By assigning different colours to the odd and even numbers in Pascal's triangle, Sierpinski's Anatomy of hesselbach triangle can be generated, as I have explained in my previous post entitled "Pascal's Triangle and Sierpinski's triangle". In this post, however, I attempt to explain why this interesting link arises between these two seemingly unrelated triangles, and how we can be sure that this pattern will always continue.

To begin with, as Pascal's Anatomy of hesselbach triangle is a series of additions and we are colouring each number by whether it is odd or even, it seems sensible to look at some basic rules of adding odd and even numbers:

Rule 1: odd + odd = even,

Rule 2: 0dd + even or even + odd = odd

Rule 3: even + even = even

So, now to the main stage of the proof. Assume that a finite number of rows of Pascal's triangle did correspond to Sierpinski's triangle after a finite number of iterations. The bottom of Sierpinski's triangle is always a row of triangles all the same colour which correspond to odd numbers in Pascal's triangle. By rule number 1, the next row of Pascal's triangle will be all even (except outer 1s which are of course odd).

From the third rule, we can see that a row such as row eight, 1,8,28,56,70,56,28,1, which has many even numbers in a row will create an upside down triangle of even numbers below it - here, we have 6 even numbers in a row, and the 5 numbers in between them in the row below will be even, by rule 3, and the 4 in between those in the row below will be even and so on. So after our row of all odd numbers, we get a triangle of even numbers.

If you try this out, you should notice that the new triangle we just created is the same size as the section of Sierpinski's triangle we were just dealing with.

We now must look at what is going to happen due to those 1s on the edge of our otherwise even row. Will they not change what goes on in the rows below them? Yes, but the fact is that the very first number in Pascal's triangle is a 1, so why would the pattern below these 1s be any different to that first section of Sierpinski's triangle we had? Sure, the actual numbers will be different, but the parity (oddness/evenness) will be identical to before, as from rule 2, the parity of a number does not change by adding an even number to it.

Let's take a step back and look at what we have achieved so far. we now have a much bigger Sierpinski's triangle which is made up of 4 parts:

1. the initial section of Sierpinski's triangle that we started with
2.  

2. the all-even upside-down triangle below it.

3. The left hand side equilateral triangle that is identical in colouring to that in 1.

4. The right hand side triangle, again identical to 1.

Therefore, we have made the next iteration of the rule generating Sierpinski's triangle! You can check it works for the first iteration, then we could do this again and again and again and create whatever iteration of Sierpinski's we wanted. Do it infinitely, and we will thus create an infinite Pascal's triangle that yields the actual infinite fractal of Sierpinski's triangle itself. We are done! We have explained the link between Pascal's triangle and Sierpinski's triangle!

I know that these kinds of proofs can sometimes be rather hard to visualise. Please visit my site shown in the resource box below, which contains three articles each (with lots of pretty pictures and diagrams) all about Sierpinski's triangle and other fractals and their link with Pascal's triangle: