How to highlight the tetrahedral numbers in Pascal's triangle

Question

The tetrahedral numbers count the number of balls with equal radii needed to create a triangular pyramid with n layers.  The nth tetrahedral number is given by the formula C(n + 2, 3).  Accordingly, the tetrahedral numbers appear as a diagonal in Pascal's triangle.  I wish to highlight those numbers in Pascal's triangle.  However, what my attempt produced was Pascal's triangle with a separate row in which the highlighted numbers appeared in red.  How do I modify my code so that the tetrahedral numbers are highlighted in the triangle itself?

Here is the code, which is based on Caramdir's answer to this question about how to present the binomial coefficients in Pascal's triangle in numerical form:
documentclass{article}
usepackage{tikz}

%calculate binomial coefficients 
makeatletter
newcommandbinomialcoefficient[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%
    advancec@pgf@countc by-c@pgf@countb%
    ifnumc@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
        ifnumc@pgf@countd<c@pgf@countb%
        multiplyc@pgf@countc byc@pgf@counta%
        advancec@pgf@counta by-1%
        advancec@pgf@countd by1%
        dividec@pgf@countc byc@pgf@countd%
    repeatpgfmathloop%
    thec@pgf@countc%
}
makeatother

begin{document}

begin{figure}[h]
centering
begin{tikzpicture}
        foreach n in {0, ..., 7} {
            foreach k in {0,...,n} {
                node (nk) at (k-n/2,-n) {(binomialcoefficient{n}{k})};
                foreach n in {3, 4, ..., 7} node[color = red] at (n, 3) {(binomialcoefficient{n}{3})};
            }
            pgfmathtruncatemacro{x}{(n+1)/2}
            pgfmathtruncatemacro{y}{n/2}
        }
end{tikzpicture}
end{figure}

end{document}

AndréC · Accepted Answer

The tetrahedral numbers in your algorithm are identified by the fact that the difference between the n index and the k index is equal to 3.
So, just calculate this difference and color them as here :
pgfmathparse{int(n-k)}
ifnum pgfmathresult=3
node[red,node font=bf] (nk) at (k-n/2,-n) {binomialcoefficient{n}{k}};
else
node (nk) at (k-n/2,-n) {binomialcoefficient{n}{k}};
fi

documentclass{article}
usepackage{tikz}

%calculate binomial coefficients 
makeatletter
newcommandbinomialcoefficient[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%
    advancec@pgf@countc by-c@pgf@countb%
    ifnumc@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
        ifnumc@pgf@countd<c@pgf@countb%
        multiplyc@pgf@countc byc@pgf@counta%
        advancec@pgf@counta by-1%
        advancec@pgf@countd by1%
        dividec@pgf@countc byc@pgf@countd%
    repeatpgfmathloop%
    thec@pgf@countc%
}
makeatother

begin{document}

begin{figure}[h]
centering
begin{tikzpicture}
        foreach n in {0, ...,7} {
            foreach k in {0,...,n} {
             pgfmathparse{int(n-k)}
             ifnum pgfmathresult=3
             node[red,node font=bf] (nk) at (k-n/2,-n) {binomialcoefficient{n}{k}};
             else
             node (nk) at (k-n/2,-n) {binomialcoefficient{n}{k}};
             fi
%                foreach n in {3, 4, ..., 7} node[color = red] at (n, 3) {(binomialcoefficient{n}{3})};
            }
            pgfmathtruncatemacro{x}{(n+1)/2}
            pgfmathtruncatemacro{y}{n/2}
        }
end{tikzpicture}
end{figure}

end{document}

JeT · Answer

AndreC was too fast for me :)
Just an alternative to evaluate directly x, y and n-k directly in the loop and avoid pgfmathtruncatemacro
documentclass{standalone}
usepackage{tikz}

%calculate binomial coefficients 
makeatletter
newcommandbinomialcoefficient[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%
    advancec@pgf@countc by-c@pgf@countb%
    ifnumc@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
        ifnumc@pgf@countd<c@pgf@countb%
        multiplyc@pgf@countc byc@pgf@counta%
        advancec@pgf@counta by-1%
        advancec@pgf@countd by1%
        dividec@pgf@countc byc@pgf@countd%
    repeatpgfmathloop%
    thec@pgf@countc%
}
makeatother

begin{document}

begin{tikzpicture}
    foreach n in {0, ..., 11} 
    {
        foreach [evaluate ={
                x = int(0.5*(n+1));
                y = int(0.5*n) ;
                NmK = int(n-k) ; %n minus k, NmK
            }] k in {0,...,n} 
        {
            ifnum  NmK=3          
                node[red,node font=bf] (nk) at (k-n/2,-n) {binomialcoefficient{n}{k}};
            else
                node (nk) at (k-n/2,-n) {binomialcoefficient{n}{k}};
            fi

}
    }
end{tikzpicture}

end{document}

N. F. Taussig · Answer

The nice solutions by AndreC and JeT both highlight the C(n + 2, n - 1) entries, with n being at least 1.  Since I was interested in highlighting the C(n + 2, 3) entries, again with n being at least 1, I modified AndreC's code to highlight the opposite diagonal as follows:
documentclass{article}
usepackage{tikz}

%calculate binomial coefficients 
makeatletter
newcommandbinomialcoefficient[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%
    advancec@pgf@countc by-c@pgf@countb%
    ifnumc@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
        ifnumc@pgf@countd<c@pgf@countb%
        multiplyc@pgf@countc byc@pgf@counta%
        advancec@pgf@counta by-1%
        advancec@pgf@countd by1%
        dividec@pgf@countc byc@pgf@countd%
    repeatpgfmathloop%
    thec@pgf@countc%
}
makeatother

begin{document}

begin{figure}[h]
centering
begin{tikzpicture}
        foreach n in {0, ...,7} {
            foreach k in {0,...,n} {
             ifnum k=3
             node[red,node font=bf] (nk) at (k-n/2,-n) {binomialcoefficient{n}{k}};
             else
             node (nk) at (k-n/2,-n) {binomialcoefficient{n}{k}};
             fi
%                foreach n in {3, 4, ..., 7} node[color = red] at (n, 3) {(binomialcoefficient{n}{3})};
            }
            pgfmathtruncatemacro{x}{(n+1)/2}
            pgfmathtruncatemacro{y}{n/2}
        }
end{tikzpicture}

end{figure}

end{document}

How to highlight the tetrahedral numbers in Pascal's triangle

3 Answers

Add your own answers!

Ask a Question