8.4 The sequence of numbers $\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \ldots, \mathrm{a}_{\mathrm{n}}, \ldots$ satisfies the relations $\mathrm{a}_{\mathrm{n}}=\mathrm{a}_{\mathrm{n}-1} \cdot \mathrm{a}_{\mathrm{n}-3}$ for $\mathrm{n}=4,5,6, \ldots$ Find $\mathrm{a}_{2019}$ if it is known that $\mathrm{a}_{1}=1, \mathrm{a}_{2}=1$, $\mathrm{a}_{3}=-1$
Solution. It is clear that all members of this sequence are equal to $\pm 1$. We find:

$$
\begin{aligned}
& a_{n}=\left(a_{n-1}\right) \cdot a_{n-3}=\left(a_{n-2} \cdot a_{n-4}\right) \cdot a_{n-3}=\left(a_{n-2}\right) \cdot a_{n-4} \cdot a_{n-3}= \\
& =\left(a_{n-3} \cdot a_{n-4}\right) \cdot a_{n-4} \cdot a_{n-3}=a_{n-3}^{2} \cdot a_{n-4} \cdot a_{n-5}=a_{n-4} \cdot a_{n-5}= \\
& =\left(a_{n-4}\right) \cdot a_{n-5}=\left(a_{n-5} \cdot a_{n-7}\right) \cdot a_{n-5}=a_{n-5}^{2} \cdot a_{n-7}=a_{n-7}
\end{aligned}
$$

That is, the sequence is periodic with a period of 7. Therefore,

$$
\mathrm{a}_{2019}=\mathrm{a}_{288 \cdot 7+3}=\mathrm{a}_{3}=-1
$$