Number of ways to tile a room with $I$-Shaped and $L$-Shaped Tiles There is a room of dimensions $2\times n$. You have to tile it using $2$ types of tiles:


*

*I-Shaped Tile ($2\times1$)

*L-Shaped Tile ($2\times1 + 1$)



However, you are forbidden to use any tiling where any four corners of the tiles meet.
For example for a $2\times4$ room, the first three will be counted and the last one will not be counted.

My Attempt
If the condition that four corners cannot meet wasn't given, a pretty neat recurrence can be formed.
$$f(n) = f(n-1) + f(n-2) + 2g(n-1)$$
$$g(n) = f(n-2) + g(n-1)$$
with $g(0) = g(1) = 0$ and $f(0) = f(1) = 1$ where $f(n) = $ number of ways of tiling a $2\times n$ rectangle and $g(n)=$ number of ways of tiling  a $2\times n$ rectangle with a square missing on top.
Hence, we multiply $g(n-1)$ by $2$ when calculating $f(n)$ because the missing square can be at the top or the bottom.
I am unable to find such a recurrence with the extra given condition.
 A: 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.
