Happy (Belated) 2026

It is customary in the new year to write a blog post containing mathematical facts about the year. For us maths writers, 2025 truly was a gift, and you can see just why in Katie’s excellent post last year. But the chances of having two consecutive years with so many fun facts are, as one might expect, small. So it shouldn’t come as a surprise that there is a dearth of interesting facts about 2026. 

2026 is not a prime number (after all, it is divisible by the prime number 1013). It is not a square number and although it is the sum of two square numbers, \( 2026 = 45^2 + 1^1 \), this feels more like a property it’s inherited from 2025. But do not fear! This is not going to be a list of anti-facts, of things that are not true about the number 2026. There is one fact that is so fun that I think it might blow all the 2025 facts out the water.

But to show you this fact, I must first introduce you to a toy that mathematicians have been playing with for over 1000 years. It may even already be familiar to you, but trust me when I say that this simple pattern holds more secrets than you or I could ever know. This tool goes by many names, but to Anglophones, this is Pascal’s Triangle.

Humble Beginnings

Pascal’s Triangle starts with what is arguably the simplest number: 1. 

In fact, every row of this triangle is bookended by a pair of 1s, so the next row is simply two ones.

The rows of the triangle are formed iteratively. Each element is formed by summing the two above it. For example, to get the middle entry of the next row, we do \(1+1 = 2\):

Now we do \(1 + 2 = 3\) and \(2 +1 = 3\):

Then \(1 +  3 =\), \(3 + 3 =6\), and \( 3 + 1 = 4\). And so on. You can keep doing this ad infinitum:

Note that if you don’t like the idea of the 1s being special and just appearing, you can imagine an infinite sea of 0s surrounding the triangle, and the 1s are actually generated by summing 1 and 0:

Some patterns in this triangle will be immediately obvious. For example, the triangle is symmetric. This makes sense because we start with a single 1, which is trivially symmetric, then every step we take to generate the next row of the triangle is done equally on the right and on the left, so of course the resulting object will be symmetric. There are much more complicated patterns and meanings hidden in the triangle though, and if you dig deep enough, 2026 crops up, and more than once too.

Patterns all the Way Down

Often, the next pattern people notice is how the “first” diagonal is the natural numbers (also known as the counting numbers). You might dispute that this is the first diagonal, but I personally count the initial diagonal with 1s in as the zeroth diagonal.

So, the expression for the nth term in the first diagonal is: \( u_n = n\), a first order (i.e. linear) equation. This pattern is to be expected – each term in this diagonal is formed by summing the previous integer (which is up and to the right of the term) to 1 (up and to the left, in our zeroth diagonal of ones).

What about the second diagonal?

If you’re well versed in common sequences,you will recognise this as the triangle numbers. The nth triangle number (let’s denote it \( t_n\)) is the sum of all the positive integers up to and including n. They are called triangle numbers because you can arrange \(t_n\) dots into a triangle of side length n:

The expression for the nth term in the second diagonal is: \( t_n = \frac{1}2(n)(n+1) \), a second order (i.e. quadratic) equation. This pattern arises because each term in this diagonal is formed by summing the previous triangle number (which is up and to the right of the term) to each natural number (up and to the left, in our first diagonal). 

What is fun is that we can actually go up another dimension. The first diagonal gave us a linear (AKA 1-dimensional) sequence, the second diagonal gives a quadratic, 2D sequence, so you might guess that the third diagonal would give a cubic, 3D sequence. And you would be right!

These are known as the tetrahedral numbers, because the nth tetrahedral number \(T_n\) spheres, they can form a tetrahedron of side length n. Note from the diagram below that each layer is a triangle, hence the tetrahedral numbers are formed by summing triangle numbers.

The formula for the tetrahedral numbers is \( T_n = \frac{1}6 n(n+1)(n+2)\}, a cubic. This pattern continues – the 4th row is a quartic (degree 4) sequence and forms 4D shapes, the 5th row is a quintic (degree 5) sequence and forms 5D shapes, and so on. 

