A classic problem using an inductive construction is to show that the $2^n \times 2^n$-square, with a missing corner, can be tiled with L-triominoes. The proof goes like this:

It is true for $n=1$, so assume it holds for $n-1$. Now, the $2^n \times 2^n$-square is subdivided into four smaller squares. In the three sub-squares not missing a corner, consider instead the three corners in the middle, which form an L-triomino. These three can be covered with an L-triomino, and the remaining figure consists of 4 squares of size $2^{n-1}$, with a corner missing. These can now be tiled by induction.

A similar problem is the equilateral triangle, which is subdivided into $4^n$ smaller congruent equilateral triangles. Again, we remove a corner triangle, and show that the remaining can be covered with the "I-triangle-triomino", the dark shaded 3 triangles in the bottom of figure.


The question: I have struggled quite a lot to generalize these two problems, that is, find tiling problems in the same spirit. The essential parts in the two above examples are that both the square and the triangle are so called rep-tiles, and can be tiled with smaller copies of themselves.

A possible approach: Start with a rep-tile $R$, remove a small 'corner' $C$ and hope that in the rep-tiling of $mR$, ($m$ is magnification factor) then all but one 'corner' can be covered by a single figure of shape $R \setminus C$. Thus, $m^n R$ with a corner missing can always be tiled by tiles of shape $R\setminus C$, for all $n\geq 1$. However, this seem rather difficult.

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    $\begingroup$ You could go up a dimension: for n=3 as an example, certain large cubes minus a unit cube are tilable by the seven cube tile which is 2by2by2 minus a corner. Gerhard "Try Some Other Semiring Geometries" Paseman, 2019.12.30. $\endgroup$ – Gerhard Paseman Dec 30 '19 at 22:59
  • $\begingroup$ One generalization is that it doesn't have to be a corner tile that is omitted in the square case. Omitting any tile is fine, again by induction. $\endgroup$ – Joel David Hamkins Dec 30 '19 at 23:49
  • $\begingroup$ However, you can't have an arbitrary 2x2 tile missing. There are (by symmetry) at least four such configurations. Are those (eight for larger tilings) all ? Gerhard "Should This Be A Colonelization?" Paseman, 2019.12.30. $\endgroup$ – Gerhard Paseman Dec 31 '19 at 6:21

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