8. 17 For every positive integer $x$, let $g(x)$ denote the largest odd divisor of $x$. Let
$$
f(x)=\left\{\begin{array}{ll}
\frac{x}{2}+\frac{2}{g(x)}, & \text { if } x \text { is even, } \\
2^{\frac{x+1}{2}}, & \text { if } x \text { is odd. }
\end{array}\right.
$$

Construct the sequence: $x_{1}=1, x_{n+1}=f\left(x_{n}\right), n=1,2, \cdots$ Prove that the number 1992 appears in this sequence, find the smallest $n$ such that $x_{n}=1992$, and determine whether such an $n$ is unique.
[Proof] For some positive integer $n$ and non-negative integer $k$, if $x_{n}=2^{k}$, then
$$
x_{n+1}=3 \cdot 2^{k-1}, x_{n+2}=5 \cdot 2^{k-2}, \cdots, x_{n+k}=2^{k+1} \text {, }
$$

thus $x_{n+k+1}=2^{k+1}$. From this, it is easy to prove by induction that
$$
x \frac{1}{2} k(k+1)+s=(2 s-1) 2^{k-s+1} \text {, }
$$

where $\quad k=0,1,2,3, \cdots, s=1,2, \cdots, k+1$. Therefore, every positive integer appears exactly once in the sequence, so there exists a unique $n$ such that
$$
x_{n}=1992 \text {. }
$$

Given $1992=249 \times 2^{3}$, we can find
$$
s=125, k=127 \text {. }
$$

Thus, $n=\frac{1}{2}(127 \times 128)+125=8253$.