A Journey Through Time and Space

Pascal’s Triangle is named after Blaise Pascal, a French mathematician, physicist and philosopher. Pascal lived from 1623 to 1662 AD, and is known not only for his work on Probability Theory, but the SI (International System of Units) unit of pressure is also named after him. He wrote about the triangle in his 1654 treatise, Traité du triangle arithmétique, which was published posthumously in 1665. 

But the triangle was in existence long before Pascal was even born. This is an Indian manuscript from 755 AD.

It is based on the work of Pingala, who was a poet alive in India in the third century BCE. Pingala stumbled across a whole array of combinatorial patterns, such as this, when considering the patterns that light (L) and heavy (H) syllables could form in a word of n syllables:

The first description of the triangle in a book was by Al-Karaji, a Persian mathematician and engineer born in 953 AD. Unfortunately, the book is now lost but we do have other publications of the triangle from not long after.

Moving east to China, Jia Xian was a Chinese mathematician born in 1010, who worked with the triangle in the 11th Century. In the 13th Century, Yang Hui (born 1238) presented the triangle, and it is still known as Yang Hui’s Triangle in China:

Even in Europe, the triangle predates Pascal. It first appeared in the 13th Century, in a work by Jordanus de Nemore. Then, Niccolò Fontana Tartaglia published six rows of the triangle in 1556, and in Italy the triangle is known as Tartaglia’s triangle. 

The fact that the triangle has been discovered and rediscovered all over the world goes to show how fundamental it is. Part of the reason it is so commonly known may be how it relates to combinatorics, the mathematical study of counting.

Counting Paths

In combinatorics, \( \begin{pmatrix} n \\ r \end{pmatrix}\) is used to denote the number of ways of choosing r items from a set of n, when the order doesn’t matter. As it happens, Pascal’s Triangle can also be written in this notation:

In maths, there are rarely coincidences, and this is no exception. We can view the top row as the origin. Then to move from one row to the next, we move either diagonally left and down, or diagonally right and down. The entry values tell us how many different paths travelling down from the origin lead to that location. 

To see this, note that to get to an entry, we have to arrive either from the entry above and to the right, or from the entry above and to the left. So, of course, we must add the number of ways to get to each of these to find the number of ways to get to our entry. There is only one path to get to the entries on the outer edge – choose only lefts for one side, and only rights for the other – which matches these numbers being 1.

Actually, with a bit of thought we can see that the entry labelled \( \begin{pmatrix} n \\ r \end{pmatrix}\) actually ends up denoting the number routes that are n steps down from the origin, r of which go right. 

2026

What happens, then, if we sum all the values in a row? Well the zeroth row (what I’m calling the top row) gives \(1\). The first row gives \(1 + 1 = 2\). The second row gives \(1 + 2 + 1 = 4\). The third row gives \( 1 + 3 + 3 +1 = 8\). Spot the pattern? The sum of the digits on the nth row is \(2^n\). This follows, as the nth row gives all possible choices of n steps down, where each step is either left or right, hence there are \(2^n\) steps.

What if we sum all the values up to and including the nth row? Well that is \(2^{n+1} -1\) and gives the number of such paths with up to n steps, sometimes called a hyperforest due to the tree-like appearance when all the paths are drawn. 

For the final variant, what about if we now say that you are not allowed to choose all right or all left paths – the path you take must have at least one right and at least one left step? Summing all such paths that are n steps long is equivalent to summing all the digits of the nth row of Pascal’s triangle excluding the 1s at either end so there are \( 2^n – 2\) such paths. 

And what if we want all such paths for up to n steps, i.e. the sum of the first n rows of Pascal’s triangle excluding the edge entries? There are \(2^{n+1} – 2n -2\) such ways. And when \(n= 10\)?

The sum of the shaded squares is 2026.

Happy New Year!