8. Let the integer sequence $a_{1}, a_{2}, \cdots, a_{10}$ satisfy $a_{10}=3 a_{1}, a_{2}+a_{8}=2 a_{5}$, and
$$
a_{i+1} \in\left\{1+a_{i}, 2+a_{i}\right\}, i=1,2, \cdots, 9 \text {, }
$$

then the number of such sequences is $\qquad$ .
Answer: 80.
Solution: Let $b_{i}=a_{i+1}-a_{i} \in\{1,2\}(i=1,2, \cdots, 9)$, then we have
$$
\begin{array}{c}
2 a_{1}=a_{10}-a_{1}=b_{1}+b_{2}+\cdots+b_{9}, \\
b_{2}+b_{3}+b_{4}=a_{5}-a_{2}=a_{8}-a_{5}=b_{5}+b_{6}+b_{7} .
\end{array}
$$

Let $t$ be the number of terms with value 2 among $b_{2}, b_{3}, b_{4}$. By (2), $t$ is also the number of terms with value 2 among $b_{5}, b_{6}, b_{7}$, where $t \in\{0,1,2,3\}$. Therefore, the number of ways to choose $b_{2}, b_{3}, \cdots, b_{7}$ is $\left(\mathrm{C}_{3}^{0}\right)^{2}+\left(\mathrm{C}_{3}^{1}\right)^{2}+\left(\mathrm{C}_{3}^{2}\right)^{2}+\left(\mathrm{C}_{3}^{3}\right)^{2}=20$. After fixing $b_{2}, b_{3}, \cdots, b_{7}$, we can arbitrarily assign values to $b_{8}, b_{9}$, which has $2^{2}=4$ ways.
Finally, by (1), we should choose $b_{1} \in\{1,2\}$ such that $b_{1}+b_{2}+\cdots+b_{9}$ is even, and such a $b_{1}$ is unique, and it determines the value of the integer $a_{1}$, and thus the sequence $b_{1}, b_{2}, \cdots, b_{9}$ uniquely corresponds to a sequence $a_{1}, a_{2}, \cdots, a_{10}$ that satisfies the conditions.
In summary, the number of sequences that satisfy the conditions is $20 \times 4=80$.