IRL ABooklet

Let $b_n$ be the number of ways of covering a $3 \times (2n + 1)$ chessboard with one corner square removed with $3n + 1$ dominoes. We are interested in the value of $b_{1010}$.
Let $a_n$ be the number of ways of covering a $3 \times 2n$ chessboard with $3n$ dominoes. We develop a recurrence relation for $a_n$ and $b_n$.
For $n \ge 1$, let us consider a covering of the $3 \times 2n$ chessboard. Number the cells of the chessboard $(i, j)$ where $1 \le i \le 3$ is a row and $1 \le j \le 2n$ is a column. With respect to the first column, there are three cases:
* All three dominoes in this column are horizontal.
* There is a vertical domino covering $(1, 1)$ and $(2, 1)$ which implies there is a horizontal domino covering $(3, 1)$ and $(3, 2)$.
* There is a vertical domino covering $(2, 1)$ and $(3, 1)$ which implies there is a horizontal domino covering $(1, 1)$ and $(1, 2)$.
The number of ways of covering the remaining parts of the chessboard are $a_{n-1}$, $b_{n-1}$ and $b_{n-1}$ respectively. Thus we have:
$$a_n = a_{n-1} + 2b_{n-1}.$$
Now consider a covering of the $3 \times (2n + 1)$ chessboard with the square $(3, 1)$ removed. There are two cases:
* The cells $(1, 1)$ and $(2, 1)$ are covered by a single vertical domino.
* There is a horizontal domino covering $(1, 1)$ and $(1, 2)$ and another horizontal domino covering $(2, 1)$ and $(2, 2)$. This implies there must also be a horizontal domino covering $(3, 2)$ and $(3, 3)$.
Adding together these two cases, we deduce that:
$$b_n = a_n + b_{n-1}.$$
We can calculate the first few here:

Let $b_n$ be the number of ways of covering a $3 \times (2n+1)$ chessboard with one corner square removed with $3n+1$ dominoes. We are interested in the value of $b_{1010}$.
Suppose the corner square is removed from the first column of the $(2n+1) \times 3$ chessboard.
If $n \ge 1$, the dominoes that cover the squares of the last column can be placed in three different ways:

In type 1, the squares of the second last column are already covered. In types 2 and 3, there are two possibilities how the remaining two squares in the second last column can be covered, provided that $n \ge 2$:

When $n=1$ and the bottom square of the first column is removed, type 3b is not possible. Using the obvious $b_0=1$, we see now that $b_1=4$; the four possibilities are obtained from types 1, 2a, 2b and 3a.

Therefore, the contribution to $b_{n+1}$ from types 2b and 3b is equal to the number of possibilities to tile the $(2n+1) \times 3$ chessboard so that the last two columns are not of type 1. As we have seen above, the number of possibilities to tile a $(2n+1) \times 3$ chessboard with the last two columns of type 1 is equal to $b_{n-1}$, hence the contribution to $b_{n+1}$ from types 2b and 3b together is equal to $b_n - b_{n-1}$, and finally
$$b_{n+1} = 3b_n + b_n - b_{n-1} = 4b_n - b_{n-1}.$$
Starting with $b_0 = 1$ and $b_1 = 4$ we can now calculate $b_n$ (mod 19):