i)=2k$ if there are $k$ touching points of ray 1.
$$
\sum _{-i_3\le i_1\le i_2<0\le i_3\le \dots\le i_{2k+1}\le n-2}
\bigg(\prod _{j=4} ^{2k+2}\binom {i_j-i_{j-1}+i_{j-2}-i_{j-3}-1}
{i_{j-2}-i_{j-3}}\bigg) 2^{i_3+i_1}.$$
\medskip\noindent
{\it Case 5}: $n(\i)=2k-1$ if there are $k$ touching points of ray 1.
$$\underset -i_3\le i_1\le -i_2-2\to
{\sum _{0\le i_2\le \dots\le i_{2k}\le n-2} ^{}}
\bigg(\prod _{j=4} ^{2k+1}\binom {i_j-i_{j-1}+i_{j-2}-i_{j-3}-1}
{i_{j-2}-i_{j-3}}\bigg) 2^{i_3+i_1}.$$
\medskip\noindent
{\it Case 6}: $n(\i)=2k-1$ if there are $k-1$ touching points of ray 1.
$$
\sum _{-i_3\le i_1\le i_2<0\le i_3\le \dots\le i_{2k}\le n-2}
\bigg(\prod _{j=4} ^{2k+1}\binom {i_j-i_{j-1}+i_{j-2}-i_{j-3}-1}
{i_{j-2}-i_{j-3}}\bigg) 2^{i_3+i_1}.$$
\medskip\noindent
{\it Case 7}: $n(\i)=2k$ if there are $k$ touching points of ray 2.
$$\sum _{0\le i_1\le i_2\le \dots\le i_{2k+1}\le n-2} ^{}
\bigg(\prod _{j=4} ^{2k+2}\binom {i_j-i_{j-1}+i_{j-2}-i_{j-3}-1}
{i_{j-2}-i_{j-3}}\bigg)\kern-4pt \sum _{\ell=0} ^{i_3-i_2-1}\binom {i_3+i_1}\ell.$$
\medskip\heading\S7 The exceptional types\endheading
It is not difficult to write down a computer program which determines explicitly the descending central series 
of a given ideal. The final results are the following.
$$\alignat6
&n(\i)\qquad\qquad &&E_6\qquad\quad &&E_7\qquad\quad &&E_8\qquad\quad  &&F_4\qquad\quad  &&G_2\\
&0&&1&&1&&1&&1&&1\\
&1&&63&&127&&255 &&15&&3\\
&2&&210&&662&&2200 &&28&&2\\
&3&&217&&894&&3804 &&21&&1\\
&4&&150&&766&&3872 &&14&&0\\
&5&&92&&576&&3372 &&12&&1\\
&6&&51&&403&&2752 &&5\\
&7&&28&&279&&2182 &&4\\
&8&&12&&175&&1656 &&2\\
&9&&6&&115&&1277 &&2\\
&10&&2&&68&&955 &&0\\
&11&&1&&44&&737&&1\\
&12&& &&23&& 536&&\\
&13&& &&14&&412 &&\\
&14&& &&7&&300 &&\\
&15&& &&4&&227 &&\\
&16&& &&1&&157 &&\\
&17&& &&1&&123 &&\\
&18&& && &&81&&\\
&19&& && &&61 &&\\
&20&& && &&40&&\\
&21&& && &&30 &&\\
&22&& && &&18 &&\\
&23&& && &&14 &&\\
&24&& && &&7 &&\\
&25&& && &&5&&\\
&26&& && &&3 &&\\
&27&& && &&2 &&\\
&28&& && && 0&&\\
&29&& && &&1&&\\
\endalignat$$

Just a few remarks on how getting the previous data. Clearly we can work at the level of the root system: if $\i\in \I$ is
encoded  by $\p\subseteq\Dp$, then $n(\i)$ equals the maximal integer $k$ such that    $\theta\in \p^k$ (here
$\theta$ is the highest root of $\D$ and $\p^k$ is inductively defined as $\p^1=\p,\ \p^k=(\p^{k-1}+\p)\cap \D$).
The input for the program (i.e., the $\p$'s) can be  reduced to the determination of the  antichains 
of the root poset (indeed, for any finite poset, there is a canonical bijection mapping the antichain $\{a_1,\ldots,a_k\}$ to
the dual  order ideal which is the union of the principal dual order ideals $V_{a_1},\ldots,V_{a_k}$).  
In turn, the calculation of the antichain has  been done using {\it Coxeter} \cite{St}.
\bigskip

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\endRefs
\enddocument
\bye
