9. Now, use identical $1 \times 1$ tiles and identical $2 \times 1$ tiles to cover a $2 \times 7$ wall. Before laying, the $2 \times 1$ tiles can be rotated. It is known that there are enough of both types of tiles. How many different ways are there to cover the wall completely?
9.2356 .

For a positive integer $n$, define $T_{n}$ as the number of ways to tile a horizontal $2 \times n$ rectangle with specified tiles, and define $S_{n}$ as the number of ways to tile a $2 \times n$ rectangle with the top-left corner removed.
First, consider the covering of the top-left corner of the $2 \times n$ rectangle.
If it is covered by a $1 \times 1$ tile, the remaining tiles correspond to the covering of $S_{n}$;

If it is covered by a $2 \times 1$ tile, the remaining tiles correspond to the covering of $T_{n-1}$;

If it is covered by a $1 \times 2$ tile, further consider the covering of the bottom-left corner: a $1 \times 1$ tile covering corresponds to $S_{n-1}$, and a $1 \times 2$ tile covering corresponds to $T_{n-2}$.
Thus, $T_{n}=T_{n-1}+S_{n}+S_{n-1}+T_{n-2}(n \geqslant 2)$.
Next, consider the covering of the $2 \times n$ rectangle with the top-left corner removed.
According to the covering method of the bottom-left corner, similarly we get
$$
\begin{array}{l}
S_{n}=T_{n-1}+S_{n-1}(n \geqslant 2) . \\
\text { Substituting } T_{n-1}=S_{n}-S_{n-1} \text { into equation (1) yields } \\
S_{n+1}-S_{n}=S_{n}-S_{n-1}+S_{n}+S_{n-1}+S_{n-1}-S_{n-2} \\
\Rightarrow S_{n+1}=3 S_{n}+S_{n-1}-S_{n-2} .
\end{array}
$$

Since $T_{n}$ can be obtained from $S_{n}$ through index shifting and linear combination, $T_{n}$ satisfies the same recurrence relation
$$
\begin{array}{l}
T_{n+1}=3 T_{n}+T_{n-1}-T_{n-2} . \\
\text { Given } T_{1}=2, T_{2}=7, T_{3}=22, \text { we calculate } \\
T_{4}=71, T_{5}=228, T_{6}=733, T_{7}=2356 .
\end{array}
$$