8. Given a sequence $\left\{a_{n}\right\}$ with 9 terms, where $a_{1}=a_{9}=1$, and for each $i \in\{1,2, \cdots, 8\}$, we have $\frac{a_{i+1}}{a_{i}} \in\left\{2,1,-\frac{1}{2}\right\}$, then the number of such sequences is $\qquad$ .
Answer 491.
Solution Let $b_{i}=\frac{a_{i+1}}{a_{i}}(1 \leq i \leq 8)$, then for each sequence $\left\{a_{n}\right\}$ that meets the conditions, we have
$$
\prod_{i=1}^{8} b_{i}=\prod_{i=1}^{8} \frac{a_{i+1}}{a_{i}}=\frac{a_{9}}{a_{1}}=1 \text {, and } b_{i} \in\left\{2,1,-\frac{1}{2}\right\}(1 \leq i \leq 8) \text {. }
$$

Conversely, a sequence $\left\{b_{n}\right\}$ of 8 terms that meets condition (1) can uniquely determine a sequence $\left\{a_{n}\right\}$ of 9 terms that meets the problem's conditions.

Let the number of sequences $\left\{b_{n}\right\}$ that meet condition (1) be $N$. Clearly, $b_{i}(1 \leq i \leq 8)$ must contain an even number of $-\frac{1}{2}$, i.e., $2 k$ instances of $-\frac{1}{2}$; and consequently, $2 k$ instances of $2$, and $8-4 k$ instances of 1. When $k$ is given, the number of ways to choose $\left\{b_{n}\right\}$ is $\mathrm{C}_{8}^{2 k} \mathrm{C}_{8-2 k}^{2 k}$. It is easy to see that the possible values of $k$ are only $0,1,2$, so
$$
N=1+\mathrm{C}_{8}^{2} \mathrm{C}_{6}^{2}+\mathrm{C}_{8}^{4} \mathrm{C}_{4}^{4}=1+28 \times 15+70 \times 1=491 .
$$

Therefore, according to the principle of correspondence, the number of sequences $\left\{a_{n}\right\}$ that meet the conditions is 491.