Example 7 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 and 8. Find the number of different ways to select such a group of eight points.
【Analysis】First analyze the essential structure of the data: Label 24 points in a clockwise direction as $1,2, \cdots, 24$. Then arrange them into the following $3 \times 8$ number table:
$$
\left(\begin{array}{cccccccc}
1 & 4 & 7 & 10 & 13 & 16 & 19 & 22 \\
9 & 12 & 15 & 18 & 21 & 24 & 3 & 6 \\
17 & 20 & 23 & 2 & 5 & 8 & 11 & 14
\end{array}\right) .
$$

In the number table, the arc length between two points represented by adjacent numbers in the same column is 8, and the arc length between two points represented by adjacent numbers in the same row is 3 (the first column and the eighth column are also adjacent, as are the first row and the third row). Therefore, any two adjacent numbers in the table cannot be taken simultaneously, meaning exactly one number is taken from each column, and the numbers taken from adjacent columns cannot be in the same row.
Generalize and construct a recursive relationship.
From a $3 \times n(n \geqslant 3)$ number table, take one number from each column such that the numbers taken from adjacent columns are not in the same row (the first column and the $n$-th column are considered adjacent), and let the number of ways to do this be $a_{n}$.
Then $a_{3}=6$.
When $n \geqslant 4$, we have
$$
a_{n}+a_{n-1}=3 \times 2^{n-1} \Rightarrow a_{n}=2^{n}+2(-1)^{n} \text {. }
$$

Recursive calculation yields $a_{8}=2^{8}+2(-1)^{8}=258$.