I’ve recently been looking at closed walks on tilings of the plane in which the “player” can move from one tile to any of its edge-adjacent neighbors. In particular, I’m trying to find asymptotic formulae for various tilings giving the number of n-step closed paths starting and ending at an arbitrary tile. For an infinite square checkerboard we have $$\sim \frac{4^n}{n}\cdot\frac{2}{\pi}$$ and for an infinite hexagonal honeycomb we have $$\sim \frac{6^n}{n}\cdot \frac{\sqrt{3}}{2\pi}$$ and for an infinite tiling of equilateral triangles: $$\sim \frac{3^n}{n}\cdot\frac{3}{\pi}$$ where the first and third formulae are applicable only for even $n$, since there are no closed paths of odd length on the square and triangular tesselations. All of these can be derived using the method employed in this answer.

**QUESTION:** Can anybody derive analogous formulae for less “nice” tilings, like the cairo pentagonal tiling, the floret pentagonal tiling, the tetrakis square tiling, the snub square tiling, or any others? Clearly they should be in the form of a constant times $c^n/n$, where $c$ is a constant depending on the tiling that is more easily calculated ($c=5$ for cairo and floret, $c=3$ for tetrakis square, and $c=\frac{1+\sqrt{33}}{2}$ for the snub square).

**WHY I CAN’T SOLVE IT:** The method I used to derive these 3 formulae relied heavily on the fact that all tiles in the tiling had the same *orientation*, and that the same “moves” could be made no matter what tile is occupied (for the square tiling, the player can always move N,E,S,W, and for the hexagonal tiling, the player can always move in the same 6 directions). This isn’t actually true for the triangular tiling, but luckily the set of moves that the player can make alternates from one move to the next, so I was able to use the same method with a slight modification. This is because there are two orientations of equilateral triangles in this tiling, and all triangles edge-adjacent to a triangle of one orientation are of the opposite orientation.

This problem has been slowly killing me for a long time, partly because it seems like it should be easy but isn’t (for me, at least). Help?

**NOTE:** Since posting the question, I have come up with the formula for another tiling that uses only regular hexagons and equilateral triangles. The formula applies to walks starting in a hexagon and is valid only for even $n$:

$$\sim \frac{(3\sqrt{2})^n}{n}\cdot \frac{9}{4\pi}$$