Let's figure out how many actions are needed to place a piece on square $k$. For convenience, let's denote the action of placing a piece on square $k$ as k+ and removing a piece as k-.

• Square 2: 1+ 2+
• Square 3: 1+ 2+ 1- 3+
• Square 4: 1+ 2+ 1- 3+ 1+ 2- 1- 4+
• Square 5: 1+ 2+ 1- 3+ 1+ 2- 1- 4+ 1+ 2+ 1- 3- 1+ 2- 1- 5+

From this pattern, we can infer that the number of actions needed to place a piece on square $k$ is $2^{k-1}$.

To explain more logically: to place a piece on square $k$, we need to create a state where there is a piece on square $k-1$ and no pieces on squares 1 through $k-2$. Then, to place a piece on square $k+1$, we need to remove the piece from square $k-1$, which involves almost the same process as placing a piece on square $k-1$ and removing pieces in front of it. The only difference is that the action of placing a piece on square $k-1$ is replaced with removing a piece from square $k-1$.

Now, returning to the original problem, the number of actions needed to place a piece on square 20 is $2^{19}$, and at this point, there is also a piece on square 19, so the next square to fill is square 18. Repeating this process, the answer becomes $2^{19} + 2^{17} + \cdots + 2^1 = \frac{2(2^{20} - 1)}{3}$.