Five, (20 points) How many $n$-digit numbers can be formed by the digits $1,2, \cdots, 6$ such that 1 and 6 are adjacent?
Let the number of $n$-digit numbers be $a_{n}$.
Let the number of $n$-digit numbers among $a_{n}$ that start with the digit 1 be $b_{n}$.
By swapping the positions of 1 and 6 in each of the $b_{n}$ numbers, we get $b_{n}$ numbers among $a_{n}$ that start with the digit 6.
The $a_{n}$ numbers can be divided into three categories:
(1) $n$-digit numbers among $a_{n}$ that start with the digit 1, such numbers are $b_{n}$ in number;
(2) $n$-digit numbers among $a_{n}$ that start with the digit 6, such numbers are $b_{n}$ in number;
(3) $n$-digit numbers among $a_{n}$ that start with one of the digits 2, 3, 4, 5, the last $n-1$ digits of these numbers are still composed of the digits 1, 2, ..., 6 and contain adjacent 1 and 6, there are $4 a_{n-1}$ such numbers.
Thus, $a_{n}=2 b_{n}+4 a_{n-1}$.
The $n$-digit numbers among $a_{n}$ that start with the digit 1 can also be divided into three categories:
(1) $n$-digit numbers where the second digit is 6, the last $n-2$ digits of these numbers can be any of the digits 1, 2, ..., 6, there are $6^{n-2}$ such numbers;
(2) $n$-digit numbers where the second digit is still 1, the last $n-1$ digits of these numbers are composed of the digits 1, 2, ..., 6 and start with 1 and contain adjacent 1 and 6, there are $b_{n-1}$ such numbers;
(3) $n$-digit numbers where the second digit is one of 2, 3, 4, 5 and contain adjacent 1 and 6, the last $n-2$ digits of these numbers are composed of the digits 1, 2, ..., 6 and contain adjacent 1 and 6, there are $4 a_{n-2}$ such numbers.
Thus, $b_{n}=6^{n-2}+b_{n-1}+4 a_{n-2}$.
From equations (1) and (2), we get
$a_{n}=5 a_{n-1}+4 a_{n-2}+2 \times 6^{n-2}$.
Also, $a_{1}=0, a_{2}=2$, thus,
$\frac{a_{n}}{6^{n}}=\frac{5}{6} \cdot \frac{a_{n-1}}{6^{n-1}}+\frac{1}{9} \cdot \frac{a_{n-2}}{6^{n-2}}+\frac{1}{18}$.
Let $c_{n}=\frac{a_{n}}{6^{n}}$. Then
$$
c_{n}=\frac{5}{6} c_{n-1}+\frac{1}{9} c_{n-2}+\frac{1}{18},
$$

where $c_{1}=0, c_{2}=\frac{1}{18}$.
Thus, $c_{n}-1=\frac{5}{6}\left(c_{n-1}-1\right)+\frac{1}{9}\left(c_{n-2}-1\right)$.
Let $d_{n}=c_{n}-1$. Then,
$d_{n}=\frac{5}{6} d_{n-1}+\frac{1}{9} d_{n-2}, d_{1}=-1, d_{2}=-\frac{17}{18}$.
Then $d_{n}=\frac{-19 \sqrt{41}-123}{246}\left(\frac{5+\sqrt{41}}{12}\right)^{n-1}+$
$$
\frac{19 \sqrt{41}-123}{246}\left(\frac{5-\sqrt{41}}{12}\right)^{n-1} \text {. }
$$

Thus, $a_{n}=6^{n}\left[\frac{-19 \sqrt{41}-123}{246}\left(\frac{5+\sqrt{41}}{12}\right)^{n-1}+\right.$
$\left.\frac{19 \sqrt{41}-123}{246}\left(\frac{5 \sqrt{41}}{12}\right)^{n-1}+1\right]$.
(Provided by Shi Dongyang)