$$8 \cdot 30 \text { Let } \alpha=\frac{3-\sqrt{5}}{2}, f(n)=[\alpha n] \text {, and let }$$
$F(k)=\min \left\{n \in N ; f^{(k)}(n)>0\right\}$, for any $k \in N$, where $f^{(k)}=f \circ f \circ \cdots \circ f$ is the $k$-fold composition of $f$, find the expression for $F(k)$.
[Solution] Clearly, $f(1)=[\alpha]=0, f(2)=[3-\sqrt{5}]=0, f(3)=1$, so $F(1)=3$. Therefore, we have
$$f(f(n))>0 \Leftrightarrow f(n)=[\alpha n] \geqslant 3 .$$

Also, $[\alpha n] \geqslant 3 \Leftrightarrow n \alpha \geqslant 3 \Leftrightarrow n \geqslant \frac{3}{\alpha}=\frac{3(3+\sqrt{5})}{2} \Leftrightarrow n \geqslant 8$, hence $F(2)=8$. Suppose $F(1), F(2), \cdots, F(k-1), F(k)$ have all been found, then
$$f^{(k+1)}(n)>0 \Leftrightarrow f(n)=[\alpha n] \geqslant F(k)$$

Also, $f(n)=[\alpha n] \geqslant F(k) \Leftrightarrow n \geqslant \frac{3+\sqrt{5}}{2} F(k)$, thus $F(k+1)$ is the smallest integer not less than $\frac{3+\sqrt{5}}{2} F(k)$, or in other words, $F(k)$ is the largest integer not exceeding $\frac{2}{3+\sqrt{5}} F(k+1)=\frac{3-\sqrt{5}}{2} F(k+1)$. Noting that $\frac{3+\sqrt{5}}{2} F(k)$ is an irrational number, the above conclusion can be expressed as
$$\begin{array}{l}
F(k+1)=\left[\frac{3+\sqrt{5}}{2} F(k)\right]+1 \\
F(k)=\left[\frac{3-\sqrt{5}}{2} F(k+1)\right]
\end{array}$$

where $k=1,2,3, \cdots$ When $k \geqslant 2$, from (1) we get
$$F(k+1)=\left[3 F(k)-\frac{3-\sqrt{5}}{2} F(k)\right]+1$$

Since $\frac{3-\sqrt{5}}{2} F(k)$ is also an irrational number, we have
$$F(k+1)=3 F(k)-\left[\frac{3-\sqrt{5}}{2} F(k)\right] .$$

From (2), we get $\left[\frac{3-\sqrt{5}}{2} F(k)\right]=F(k-1)$, thus
$$F(k+1)=3 F(k)-F(k-1)$$

Supplementing $F(0)=1$, then (3) holds for $k \geqslant 1$. The characteristic equation of the recurrence relation (3) is $\lambda^{2}-3 \lambda+1=0$, with characteristic roots $\frac{3 \pm \sqrt{5}}{2}$, so
$$F(k)=A\left(\frac{3+\sqrt{5}}{2}\right)^{k}+B\left(\frac{3-\sqrt{5}}{2}\right)^{k}$$

Using $F(0)=1, F(1)=3$, we can calculate
$$A=\frac{1}{\sqrt{5}} \cdot \frac{3+\sqrt{5}}{2}, B=-\frac{1}{\sqrt{5}} \cdot \frac{3-\sqrt{5}}{2}$$

Thus,
$$F(k)=\frac{1}{\sqrt{5}}\left(\frac{3+\sqrt{5}}{2}\right)^{k+1}-\frac{1}{\sqrt{5}}\left(\frac{3-\sqrt{5}}{2}\right)^{k+1}$$

where $k=0,1,2, \cdots$