China Mathematical Competition

Let $b_i=\frac{a_{i+1}}{a_i}$ ($1\le i\le 8$). Then for each $\{a_n\}$ satisfying the given condition, we have $$\prod _{i=1}^8{b}_i=\prod _{i=1}^8\frac{a_{i+1}}{a_i}=\frac{a_9}{a_1}=1,\text{ with }{b}_i\in \{2,1,-\frac{1}{2}\}(1\le i\le 8).\stackrel{◯}{1}$$ Conversely, a sequence of eight terms $\{b_n\}$ satisfying 1 can uniquely determine a sequence $\{a_n\}$ in the question.

In each $\{b_n\}$, there are obviously even number of $-\frac{1}{2}$ and the same number of $2$, with the remainder being $1$. Or, in other words, the numbers of $-\frac{1}{2}$ and $2$ are both $2k$, while the number of $1$ is $8-4k$. Here, it is easy to check that $k$ can only be $0,1,2$. Once $k$ is given, there are $C_8^{2k}{C}_{8-2k}^{2k}$ ways to construct $\{b_n\}$.

Therefore, the total number of $\{b_n\}$ satisfying ① is $$N=1+C_8^2{C}_6^2+C_8^4{C}_4^4=1+28\times 15+70\times 1=491.$$ The answer is 491.