### Different ways of counting

To introduce the topic of my next paper, I want to step back a little bit to something classically studied by mathematicians. In fact, it has a pretty rich and interesting history going back to a major result due to Leonhard Euler, which I will get to in time.

So suppose that you’re trying to organize some boxes (of books, naturally) in the back corner of the room. For stability, you put them all up in one corner, maybe like so.

Of course, this isn’t the only way that you could have stacked these boxes. You also could have done like so.

Or more inspired by Ikea, even like the following.

A natural question is, how many different ways can you do this? For the 12 boxes shown above, it turns out that there are 77 different ways of stacking them (If you want a painful exercise, try and draw all of them!).

By adding up the heights of the columns, we can view these is as different ways of writing sums that total 12. The three above are, respectively,

5 + 3 + 2 + 2 = 12

4 + 3 + 3 + 1 + 1 = 12

3 + 3 + 3 + 3 = 12.

We mathematicians call these sums (or diagrams—which we call Young diagrams) *partitions* of 12. As stated above, you could enumerate somewhat painfully that there are 77 different partitions of 12.

Here are all of the partitions of the numbers 1 through 4:

is the loneliest partition of 1;

and are better than one;

,, and are a bit of a crowd;

, , , , with being the fifth wheel;

and they go on after that. Note that while in the first three cases the number of boxes is the same as the number of diagrams, this is coincidental—what we are interested in is the number of diagrams for a fixed number of boxes.^{[1]}

Now as you may expect from my previous post, we probably want to make a generating function for these things. Euler’s theorem is that this has a very nice form as an infinite product (which isn’t as terrible as it sounds), which is essentially the inverse of the Dedekind Eta function.^{[More details]}

### Rectangles, rectangles.

Anyhow, as interesting as this is, it isn’t our main point. In order to introduce this, we need to take these ideas in a slightly different direction. If you wanted to count the divisors of a number, we can represent them in similar ways—in fact, this may be how you were introduced to multiplication in the first place!—that is, we look at how many different ways we can arrange boxes into rectangles. For example, for 4 we have the following three possibilities:

, , and .

Note that we threw two of the previous five diagrams out since they did not form rectangles. These remaining ones correspond to the divisors 1, 2, and 4 (the widths of the rectangles^{[3]}). So the number of divisors of some integer are just the number of ways of arranging that many boxes in a rectangle.

There are of course other things that you could do with this idea, the simplest of which is *summing* the divisors, which in this description just amounts to adding up the widths of the boxes. That is,

+ + = 1 + 2 + 4 = 7.

The function in question we call the “sum-of-divisors” function, and we write it as (a lower-case Greek ‘sigma’). So this above says that . It’s a fun function to play with, and it satisfies the interesting property that some of the time (can you figure out when?^{[4]}), that . Functions like this are rather rare, and so mathematicians’ ears tend to perk up when we find them.

### Yet more rectangles!

This brings us back to Percy MacMahon, and a little bit more information about my previous post. What he studied was how many ways you can partition a number into a row of rectangles, all of whom have different heights (within a single diagram, that is). If we say that we want to have exactly two different heights, then we see that for 4 total boxes we have the following two possibilities:

and ,

while for 5 we have the following 5 possibilities:

, , , , and .

As usual, there are a number of ways you can study these diagrams; you can count the number of such possibilities, or you can “add up their widths” (in a certain sense), which is what MacMahon did. In fact, he did a little bit more than that—in his paper (with the catchy title “Divisors of Numbers and their Continuations in the Theory of Partitions“) he studied a total of 8 variations of this theme, and played around with them and showed a bunch of different properties.

And now here is where I come in. In the paper that I discussed last time, two of these functions had shown up naturally in some of my other research. George Andrews and I were able to show that these functions were modular forms, and that was that. However…

As I said above, MacMahon had studied 8 variations on a theme and I could only show that two of them had some nice properties. I found this a little intellectually unsatisfying: what about the rest of them? There arose two natural questions to ask at this point.

- Are the rest of them modular?
- Given that the first two variations occurred in my research (as will be discussed in a later post), do the rest of them arise in a similar way?

The second question is still one that I have not answered, and would love to understand better. The first, however, was the topic of my next-to-most-recent paper, which I will say more about next time.

[1] This sequence goes 1, 2, 3, 5, 7, 11, 15, 22, and so on. As usual, see my bestie the OEIS. Not surprisingly, this is one of the first sequences to show up on there.

[2] For those that are curious, we write it as

.

The reasoning isn’t too hard to follow, and is kind of fun to think about: it relies on one fact (that you probably learned in high school!) about geometric series, namely that

There’s even a more interesting case if we stack boxes in the corner in a 3-d setting (with the terrible name of Plane Partition), where the generating is the famous MacMahon function. A colleague of mine essentially got his PhD by working with a variant of this problem, and describes his work as “counting coloured boxes in the corner of a room”; this is a pretty good description.

[3] Or if you’re feeling particularly contrarian, you could use the heights instead.

[4] Here’s a hint: , but .

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