6. Given the sequence $\left\{a_{n}\right\}$, where $a_{1}=a_{2}=1, a_{3}=-1, a_{n}=a_{n-1} a_{n-3}$, find $a_{1964}$.
3 items are enough to show that the sequence is a periodic sequence. Since $1964=7 \times 280+4$, we have $a_{1964}=-1$.