TITLE: Tiling a rectangle with all simply connected polyominoes of fixed size
QUESTION [15 upvotes]: For which values of $n$ can we tile some rectangle with one copy of each free simply-connected $n$-omino (that is, each polyomino with $n$ squares that has no holes)?
It appears that it is possible for $n=1$ (trivial), $n=2$ (trivial), $n=5$ (see here), and $n=7$ (see here); and impossible for $n=3$ (trivial), $n=4$ (by a parity argument), and $n=6$ (by a parity argument).  Is it known to be possible or impossible for any $n\geq 8$?

REPLY [12 votes]: Timothy Budd's construction for $n =20$ actually shows that it is impossible for $n \geq 18$. The following images show constructions for $18 \leq n \leq 23$. For each of them, the polyomino containing the red square must fill the cavity and there are not enough possible ways of attaching the remaining (at most $2$) squares to cater for all siblings.

To go from $n$ to $n+4$ we can for instance add an additional column and increase the size of the "lump" by $2$.