Fractals patterns hidden inside Pascal's triangle

July 31, 2026

Pascal's triangle is a wonderful pattern, with lots of applications in combinatorics. It looks like this:

To see the pattern, let's step through the construction of a new row.

You can continue Pascal's triangle indefinitely by using this summation pattern. Pascal's triangle has great importance in mathematics (it's entries are called binomial coefficients, and appear in expansions of binomials, in combinatorics, and in many other parts of math).

However, in today's article, we won't be exploring those uses of Pascal's triangle. Instead, we wanted to showcase a very pretty pattern.

Let's highlight each entry of the triangle according to its parity — whether it's odd or even:

Notice how the shadings seem to form a pattern, similar to the fractal Sierpinski triangle.

Why do the digits make this pattern? What can we say about it? For that, we need Lucas' theorem. But before we get to Lucas' theorem, we need to say something about Pascal's triangle.

Binomial coefficients

The numbers in Pascal's triangle are called binomial coefficients. They arise very often, because of two crucial applications.

People typically write \(\binom{n}{k}\) (pronounced "\(n\) choose \(k\)") to denote the \(k^{\text{th}}\) entry on the \(n^{\text{th}}\) row of Pascal's triangle. For example, below we recreated the first figure of Pascal's triangle, but labelling each cell with \(\binom{n}{k}\), instead of with the actual value; compare this to the figure at the start of the page.

The quantity \(\binom{n}{k}\) appearing in Pascal's triangle is equal to the number of ways to choose \(k\) items from a group of \(n\) items. For example, we can see from Pascal's triangle (look at the figure from the start of the article) that \(\binom{4}{2} = 6.\) And we can interpret that as follows.

I have four neighbors: Sophie, Kenta, Ines, and Xiaoyu. I want to choose two of them to ask for help moving into my new apartment. How many ways are there to do this? If you enumerate all the possible pairings, there are six different groups of two I can form from four people:

This is why \(\binom{4}{2} = 6\): if you have a group of four entities, and want to choose two of them, then there are six ways to do so. This is also why \(\binom{4}{2}\) is pronounced as "four choose two": it counts the number of ways to choose two things from a group of four.

The numbers in Pascal's triangle have a separate interpretation, which is why they are called binomial coefficients. Recall that \[(x+y)^2 = x^2 + 2xy + y^2.\] We can rewrite this as \[(x+y)^2 = \binom{2}{0}x^2 + \binom{2}{1}xy + \binom{2}{2}y^2.\] Similarly, \[(x+y)^3 = \binom{3}{0}x^3 + \binom{3}{1}x^2y + \binom{3}{2}xy^2 + \binom{3}{3}y^3,\] and \[(x+y)^{5} = \binom{5}{0}x^{5} + \binom{5}{1}x^4y + \binom{5}{2}x^3y^2 + \binom{5}{3}x^2y^3 + \binom{5}{4}xy^4 + \binom{5}{5}y^5.\] This holds for any exponent you want: to raise a sum of two variables to a large power, you can always use Pascal's triangle.

We'll remark here that, while we defined the binomial coefficients via a certain recursive procedure (where you generate Psacal's triangle one row at a time), there is a closed form for binomial coefficients, in terms of factorials: \[\binom{n}{k} = \frac{n!}{k!(n-k)!}.\] This way, you don't need to generate rows \(1, 2, ..., n\) of Pascal's triangle just to compute row \((n+1)\) -- you can use this closed form instead.

Lucas' theorem

Édouard Lucas was a ninteenth century French mathematician, who was very interested in number theory. As the binomial coefficients \(\binom{n}{k}\) -- that is, the entries of Pascal's triangle -- arise so often, Lucas was interested in exploring their number theoretic properties. One of the most basic number theoretic properties of binomial coefficients is asking when \(\binom{n}{k}\) is even, and when it is odd.

Lucas observed that this depended only on the binary expansions of \(n\) and \(k.\) Let's look at a few examples.

Lucas observed a pattern: \(\binom{n}{k}\) is even exactly when there is some spot where the binary expansion of \(n\) has a 0, but the binary expansion of \(k\) has a 1. The activity above highlights the bits where this happens; try playing with it to see the pattern for yourself!

The fractal pattern

Using Lucas' theorem, let's attempt to understand the pattern in Pascal's triangle colored according to parity, as we saw above.

Let's look at our colored Pascal's triangle again, but this time with the cells labelled differently. Instead of writing the numerical value of \(\binom{n}{k},\) like we did before, each cell is labelled with the binary expansions of \(n\) and \(k.\)

Notice the self-similarity of this figure. For example, at the top of our colored Pascal's triangle, we see one black triangle surrounded by three identical looking yellow triangles. Let's reproduce those three triangles side by side (you can also look back at our original graphic, to see these pieces sitting inside the full Pascal's triangle).

The similarity of these triangles are explained by a certain self-similarity within binary. In binary, the numbers 0 through 7 become \[000, 001, 010, 011, 100, 101, 110, 111.\] And the numbers 8 through 15 becomes \[1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111.\]

Notice how the numbers 8 through 15 are identical to the numbers 0 through 7, but with 1's added at the start. And indeed, if you look at our three triangles above, and examine the labels on their columns closely, you'll notice that the triangles are almost identical: the only difference is the first bit of each label. In the top copy, every bit string begins with \(0\); in the bottom-left copy, the \(n\) bit strings begin with 1 and the \(k\) bit strings begin with 0; and in the bottom-right copy, every bit string begins with \(1.\)

If we think back to Lucas' theorem, changing the first bit of \(n\) from \(0\) to \(1\) will have no impact on whether \(\binom{n}{k}\) is even or odd (see for yourself by playing with the Lucas' theorem widget!). So, these three pieces of Pascal's triangle, colored by parity, end up looking identical! This remarkable self-similarity in binary explains the self-similarity in our colored Pascal's triangle.

More on Lucas' theorem