, 2023

Solution:
Consider the 4 largest plots Roger will fence off. We will prove they will comprise the border of the garden.

Consider a vertical $1 \times 2k$ piece. Clearly, one of its short (horizontal) edges must touch the border, because otherwise there would be a narrow margin of width smaller than $1$ on either side of the piece which cannot belong to any of the other rectangles.

We will show that a long (vertical) edge also touches the border. If this is not the case, then the horizontal space on either side of the rectangle is strictly less than $2n$, meaning that both horizontal plots would need to be situated in the remaining $(2n+1) \times 1$ strip above or below our rectangle. This is clearly impossible since $2n + 2n > 2n+1$.

Therefore it is clear that a $1 \times 2k$ piece must touch both a vertical and horizontal border, and therefore a corner; the same is true by symmetry for a $2k \times 1$ piece. We therefore have one such piece for every corner, and it is simple to see there are only two configurations possible:

After removing these 4 pieces, we are now left with a $(2n-1) \times (2n-1)$ square in the middle, which has to be subdivided exactly like in the initial problem statement (for $n-1$ instead of $n$). Iterating the same argument should give us the answer $2^{n}$. We prove this more formally with induction.

Induction hypothesis: There are exactly $2^{k}$ possibilities to cover the $(2k+1) \times (2k+1)$-square.

Base case: For $n=1$ we have the $2=2^{1}$ possibilities described above for the border. The remaining area is exactly the $1 \times 1$ square, which means we don't get more possibilities.

Induction step: By the reasoning above, we first choose one of two possibilities for the border of the $(2n+1) \times (2n+1)$-square and end up with a $(2n-1) \times (2n-1)$-square in the middle, which can be covered in $2^{n-1}$ different ways by the induction hypothesis for $k=n-1$. Since we can combine both possibilities for the border with all possibilities of the interior, we obtain $2 \cdot 2^{n-1} = 2^{n}$ possibilities in total.

This proves that $2^{n}$ is indeed the answer.