How many "prime" rectangle tilings are there?

Question

Given two tilings of a rectangle by other rectangles, say that they are equivalent if there is a bijection from the edges, vertices, and faces of the tilings which preserves inclusion. For instance, the following two tilings are equivalent (some corresponding vertices colored):

The number of non-equivalent rectangle tilings with $k$ tiles is given at A049021 in the OEIS.

Say that a tiling of a rectangle by $k$ other rectangles is prime if the only sub-rectangles in the tiling are individual tiles and the whole set.

Here are the unique prime tilings for $k=1,2,5$ up to equivalence:

With some casework, one can verify that there are no prime tilings for $k=3,4,6$.

For all $k\ge 7$, there are at least two prime tilings. One follows the pattern shown below:

The other proceeds in a spiral pattern (in fact this construction yields the $k=5$ solution too):

It is not hard to check that these two constructions are distinct for all $k$.

How many non-equivalent prime tilings are there with k rectangles?

When $k=7$, I have verified that the above two constructions are the only ones possible, but the casework was extensive and I think approaches for $k$ beyond $9$ or $10$ will require new methods or computer-aided searches to be tractable.

This sequence starts $1,1,0,0,1,0,2,\ge2,\ge2,\ldots$. Even accounting for a possible initial term in the case $k=0$ and the possibility that I have missed a solution for $k=7$, there are no matches in the OEIS. I have started a draft for a new sequence, but I would like to be able to compute substantially more terms than this (and confirm my existing results) before adding anything. For comparison, A053740 gives the analogous sequence for prime dissections of a triangle into smaller triangles.

Some related discussion is at this post on /r/mathriddles.

Edit: After a laborious manual classification, I believe there are exactly $6$ prime tilings in the $k=8$ case:

(Note that the horizontal symmetry of the example in the center of the second row is merely for visual appeal; the "heights" of either the SW or the SE rectangles could be varied without affecting equivalence.)

Edit 2021-02-15: These tilings are described in Tiling Rectangles With Rectangles (Chung et al., 1982), where they are called simple tilings. In the article, it is remarked that there are at least $c\cdot 2^{n/7}$ "essentially different" simple tilings with $n$ rectangles for some $c>0$ (and presumably only for $n\ge 7$?). It is also stated without proof that the number is bounded above by $20000^n$ given "a rather natural definition of equivalence". So the sequence grows exponentially, but it seems that the rate at which it does so is not very well pinned down.

Answer 1

I was able to computationally find all tilings for $k\le11$ by using the primality property to significantly reduce the search space. Specifically, it can be shown that any prime rectangular tiling with $k\ge3$ tiles can be "embedded" in a rectangle of size at most $\lfloor(2*k-1)/3\rfloor$ x $\lfloor(2*k-1)/3\rfloor$.

We will define a "locally optimal" embedding as one in which each line between two rows or two columns contains at least one dividing edge between rectangles. It can be easily seen that any embedding which is not locally optimal can be reduced in size by combining some pair of rows or columns which breaks the aforementioned rule, which will not change its topology. It can also be seen that any rectangular tiling can be turned into a locally optimal embedding by repeated applications of the reduction operation.

Each rectangle can be considered to "contribute" to the vertical segments of an embedding exaclty twice, once with its left edge and once with its right edge. We can see that, in a locally optimal embedding, each line between two columns must be contributed to at least two times, as that line must contain at least one dividing edge contributed to by its left and right rectangles.

However, it can also be shown that, in a locally optimal embedding of a prime rectangular tiling, such a line must be contributed to at least three times. This is because, if it were only contributed to exactly twice, the left and right rectangles would have the exact same top and bottom edge locations, and could therefore be combined into one rectangle, breaking the primality requirement for $k\ge3$.

We can also show that the left and right edges of the grid must each be contributed to at least twice. If the left or right edge was contributed to exactly once, the set of all rectangles other than the contributing rectangle could be combined into a single rectangle, breaking the primality requirement for $k\ge3$.

Therefore, for a grid with n columns, there must be at least $3*(n-1)+4$, or $3*n+1$, contribution. As such, it must be comprised of at least $\lceil(3*n+1)/2\rceil$ rectangles. If we require the grid to have k rectangles, then the grid can have at most $\lfloor(2k-1)/3\rfloor$ columns and, without loss of generality, at most $\lfloor(2k-1)/3\rfloor$ rows.

From here, I was able to code a fairly rudimentary backtracking solution, in which I tried all rectangles whose top left corner is the first unused cell in reading order, cutting out whenever the primality requirement is broken. To judge topological equivalence, I generated all 8 rotations/reflections and created a canonicalized representation of the graph, taking into account the directions and orderings of adjacencies.

The numbers of prime rectangle tilings comprised of $k$ tiles for $1\le k\le11$ are as follows: $1, 1, 0, 0, 1, 0, 2, 6, 29, 119, 600$

The following are all tilings for $9\le k\le11$: