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You have an n by m (or n by n for simplicity) grid; how many tilings can be made from rectangles which are smaller or equal to the grid itself. Just to be clear, rotations and reflections (of tilings that aren't symmetric) count as distinct tilings. The same rectangle can be used multiple times. Just to sanity check and make sure we're on the same page, the 2 by 2 case should have 8 tilings.

There are similar questions like Rectangling the Rectangle, but this one asks about similar rectangles instead of all rectangles that fit within the grid.

I'd imagine this count grows extremely fast. Even the 3 by 3 case is non-trivial to do by hand in a naive pen and paper way.

An attempt was made to naively solve the problem using compositions ($2^{n-1}$ compositions can be made from any integer n). The problem can be recursively solved. For the 3 by 3 case, you have the compositions of 3 which are $\{1+1+1, 1+2, 2+1, 3\}$. For the example of the $1+1+1$ composition, the top row of the 3 by 3 grid would be occupied with 3 rectangles of width 1. For the $1+2$ composition. The top row of the grid would be occupied with 2 rectangles of width 1 and then width 2.

Back to the example of the $1+1+1$ composition: Each rectangle of width 1 can be a height of either 1,2, or 3. This can be seen to lead to $3^3$ different possible arrangements of the top row of rectangles. Each composition should lead to different sets of possible arrangements of top row rectangles. From here though I realized, trying to solve it recursively by treating the part beneath each of the top row as a separate grid would not account for all possible arrangements of rectangles. This is where I get stuck.

I'm not familiar with discrete math, and I just randomly came up with this problem to see if I could manage it as a recreational exercise. It would be nice if someone could shine some light on this problem.

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    Each internal vertex has four edges connected to it (there are $E=n(m-1)+m(n-1)$ total edges). This problem is equivalent to counting the subsets of edges such that each vertex has either $4$, $3$, $0$, or $2$ edges connected to it (and, if $2$, either NS or EW). Equivalently, you can choose the $V=(m-1)(n-1)$ vertices, each with $8$ legal states, with consistency conditions between neighboring vertices. So a simple upper bound on the answer is $2^E < 4^{mn}$. – mjqxxxx Jan 01 '25 at 01:11
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    A transfer matrix approach should work for finding the exact result when either $m$ or $n$ is small. – mjqxxxx Jan 01 '25 at 01:14

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