14. (POL 2) For any positive integer $x$ define  $$ \begin{aligned} g(x) & =\text { greatest odd divisor of } x, \\ f(x) & =\left\{\begin{array}{l} x / 2+x / g(x), \text { if } x \text { is even; } \\ 2^{(x+1) / 2}, \text { if } x \text { is odd. } \end{array}\right. \end{aligned} $$  Construct the sequence $x_{1}=1, x_{n+1}=f\left(x_{n}\right)$. Show that the number 1992 appears in this sequence, determine the least $n$ such that $x_{n}=1992$, and determine whether $n$ is unique.
14. We see that $x_{1}=2^{0}$. Suppose that for some $m, r \in \mathbb{N}$ we have $x_{m}=2^{r}$. Then inductively $x_{m+i}=2^{r-i}(2 i+1)$ for $i=1,2, \ldots, r$ and $x_{m+r+1}=$ $2^{r+1}$. Since every natural number can be uniquely represented as the product of an odd number and a power of two, we conclude that every natural number occurs in our sequence exactly once. Moreover, it follows that $2 k-1=x_{k(k+1) / 2}$. Thus $x_{n}=1992=2^{3} \cdot 249$ implies that $x_{n+3}=255=2 \cdot 128-1=x_{128 \cdot 129 / 2}=x_{8256}$. Hence $n=8253$.