Generalization of pentomino-rectangle tiling

Question

It is very well known that there are $12$ pentominos and they can tile $6 \times 10$, $5 \times 12$, $4 \times 15$ and $3 \times 20$ rectangles. Now, let's define a function for simplify this. $$t(n)=\begin{cases} 1, & \text{if $n$-ominos tile rectangle} \\ 0 & \text{else} \end{cases}$$ $t(5)=1$. Also, one can prove that $t(6)=t(4)=0$ by "checkboard coloring". It is obvious that $t(n)=0$ for $n \ge 7$ since at least one of $n$-omino contains a hole.

By simple brute force, $t(1)=t(2)=1$, $t(3)=0$.

Question. Is there any higher-dimensional researches about this? For example, is it possible to tile $4 \times 4 \times 2$ cube with $8$ possible $4$-polycubes?

Answer 1

I don't know if this is something people have studied, but after playing around for a few minutes I found this (a-h represent the eight tiles):

\begin{align} \text{a a a a} \\ \text{b d e c} \\ \text{d d d f} \\ \text{h h h f} \\ \\ \text{b b c c} \\ \text{b e e c} \\ \text{g g e f} \\ \text{h g g f} \\ \end{align}

Answer 2

This is a great question, as it allows me to mention one of my favourite combinatorial geometry results of all time, which I first read about on the great Combinatorics and More.

Consider the fact that an $n$-omino can be represented as a connected set of $n$ points in $\mathbb{Z}^2$. The result said that given any such set of points in $\mathbb{Z}^m$, for arbitrary $m>0$, there exists a $k\geq m$ such that the $n$-omino tiles $\mathbb{Z}^k$! So, any q-dimensional n-omino tiles an entire space in some dimension!

Remarkable! You can read about this here.