I will make frequent use of the words "domino" and "monomino" in this problem.
Let $a_k$ represent the number of ways to tile a $2\times k$ wall. We will seek to find a recurrence relation for $a_{k+1}$ by considering the rightmost between-column where there is a vertical line that doesn't cut through a domino.
Here will be the different summands of $a_{k+1}$:
- $2a_k$: There are two different ways to "cleanly" cover just the last column: one domino or two monominoes.
- $3a_{k-1}$: There are three different ways to cleanly cover exactly the last two columns: two dominoes one above the other, a domino on top and two monominoes on bottom, and a domino on bottom and two monominoes on top.
- $2a_{k-2}$: There are two ways to cleanly cover exactly exactly the last three columns:
- $2a_{k-3}+2a_{k-4}+\cdots+2a_1$+: Extending the same zipper pattern further, there are exactly two ways to cover any extra number of columns.
So we have $$a_{k+1}=2a_k+3a_{k-1}+2a_{k-2}+...+a_1,\ a_0=1$$
This is A030186 in OEIS. Also note that, since $$a_k=2a_{k-1}+3a_{k-2}+2a_{k-3}+...+a_1$$ we can rearrange and add the two equations together to get
$$a_{k+1} = 3a_k + a_{k-1} - a_{k-2}$$ Either way, $a=(1,2,7,22,71,228,733,2356,\cdots)$, so $a_7=2356$.
The variant can be settled by splitting up into three cases:
- $f_1(n)$, the number of ways to tile a rectangle ending with a flat line,
- $f_2(n)$, the number of ways to tile a rectangle where two corners meet in the middle,
- $g(n)$, the number of ways to tile a rectangle where the top right square is missing.
We have
$$
f_1(n) = f_1(n-1) + f_2(n-1) + 2 g(n-1)
$$
because the last tile of a rectangle ending in a flat line can either be an I tile or an L tile oriented in two ways. We just have
$$
f_2(n) = f_1(n-2)
$$
since the only way to create an $f_2$-type rectangle is to put down two horizontal dominoes at the end, and just as before, we have
$$
g(n) = f_1(n-2) + f_2(n-2) + g(n-1).
$$
From here, it's also possible to write down a linear recurrence for just the total $f(n) = f_1(n) + f_2(n)$ in terms of $f(n-1), f(n-2), \dots$. We have $f(n) = f_1(n) + f_2(n) = f_1(n) + f_1(n-2)$, so it's enough to solve for $f_1(n)$. In fact, as a linear combination of $f_1$ and its translate, $f$ will satisfy the same recurrence relation as $f_1$ with different initial conditions.
From $g(n) = f_1(n-2) + f_1(n-4) + g(n-1)$, we get the infinite series $g(n) = f_1(n-2) + f_1(n-3) + 2f_1(n-4) + 2f_1(n-5) + \dotsb$, and this gives us a recurrence
\begin{align}
f_1(n) &= f_1(n-1) + f_2(n-1) + 2g(n-1) \\
&= f_1(n-1) + f_1(n-3) + 2g(n-1) \\
&= f_1(n-1) + 3f_1(n-3) + 2f_1(n-4) + 4f_1(n-5) + 4f_1(n-6) + \dotsb \\
\end{align}
Subtracting $f_1(n-1)$ from $f_1(n)$, we get
$$
f_1(n) - f_1(n-1) = \color{red}{(f_1(n-1) + 3f_1(n-3) + 2f_1(n-4) + 4f_1(n-5) + 4f_1(n-6) + \dotsb )} - \color{blue}{(f_1(n-2) + 3f_1(n-4) + 2f_1(n-5) + 4f_1(n-6) + 4f_1(n-7) + \dotsb )}
$$
and most of the red terms cancel with blue terms, leaving us with
$$
f_1(n) - f_1(n-1) = f_1(n-1) - f_1(n-2) + 3f_1(n-3) - f_1(n-4) + 2f_1(n-5)
$$
or $f_1(n) = 2f_1(n-1) - f_1(n-2) + 3f_1(n-3) - f_1(n-4) + 2f_1(n-5)$. As observed earlier, this also means that the recurrence $$f(n) = 2f(n-1) - f(n-2) + 3f(n-3) - f(n-4) + 2f(n-5)$$ holds.
The first few nonzero terms of the sequence are $1, 1, 2, 5, 10, 22, 49, 105, 227, 494, 1071, \dots$, as computed by @PeterKagey in the comments and in the upcoming OEIS listing.
Best Answer
You've mentioned that the two tilings are identical if they give the same color pattern. Essentially you're asking the number of colorings of a $3\times3$ board with blue and red such that the blue part can be tiled with $2\times1$ or $1\times2$ tiles without overlap.
From the $2\times1$ and $1\times2$ condition therefore you must satisfy the following
I have a quick script that checks the above conditions for all possible colorings and you can eyeball that all the resulting matrices are indeed proper.
Python Code
There are $98$ such tilings (colorings)