A series of hexagons on an hexagonal lattice means that the every point in the entire area is covered by one polygon only.
A grid of octagons will not tesselate, leaving square holes such that 4/18 of the area is not covered.
Is it possible to stack multiple layers of identical, non-tesselating, regular polygons such that every point in the area underneath is covered by the same number of integer polygons? e.g. every point being covered by n - m polygons from n layers of k-sided polygons spaced on a specified lattice (m < n).
I have started by considering the case of the octagon grid. I don't think it is possible with this basis, as it would require being able to tesselate other squares around the square vacancies, which they cannot. Are other values of k viable?
If you don't require regular octagons, you can do it. In the tessellation below, let the horizontal and vertical sides of the octagons be $1$ and the diagonal sides of the octagons be $\sqrt 2$. Then the distance between squares is $3$ and the squares cover $\frac 18$ of the plane. You can stack eight of these and cover each point seven times. You can do a similar thing with the canonical dodecagon/equilateral triangle tiling-make the sides of the dodecagons that touch each other the proper value and the triangles will fill the plane.