## Aztec diamonds: How I came to learn of them

A little project I’m working on requires me to suss out the size of a square given the following: Imagine a square of width and height *x* units. The square is gridded by unit such that *x*^2 squares are visible inside it. How many of these squares’ perimeter lines exist inside the outer square? Or, put another way, if you then erase the border of the original square and break every line segment at the intersections, how many line segments do you have left?

It seems to work out to *x*^2+*x**(*x*-2). For a 2×2 square, there are 4 segments left. 3×3 yields 12, 4×4 yields 24, 5×5 yields 40, 6×6 yields 60, and so on. I confirmed this far by manually counting the segments, and everything seemed good. This wasn’t what I needed, however, I needed to be able to do it in reverse – given 24 segments, I needed to come up with my outer square’s width and height of 4. This was not an obvious solution.

I tried searching for things like ‘size of square from inner grid segments’, but I couldn’t really articulate the thing in a way that got me anywhere. Or, perhaps, nobody has ever really needed this version of this problem solved before (though I find that doubtful). I needed a new angle. I searched OEIS for my sequence, 4, 12, 24, 40, 60, and came up with A046092, ‘4 times triangular numbers’. Now, OEIS is great, but it has a way of presenting a lot of information very tersely, which can be overwhelming. So I googled A046092, and nearly every hit came back to one thing – Aztec diamonds.

Further searching revealed that Aztec diamonds are popular because of the Aztec diamond theorem and the Arctic Circle theorem, both related to domino tilings. This is all very fascinating, but unfortunately presented me with a dead end. Fortunately, I did also discover that Aztec Diamond Equestrian is a company that makes leggings with very functional looking pockets, so that was a win. But on the math side of things, I wasn’t coming up with much. I did, at least, realize that if I rotated my grid and treated each of those segments as the diagonal of a square, I was in fact dealing with an Aztec diamond. If nothing else, this allowed me to confirm that A046092 was the sequence I was dealing with, it allowed me to confirm my original equation, and therefore meant I could test any arbitrary case without manually counting.

I started noticing other patterns, too. The problem is complicated because it’s like a square, but it’s missing the corners. This is where the narrative begins to fall apart. All these thoughts of squares and right triangles, hypotenuses… I established a formula that I have tested for the first 1000 integers in A046092, they all pass. But I cannot come up with a proof that explains it. Furthermore, I am not confident that it actually holds up despite the first 1000 integers validating. The next step is scripting the formulae out to test well beyond 1000.

Now, for my project, I don’t need anywhere near 1000, so the equation is Good Enough. Hell, I could probably hardcode the sequence as far as I need it, and a lookup table would likely be faster than a bunch of math. But now I’m *really* curious. So for a square *x* by *x* units, and *y* as the *x*th integer in A046092, I’d love to prove that *x*=ceil(sqrt(2*y*)/2). I don’t know that I can, but I would really like to. Ultimately I know I’m looking at a hypotenuse, or an approximation of a hypotenuse, with the Pythagorean theorem being beautifully simple given the area of a square.

I guess, in the end, I have an equation that more than meets my needs. I learned about Aztec diamonds, and I figured out why this problem is not as simple as it originally seems. I also learned about some pretty bangin’ equestrian leggings. I’m going to keep at this, though, because it fascinates me, and I haven’t found a solution in the wild. I’ll report back, but before I do, there will likely be a game-in-a-post to explain *why* I was trying to solve it in the first place.