As a maths teacher, I often find that setting seemingly ordinary tasks for my students lead to surprisingly interesting mathematics. Recently, an investigation that on the surface seemed quite unremarkable ended with me taking a deep dive into areas of mathematics I had never seen before. It started with us thinking about the spread of a forest fire.

The trees in the forest in question are perceived to be squares on a sheet of squared paper. At the start, a random tree catches fire and grid looks this:

After a unit of time, all trees (squares) that touched a tree that is already on fire catch fire too. This leads to nine trees being on fire:

The spread of the fire repeats. After a second unit of time, there are 25 trees on fire:

If you continue, then the next numbers in the sequence are 49, 81, and 121. From immediate inspection, you can spot that these are the odd square numbers. Part of the task was to then take this and write a quadratic for the $n$th term in the sequence. Providing you know that $2n + 1$ is an algebraic way of writing odd numbers, the sequence jumps out immediately:

\begin{align*} a(n)&=(2n+1)^2\\ &=4n^2 + 4n + 1. \end{align*}

## A forest of hexagons

Following the same process as for squared paper, we can look at the fire spreading but using differently tessellated paper. Squared paper produced a quadratic sequence, but it is a little harder to predict what would happen when the tiling changes.

Looking first at the hexagonal paper, we see that, as it did for squared paper, each iteration looks like an enlarged version of the previous one:

Listing out the sequence we have 1, 7, 19, 37, 61, 91, 127, and so on. Perhaps surprisingly, this too turns out to be a quadratic sequence. Quadratic sequences are sequences that have a general rule of $$a(n)=c_1n^2+c_2n+c_3,$$ where $c_1$, $c_2$, and $c_3$ are real numbers. To calculate $c_1$, we look at the second difference between terms of the growing sequence:

The second difference is a constant, and can be halved to obtain $c_1$. Next, you can subtract the sequence $c_1n^2$ from the original sequence. If your original sequence is quadratic and you’ve done it right so far, this should leave you with a linear sequence that you can then work out the rule for. For our sequence this ends up being:

\begin{align*} a(n) &= 3n(n+1) + 1\\ &= 3n^2 + 3n + 1. \end{align*}

## A forest of triangles

Finally, the students looked at the growth on triangular paper. The spread here ends up being a little different to the previous two. For both the square and hexagonal paper the shape grows and replicates the original shape. However, for the triangular paper, as it spreads the triangle shape gets smoothed at the edges.

Continuing the sequence we get 1, 13, 37, 73, 121, 181, and so on. Once again, this is a quadratic sequence. Using a similar method as we did for hexagons, we obtain:

\begin{align*} a(n) &= 6n(n-1)+1\\ &= 6n^2-6n + 1. \end{align*}

## Polygonal numbers

These results may look like just three quadratic sequences, but there’s more that connects them: they’re all examples of a type of number sequence called the centred polygonal numbers (also called the centred $k$-gonal numbers where the choice of $k$ gives a specific sequence). These sequences are created by drawing increasingly larger polygons made of dots around a central dot, hence the name.

The centred polygonal numbers are closely related to, but not the same as, another type of number sequence: the polygonal numbers. These include the well known triangular and square numbers, and are also created by drawing increasingly large polygons of dots, but in this case the polygons all share a vertex rather than all being around a central point. You can see the difference by looking at the two sequences for a square.

The centred polygonal number can also be produced using the triangle numbers: you can start with the central dot, then make the centred $k$-gonal numbers by putting $k$ copies of the triangle numbers around the point. For the centred square and hexagonal numbers, that looks like this:

If we call the central dot the 0th centred $k$-gonal number, then we see that the $n$th $k$-gonal number is built from the $(n-1)$th triangular number. This means that the formula for the $n$th term of the centred $k$-gonal number is:

\begin{align*} a_k(n) &= k\times(\text{$n$th triangular number}) + 1\\ &= \frac{kn(n+1)}2 + 1. \end{align*} Despite our sequence of fire on squared paper being the odd square numbers, the sequence is actually the centred octagonal numbers (ie $k = 8$). This is because we can rearrange an odd square number of dots to make increasingly larger octagons from a central point. It is also interesting to note that the sum of the reciprocals of this sequence is convergent: \[ \sum_{n=0}^\infty \frac1{(2n+1)^2} = \frac{\pi^2}8. \] The sequence we observed on hexagonal paper is much more aptly named: the centred hexagonal numbers (ie $k = 6$). Hexagons are notable shapes and can be seen in the natural world in such oddities as bee honeycombs or in popular media such as the board game Settlers of Catan.

The final sequence was created from the triangular paper. This produces a sequence called the centred dodecagonal numbers (ie $k = 12$). Theses dots that make up the dodecagonal numbers can also be rearranged to produce evergrowing stars and so are also called star numbers. Like the centred hexagonal numbers, star numbers have been used in board games such as in a Chinese checkers board.

The centred polygonal numbers are interesting in their own ways for many other values of $k$, and we could talk about them for hours, but I’m going to return to my students and to something really interesting that they noticed.

The three choices of paper we used are the only three examples of regular tessellation. This is because squares, triangles and hexagons are the only three regular shapes whose interior angles are factors of 360°: *six* equilateral triangles, *four* squares, or *three* regular hexagons can fit around a single point. We’ve seen these numbers before: they are the coefficients of $x^2$ and $x$ in the general rules for the three sequences! This shows a remarkable link between the geometric property of tessellation and the algebraic representation of the sequences we produced.

This is what I love about mathematics. A task that on the surface seemed like a nice little investigation for students ended with varied classroom discussions as well as my own deep dive down a mathematical rabbit hole. Despite spending years learning and teaching, I still discover new ways of looking at things that I thought I had previously mastered. I hope that as I get older, I still get the joy of discoveries like this and can continue to share these with the students I teach.