At here Pascal's triangle in tikz we can draw Pascal's triangle. Now I want to note some properties of coefficients C_{n+3}^4 -C_{n+2}^4-C_{n+1}^4+C_{n}^4=n^2 (the first picture) as folowing pictures 
How can I draw this pictures?
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2 Answers
I'll left to you understand the code but next is a possible solution based in Paul Gaborit's example
%%%%%%%%%%%%%%%%%%%%%%%%%%% % Author : Paul Gaborit (2009) % under Creative Commons attribution license. % Title : Pascal's triangle and Sierpinski triangle % Note : 17 lines maximum \documentclass[border=2mm, tikz]{standalone} %\usepackage[landscape,margin=1cm]{geometry} %\pagestyle{empty} %\usepackage[T1]{fontenc} %\usepackage{lmodern} \usepackage{tikz} \usetikzlibrary{positioning,shadows,backgrounds,shapes.geometric} \begin{document} \centering % % x=\sqrt{3/4}*minimum size % y=3/4*minimum size % \begin{tikzpicture}[y=7.5mm,x=8.66mm] % some colors \colorlet{even}{cyan!60!black} \colorlet{odd}{orange!100!black} \colorlet{links}{red!70!black} \colorlet{back}{yellow!20!white} % some styles \tikzset{ box/.style={ regular polygon, regular polygon sides=6, minimum size=10mm, inner sep=0mm, outer sep=0mm, text centered, font=\small\bfseries\sffamily, text=#1!50!black, draw=#1, line width=.25mm, rotate=30, }, link/.style={black, shorten >=2mm, shorten <=2mm, line width=1mm}, } % Pascal's triangle % row #0 => value is 1 \node[box=even] (p-0-0) at (0,0) {\rotatebox{-30}{1}}; \foreach \row in {1,...,11} { % col #0 => value is 1 \node[box=even] (p-\row-0) at (-\row/2,-\row) {\rotatebox{-30}{1}}; \pgfmathsetmacro{\myvalue}{1}; \foreach \col in {1,...,\row} { % iterative formula : val = precval * (row-col+1)/col % (+ 0.5 to bypass rounding errors) \pgfmathtruncatemacro{\myvalue}{\myvalue*((\row-\col+1)/\col)+0.5}; \global\let\myvalue=\myvalue % position of each value \coordinate (pos) at (-\row/2+\col,-\row); % odd color for odd value and even color for even value \pgfmathtruncatemacro{\rest}{mod(\myvalue,2)} \node[box=even] (p-\row-\col) at (pos) {\rotatebox{-30}{\myvalue}}; } } \begin{pgfonlayer}{background} \foreach \i/\j in {4/1,5/1,5/2,6/2,7/6,8/6,8/7,9/7} \node[box=even,fill=odd] at (p-\i-\j) {}; \end{pgfonlayer} \draw[link] (p-4-1.center)--(p-6-2.center); \draw[link] (p-5-1.center)--(p-5-2.center); \draw[link] (p-7-6.center)--(p-9-7.center); \draw[link] (p-8-6.center)--(p-8-7.center); \node[right=5mm of p-8-8.center, align=left] {$(36-7)+(28-8)=49$}; \node[left=5mm of p-5-0.center, align=right] {$(15-4)+(10-5)=16$}; \end{tikzpicture} \end{document} 
(19/11/18) Code updated to avoid problems mentioned in Problem with Pascal triangle example
- 1hi. There may be a problem in this code, an interaction with some other package. If you have trouble compiling it, see my other post : tex.stackexchange.com/questions/460463/…Nicolas FRANCOIS– Nicolas FRANCOIS2018-11-17 18:14:20 +00:00Commented Nov 17, 2018 at 18:14
- Please consider replacing
\valueby something less dangerous (OK, this expression sounds stronger than it should sound, this is not meant to sound strong) since\valuehas a well-defined meaning in many packages/classes, so there could be problems, see e.g. this post. How about\myvalue?user121799– user1217992018-11-17 18:17:31 +00:00Commented Nov 17, 2018 at 18:17 - @NicolasFRANCOIS Code updated.Ignasi– Ignasi2018-11-19 08:12:57 +00:00Commented Nov 19, 2018 at 8:12
Not a complete answer, but just intended to show a different way of drawing the triangle and also calculating the values. Requires lualatex:
\documentclass[tikz,border=5]{standalone} \usetikzlibrary{shapes.geometric} \directlua{ function factorial (f) if f < 2 then return 1 else return f*factorial(f-1) end end function nchoosek(n, k) return factorial(n) / (factorial(n-k) * factorial(k)) end } \tikzset{hexagon/.style={ regular polygon, regular polygon sides=6, shape border rotate=30, minimum size=1cm, inner sep=0pt, draw, ultra thick, execute at begin node={\setbox0\hbox\bgroup}, execute at end node={\egroup\pgfmathparse{min(4ex,\wd0)/\wd0}% \scalebox{\pgfmathresult}{\box0}} }} \begin{document} \tikz[x=1cm*sin 60, y=1.5cm*cos 60] \foreach \n in {0,...,11} \foreach \k in {0,...,\n} \node [hexagon] at (-\n/2+\k, -\n) {\directlua{tex.print("" .. nchoosek(\n,\k))}}; \end{document} 
The advantage with using lualatex is that it is possible to bypass the limits on mathematical calculations in PGF (actually in TeX):
- This won't compile at my place as of today. The fix is explained here : add \RequirePackage{luatex85} as the very first line, a temporary workaround until standalone is updated.marsupilam– marsupilam2016-10-25 16:29:44 +00:00Commented Oct 25, 2016 at 16:29
- Hi @mark-wibrow, I would like to use your code with a little change: I would like to be able to change the background color of some hexagons, inserting some conditional code. I don't know where to insert it, though. A simple line like "\ifthenelse{\n=2}{fill=blue!30,}{}" won't work. How could I do?zar– zar2016-11-02 17:31:22 +00:00Commented Nov 2, 2016 at 17:31