I have seen this question, but I'm thinking about why the following solution can't be true?
As we now for a $2\times n$ board, the solution is:
$$ f_2(n) =
\begin{cases}
f_2(1) = 1 \\
f_2(2) = 2 \\
f_2(n)=f_2(n-1)+f_2(n-2)
\end{cases}$$
And for $3\times n$ board we have:
$$\begin{cases}
f_3(n) = f_3(n-2)+2g_3(n-1) \\
g_3(n) = f_3(n-1)+g_3(n-2) \\
f_3(1)=0, f_3(2)=3, g_3(1)=1, g_3(2)=0
\end{cases}$$
now for a $4\times n$ board
1- We can think of $4\times 1= 1\times 4$ board which has just 1 way to tile and a $4\times (n-1)$ board which has $f_4(n-1)$ ways to tile
2- We can think of $4\times 2= 2\times 4$ board which has
$$f_2(4)= f_2(3)+f_2(2)=f_2(2)+f_2(1)+f_2(2)=5$$
ways to tile and a $4\times (n-2)$ board which has $f_4(n-1)$ ways to tile
3- We can think of $4\times 3= 3\times 4$ board which has
$$\begin{cases}
g_3(3) = f_3(2)+g_3(2)=3+1=4 \\
f_3(4) = f_3(2)+2g_3(3)=3+2(4)=11 \\
\end{cases}$$
ways to tile and a $4\times (n-3)$ board which has $f_4(n-3)$ ways to tile
So for the $4\times n$ board and $f_4(n)$, we'll have:
$$ f_4(n) =
\begin{cases}
f_4(1) = 1 \\
f_4(2) = 5 \\
f_4(3) = 11 \\
f_4(n)=f_4(n-1)+5f_4(n-2)+11f_4(n-3)
\end{cases}$$