Triangular numbers (again) in TikZ

Question

I am trying to recreate the following diagram for triangular numbers (smaller triangles within one larger triangle) using TikZ.

The colour is not important but white and grey is probably preferred.

I have tried to modify the code used here which produces a triangular grid but without success (I do not know how to produce more than one grid in a single figure and could not shade the triangles in the alternating pattern shown) and is the reason why I have not included any MWE.

Ideally the code would be flexible enough to choose the number of larger triangles required (6 are shown in the above figure).

Answer 1

\documentclass{standalone}
\usepackage{tikz}
\newcommand\triangles[2]{\def\w{#1}\foreach\n in{1,...,#2}{\path
    ([xshift=\w]current bounding box.south east)coordinate(O);
    \foreach\s[count=\c]in{\n,...,1}{\foreach\m in{1,...,\s}{
        \fill[green!50!black](O)++(60:{(\c-1)*\w})++(0:{(\m-1)*\w})--
        ++(60:\w)--++(-60:\w)--cycle;}}}}
\begin{document}
\begin{tikzpicture}\triangles{3mm}{8}\end{tikzpicture}
\end{document}

Answer 2

A small modification of Ignasi's answer on the question you refer to:

\documentclass[tikz,border=5mm]{standalone}
\usetikzlibrary{shapes}

\tikzset{
  tri/.style={
    draw=none,
    fill=black!50,
    regular polygon,
    regular polygon sides=3, 
    minimum size=2cm,
    inner sep=0pt,
    outer sep=0pt
 }
}

\newcommand{\grid}[1]{
\foreach \i [count=\row from 0, remember=\row as \lastrow (initially 0)] in {0,...,#1}{
    \foreach \j [count=\column from 0, remember=\column as \lastcolumn (initially 0)] in {0,...,\i}{
        \ifnum\row=0
            \node[tri](0-0){};
        \else
            \ifnum\column=0
                \node[tri, anchor=corner 2](\row-0) at (\lastrow-0.corner 3) {};
            \else
                \node[tri, anchor=corner 2](\row-\column) at (\lastrow-\lastcolumn.corner 1) {};
            \fi
        \fi}}
}

\begin{document}
\begin{tikzpicture}

\foreach \x in {0,...,5}
{
\begin{scope}[xshift=2*\x cm + \x*\x cm]
 \grid{\x}
\end{scope}
}
\end{tikzpicture}
\end{document}

Answer 3

Almost as compact in Metapost...

\RequirePackage{luatex85}
\documentclass[border=5mm]{standalone}
\usepackage{luamplib}
\begin{document}
\mplibtextextlabel{enable}
\begin{mplibcode}
beginfig(1);
picture trig; 
trig = image(fill origin -- right -- right rotated 60 -- cycle withcolor 1/2 white);
for k=1 upto 7:
    for j=1 upto k:
        for i=1 upto j:
            draw trig 
                 shifted (1/2(k+1)*k + 1/2(k-j) + i, sqrt(3/4)*(k-j))
                 scaled 5mm;
        endfor
    endfor
endfor
endfig;
\end{mplibcode}
\end{document}

The trick here is that the scaled 5mm scales both the triangle picture and the amount it's shifted.

Answer 4

Here's a one path approach which allows one to use the whole construct for shadings or clipping.

Code

\documentclass[tikz]{standalone}
\usetikzlibrary{calc}
\tikzset{
  triangles/.style 2 args={
    /utils/exec=\pgfmathtruncatemacro \n         {#2}%
                \pgfmathsetlengthmacro\size      {(#1)*\n}%
                \pgfmathtruncatemacro \iterations{\n/2},
    line join=round, insert path={
      (210:\size) coordinate(x) (-30:\size) coordinate(y) (90:\size)coordinate (z)
      foreach[evaluate={\f=\i/\n;}]\i in {0, ..., \iterations}{
          ($(x)!\f!(y)$) -- ($(x)!\f!(z)$) -- ($(y)!\f!(z)$)
        \unless\ifnum\pgfinteval{2*\i}=\n\relax
        --($(y)!\f!(x)$) -- ($(z)!\f!(x)$) -- ($(z)!\f!(y)$)
        \fi -- cycle}}}}
\begin{document}
\tikz\path[ball color=blue, triangles={5mm}{6}];
\tikz\matrix[
  row sep=5mm, column sep=5mm, execute at empty cell={
    \draw[fill=green!50!black, triangles=
      {5mm}{\pgfmatrixcurrentcolumn+5*\pgfmatrixcurrentrow-5}];}]{
   & & & & \\ & & & & \\ & & & & \\};
\end{document}

Output

Answer 5

With Asymptote: for a specific triangle

// http://asymptote.ualberta.ca/
size(6cm);
int n=8;
path p=(0,0)--(1,0)--dir(60)--cycle;
for (int i=0; i<n; ++i)
for (int j=0; j<n-i; ++j)
fill(shift(i*dir(60)+(j,0))*p,purple);

and for a series of triangle: we can make a function returning a picture with a parameter for its size and add to the current picture by add(alttriangle()); (default n=8), or add(alttriangle(5));, or add(alttriangle(n=5));

// http://asymptote.ualberta.ca/
size(12cm);
picture alttriangle(int n=8){
picture temp;
path p=(0,0)--(1,0)--dir(60)--cycle;
for (int i=0; i<n; ++i)
for (int j=0; j<n-i; ++j)
fill(temp,shift(i*dir(60)+(j,0))*p,purple);
return temp;
}
int n=7;
for (int i=1; i<n; ++i) 
add(shift(i(i+1)/2,0)alttriangle(i));
//shipout(bbox(5mm,invisible));