Even Odd Pascals Triangle

\documentclass{article}
\usepackage{tikz}
 \makeatletter
 \newcommand\binomialCoefficient[2]{%
     % Store values
     \c@pgf@counta=#1% n
     \c@pgf@countb=#2% k
     %
     % Take advantage of symmetry if k > n - k
     \c@pgf@countc=\c@pgf@counta%
     \advance\c@pgf@countc by-\c@pgf@countb%
     \ifnum\c@pgf@countb>\c@pgf@countc%
         \c@pgf@countb=\c@pgf@countc%
     \fi%
     %
     % Recursively compute the coefficients
     \c@pgf@countc=1% will hold the result
     \c@pgf@countd=0% counter
     \pgfmathloop% c -> c*(n-i)/(i+1) for i=0,...,k-1
         \ifnum\c@pgf@countd<\c@pgf@countb%
         \multiply\c@pgf@countc by\c@pgf@counta%
         \advance\c@pgf@counta by-1%
         \advance\c@pgf@countd by1%
         \divide\c@pgf@countc by\c@pgf@countd%
     \repeatpgfmathloop%
     \xdef\theresult{\the\c@pgf@countc}%
 }
 \makeatother

 \begin{document}
 \begin{tikzpicture}
 \foreach \n in {0,...,15} {
   \foreach \k in {0,...,\n} {
     \binomialCoefficient{\n}{\k}%
     \ifodd\theresult
       \node [fill=green] at (\k-\n/2,-\n){$\theresult$};
     \else
       \node at (\k-\n/2,-\n) {$\theresult$};
     \fi
   }
 }
 \end{tikzpicture}
\end{document}