Example 2 Divide a circle with a circumference of 24 into 24 equal segments, and select eight points from the 24 points, such that the arc length between any two points is not equal to 3 or 8. How many different ways are there to select such a group of eight points? Explain your reasoning.
(2001, China Mathematical Olympiad)
Solution: Number the points in sequence as $1, 2, \cdots, 24$, and list a $3 \times 8$ number table (see Table 1).
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline 1 & 4 & 7 & 10 & 13 & 16 & 19 & 22 \\
\hline 9 & 12 & 15 & 18 & 21 & 24 & 3 & 6 \\
\hline 17 & 20 & 23 & 2 & 5 & 8 & 11 & 14 \\
\hline
\end{tabular}

In this table, the arc length between adjacent points in the same row is 3 (the first and last numbers are considered adjacent), and the arc length between any two points in the same column is 8. To satisfy the requirements, we must select one number from each column, and the numbers selected from adjacent columns must be in different rows (the first and eighth columns are considered adjacent).

Generalizing, consider a $3 \times n$ number table (see Table 2).

We need to select $n$ numbers such that one number is chosen from each column, and the numbers selected from adjacent columns are in different rows (the first and $n$-th columns are considered adjacent).
Table 2
\begin{tabular}{|c|c|c|c|c|}
\hline$a_{1}$ & $a_{2}$ & $a_{3}$ & $\cdots$ & $a_{n}$ \\
\hline$b_{1}$ & $b_{2}$ & $b_{3}$ & $\cdots$ & $b_{n}$ \\
\hline$c_{1}$ & $c_{2}$ & $c_{3}$ & $\cdots$ & $c_{n}$ \\
\hline
\end{tabular}

The numbers in the first, second, and third rows $\left\{a_{k}\right\}, \left\{b_{k}\right\}, \left\{c_{k}\right\}$ are colored red, yellow, and blue, respectively.

Make a disk and divide it into $n$ sectors, numbered clockwise as $1, 2, \cdots, n$.

Define a mapping: if the number selected from the $i$-th column in the table is of a certain color, then the $i$-th sector of the disk is colored that color. It is easy to see that this correspondence is one-to-one. Thus, the problem is transformed into:

Find the number of ways to color the $n$ sectors of the disk with three colors such that adjacent sectors have different colors, denoted as $x_{n}$.
It is easy to see that $x_{1}=3, x_{2}=x_{3}=6$.
When $n \geqslant 3$, $x_{n}+x_{n-1}=3 \times 2^{n-1}$, i.e.,
$$
\begin{array}{l}
(-1)^{n} x_{n}-(-1)^{n-1} x_{n-1}=-3(-2)^{n-1} . \\
\text { Let } y_{n}=(-1)^{n} x_{n}, \text { then } \\
y_{n}-y_{n-1}=-3(-2)^{n-1}, \\
y_{n-1}-y_{n-2}=-3(-2)^{n-2}, \\
\cdots \cdots . \\
y_{3}-y_{2}=-3(-2)^{2}, \\
y_{2}=-3(-2)^{1} .
\end{array}
$$

Adding these equations, we get $y_{n}=(-2)^{n}+2(n \geqslant 2)$.
Therefore, $x_{n}=2^{n}+2(-1)^{n}(n \geqslant 2)$.
When $n=8$, $x_{8}=2^{8}+2=258$, which is the number of different ways to select the eight points.