Five. (20 points) How many $n$-digit numbers can be formed from the digits $1,2, \cdots, 6$ such that $1$ and $6$ are adjacent?
Let the number of $n$-digit numbers we are looking for 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 the $a_{n}$ numbers that start with the digit 6.
The $a_{n}$ numbers can be divided into three categories:
(1) $n$-digit numbers among the $a_{n}$ numbers that start with the digit 1, such numbers are $b_{n}$ in number;
(2) $n$-digit numbers among the $a_{n}$ numbers that start with the digit 6, such numbers are $b_{n}$ in number;
(3) $n$-digit numbers among the $a_{n}$ numbers 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.
Therefore, $a_{n}=2 b_{n}+4 a_{n-1}$.
The $n$-digit numbers among the $a_{n}$ numbers 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;