9. In the sequence $\left\{a_{n}\right\}$, $a_{4}=1, a_{11}=9$, and the sum of any three consecutive terms is 15, then $a_{2016}=$ $\qquad$
Solution: 5.
According to the problem, for any $n \in \mathbf{N}, a_{n}+a_{n+1}+a_{n+2}=a_{n+1}+a_{n+2}+a_{n+3}=15$.
Therefore, $a_{n+1}=a_{n}$.
Thus, $a_{1}=a_{4}=1, a_{2}=a_{11}=9, a_{3}=15-a_{1}-a_{2}=5$.
Hence, $a_{3016}=a_{3 \times 672}=a_{3}=5$.