%This command provides the identities in the third
%horizontal area in page 1
%The macro has oe parameter:
%      1) The width of the text
\newcommand\TOneIdentitiesB[1]{%
   \parbox[t]{#1}{%
      \TOneFormulaeFontSize
      \vspace{\TOnePushFormulaeB}%
      \begin{DisplayFormulae}{19}{0pt}{\TOneInterlineFormulaeB}{\BigChar}{\StyleBold}
         \Fm{\SousEnsemble{n}{n-1} = \Cycle{n}{n-1} = \binom{n}{2}}
         \Fm{\sum_{k=0}^n\Cycle{n}{k} = n!}
         \Fm{C_n = \frac{1}{n+1}\binom{2n}{n}}
         \Fm{\Eul{n}{0} = \Eul{n}{n-1} = 1}
         \Fm{\Eul{n}{k} = \Eul{n}{n-1-k}}
         \Fm{\Eul{n}{k} = (k+1)\Eul{n-1}{k} + (n-k)\Eul{n-1}{k-1}}
         \Fm{\Eul{0}{k} = \left\{%
                            \begin{array}{lr}%
                                1 & \mbox{if }$k=0$, \\
                                0 & \mbox{otherwise} \\
                            \end{array}%
                          \right.}
         \Fm{\Eul{n}{1} = 2^n - n - 1}
         \Fm{\Eul{n}{2} = 3^n - (n+1)2^n + \binom{n+1}{2}}
         \Fm{x^n = \sum_{k=0}^n \Eul{n}{k} \binom{x+k}{n}}
         \Fm{\Eul{n}{m} = \sum_{k=0}^m \binom{n+1}{k} (m+1-k)^n(-1)^k}
         \Fm{m! \SousEnsemble{n}{m} = \sum_{k=0}^n \Eul{n}{k} \binom{k}{n-m}}
         \Fm{\Eul{n}{m} =\sum_{k=0}^n \SousEnsemble{n}{k} \binom{n-k}{m} (-1)^{n-k-m}k!}
         \Fm{\Euls{n}{0} = 1}
         \Fm{\Euls{n}{n} = 0 \MathRemark{\text{for }n\neq 0}}
         \Fm{\Euls{n}{k} = (k+1)\Euls{n-1}{k} + (2n-1-k)\Euls{n-1}{k-1}}
         \Fm{\sum_{k=0}^n \Euls{n}{k} = \frac{(2n)^{\underline{n}}}{2^n}}
         \Fm{\SousEnsemble{x}{x-n} = \sum_{k=0}^n \Euls{n}{k} \binom{x+n-1-k}{2n}}
         \Fm{\SousEnsemble{n+1}{m+1} = \sum_k \binom{n}{k} \SousEnsemble{k}{m} = \sum_{k=0}^n \SousEnsemble{k}{m} (m+1)^{n-k}}
      \end{DisplayFormulae}
      %To keep a small horizontal white space after pushing the block
      \vspace{1.1\TOnePushFormulaeB}
   }%
}