9. Let the sequence $a_{1}, a_{2}, a_{3}, \cdots, a_{21}$ satisfy: $\left|a_{n+1}-a_{n}\right|=1(n=1,2,3, \cdots, 20), a_{1}, a_{7}, a_{21}$ form a geometric progression. If $a_{1}=1, a_{21}=9$, then the number of different sequences that meet the conditions is $\qquad$
Let $a_{n+1}-a_{n}=b_{n}$, then we have $b_{n}= \pm 1$. If $a_{7}=3$, and $b_{1}, b_{2}, \cdots, b_{6}$ have $x_{1}$ occurrences of $+1$ and $y_{1}$ occurrences of -1, then $\left\{\begin{array}{l}x_{1}+y_{1}=6, \\ x_{1}-y_{1}=2\end{array} \Rightarrow\left\{\begin{array}{l}x_{1}=4, \\ y_{1}=2\end{array} \Rightarrow\left(b_{1}, b_{2}, \cdots, b_{6}\right)\right.\right.$ has $C_{6}^{2}=15$ solutions. Similarly, for $b_{7}, b_{8}, \cdots, b_{20}$, if there are $x_{2}$ occurrences of $+1$ and $y_{2}$ occurrences of -1, then $\left\{\begin{array}{l}x_{2}+y_{2}=14, \\ x_{2}-y_{2}=6\end{array} \Rightarrow\left\{\begin{array}{l}x_{2}=10, \\ y_{2}=4\end{array}\right.\right.$ $\Rightarrow\left(b_{7}, b_{8}, \cdots, b_{20}\right)$ has $C_{14}^{4}=1001$ solutions. Therefore, the sequence $a_{1}, a_{2}, a_{3}, \cdots, a_{21}$ has $15 \times 1001=15015$ sequences. If $a_{7}=-3$, similarly, the sequence $a_{1}, a_{2}, a_{3}, \cdots, a_{21}$ has $C_{6}^{1} \cdot C_{14}^{1}=84$ sequences. In summary, the number of different sequences that satisfy the conditions is $15015+84=15099$.