Inspired by some other questions, (this and this), I wonder what numbers $n$ there are that satisfy

$$p(n)=\text{there is no region that admits exactly } n \text{ domino tilings}.$$
If this is true, $n$ is *non-realizable*, otherwise, it is *realizable*.

A region in the plane is a union of squares in the unit square grid, and a domino is of course two adjacent unit squares.

Clearly, if $m$ and $n$ *are* realizable, then so is $mn$.

Note that all powers of $2$ are easily realizable. $2 \times k$-regions have a Fibonacci-number of tilings, so all Fibonacci-numbers are realizable.

Note that there is no restriction on the subset of the plane (it can be disconnected, or have holes). Will the answer be different if the region is simply connected? In the latter case, there is a theorem stating that we can reach each tiling from any other using "flips".

**Update:** Let $F_k$ be the number of ways to tile a $2 \times k$-rectangle. This is a Fibonacci number, $F_1=1,F_2=2,F_3=3,\dots$.

Consider the Young diagram given by the partition $(k,k,2)$. We can either choose to have a horizontal domino in the third row, which give $F_k$ tilings of the remaining, or the third row is covered by two vertical dominos. The remaining part can then be tiled in exactly $F_{k-2}$ ways. Hence, all numbers of the form $F_k + F_{k-2}$ are realizable. In particular, $7 = 2+5$, and $11 = 3+8$.

We can do a more general construct on the "other end" of a long $2\times k$-shape, and see that all $F_k + F_{k-2}+F_{k-4}$ are realizable whenever $k\geq 4$.

Thus, we are now incredibly close to applying Zeckendorf's theorem.