Periodic sequence problem

As you've noticed, since $3\mid a_1$ and $3\mid 1983$, it follows that $3\mid a_n$ for all $n$. This allows us to simplify the problem by considering the associated sequence defined by $b_n = a_n/3$. We can easily prove by induction that we have $1 \le b_n \le 660$ for all $n$. Since the admissible range of values for $b_n$ is finite, the sequence must be eventually periodic.

Your conjecture that the period is $660$ is in fact true. Showing that the period is $660$ will show that the sequence is not just eventually periodic, but fully periodic (alternatively, as you've noted, this follows from the fact that $b_n$ uniquely determines $b_{n-1}$). The conjecture that the period is $660$, together with the fact that $1 \le b_n \le 660$, motivates looking at the values of the sequence modulo $661$.

Let $[k]$ denote the remainder of $k\in \mathbb{Z}$ modulo $661$, i.e., the unique integer $0 \le [k] < 661$ such that $[k] \equiv k \pmod{661}$. Since $1 \le b_n < 661$, it follows that $b_n = [b_n]$ for all $n\in \mathbb{N}$.

Lemma 1: Let $m \in \mathbb{Z}$ be an even integer. Then $[m/2] = [331m]$.

Proof: Note that $2$ is a unit in $\mathbb{Z}/661\mathbb{Z}$. Indeed, we have $2^{-1} \equiv 331 \pmod{661}$. Therefore we have $$331m \equiv 331 \cdot \left[2\cdot \left(\frac{m}{2}\right)\right] \equiv [331 \cdot 2]\left(\frac{m}{2}\right)\equiv \frac{m}{2} \pmod{661}.$$ It follows that $[m/2] = [331m]$. $\square$

Lemma 2: For all $n\ge 1$, we have $b_n = [331^{(n-1)}]$.

Proof: Consider the defining recursion $$b_{n+1} = \begin{cases}b_n/2 & 2 \mid b_n,\\ (b_n + 661)/2 & 2\not\mid b_n.\end{cases}$$ In the first case, we have $$b_{n+1} = [b_{n+1}] = [b_n/2] = [331b_n].$$ In the second case, we have $$b_{n+1} = [b_{n+1}] = [(b_n + 661)/2] = [331(b_n + 661)] = [331b_n].$$ In either case, we have $b_{n+1} = [331b_n]$. Starting with $b_1 = 1$, it follows that $b_n = [331^{(n-1)}]$. $\square$

The period of the sequence is therefore the order of $331$ mod $661$. The result then follows by noting $661$ is prime, so that $(\mathbb{Z}/661\mathbb{Z})^{\times} \cong \mathbb{Z}_{660}$ is cyclic, and moreover that $331$ (or equivalently, $2$) is a primitive root modulo $661$. This last fact can be verified with a quick (albeit tedious) calculation.