Creating a triangular table

TikZ has a rotate option. Here is an example:

% without rotate
\begin{tikzpicture}[y={(0, -1)}]
    \path (0.5, -0.5) node{1} ++(1, 0) node{2};
    \path (-0.5, 0.5) node{1} ++(0, 1) node{2};
    \draw (0, 2) grid (2, 0);
    \node at (0.5, 0.5) {$12.34$};
    \node at (2.5, 2.5) {$A_1$};
\end{tikzpicture}

% with rotate
\begin{tikzpicture}[y={(0, -1)}, rotate=-45]
    \path (0.5, -0.5) node{1} ++(1, 0) node{2};
    \path (-0.5, 0.5) node{1} ++(0, 1) node{2};
    \draw (0, 2) grid (2, 0);
    \node at (0.5, 0.5) {$12.34$};
    \node at (2.5, 2.5) {$A_1$};
\end{tikzpicture}

And the results:

---

Update:

\begin{tikzpicture}[y={(0, -1)}, rotate=-45]
    \path (0.5, -0.5) node{1} ++(1, 0) node{2};
    \path (-0.5, 0.5) node{1} ++(0, 1) node{2};
    \foreach \y in {0,...,2} {
        \foreach \x in {0,...,\y} {
            \draw (\x, \y - \x) rectangle +(1, 1);
        }
    }
    \node at (0.5, 0.5) {$12.34$};
    \node at (2.5, 2.5) {$A_1$};
\end{tikzpicture}

Result:

The diamond shape is only there for drawing and filling, the text is actually only a label (which is actually a node too) to the diamond node.

This works best only with an angle of 45.
One could also solve this with a custom coordinate system (x going ↗, y going ↘) instead of rotation, the squares/diamonds could have been drawn also with a rectangular path.

The size of the shape is manually set to

minimum size=1.414cm+0.4\pgflinewidth

The co-efficient of \pgflinewidth is found empirical and is chosen so that the lines overdraw eachother, as a grid would do that.

Code

\documentclass[tikz]{standalone}
\usetikzlibrary{shapes.geometric}
\begin{document}
\begin{tikzpicture}[
    rotate=-45,
    every label/.append style={text depth=+0pt},
    label position=center,
    every cell/.style={fill=gray!25},
    column 3/.style={fill=red!25},
    row 5/.style={fill=green!25},
    cell 2-2/.style={fill=gray},
    cell 3-2/.style={fill=gray!50},
    ]
\foreach \jRow[count=\jCount from 1, remember=\mCount] in {%
        0,%
        {15750,0},%
        {7875,2625,0},%
        {9375,4375,750,0},%
        {11875,7125,2500,1000,0},%
        {15125,10500,5375,3500,5000,0}%
    } {
    \foreach \mCell[count=\mCount from 1, remember=\mCount] in \jRow {
        \node[
            diamond,
            minimum size=1.414cm+0.4\pgflinewidth,
            draw,
            every cell/.try,
            row \jCount/.try,
            column \mCount/.try,
            cell \jCount-\mCount/.try,
            label={\pgfmathprintnumber{\mCell}},
            alias=@lastnode,
            alias=@lastrow-\mCount
        ] at (\mCount-.5,\jCount-.5) {};
        \ifnum\mCount=1
            \path [late options={name=@lastnode, label=above left:$\jCount$}];
        \fi
    }
        \path [late options={name=@lastnode, label=below:$A_\jCount$}];
    }
    \foreach \mCountExtra in {1,...,\mCount}
        \path [late options={name=@lastrow-\mCountExtra, label=above right:$\mCountExtra$}];
\end{tikzpicture}
\end{document}

Output