Example 11 Let $\alpha=\frac{3-\sqrt{5}}{2}, f(n)=[\alpha n], f^{(k)}=f \cdot f \cdot \cdots \cdot f$ be the $k$-th composition of $f$, find the expression for $F(k)$.
Solution: Clearly, by the properties of the floor function, we have
$$
f(1)=[\alpha]=0, f(2)=[3-\sqrt{5}]=0, f(3)=\left[3 \cdot \frac{3-\sqrt{5}}{2}\right]=1 \text {. }
$$

Thus, we can obtain $f[f(n)]>0 \Leftrightarrow f(n)=[\alpha n] \geqslant 3$.
Also, $[\alpha n] \geqslant 3 \Leftrightarrow n \alpha>3 \Leftrightarrow n \geqslant \frac{3}{\alpha}=\frac{3(3+\sqrt{5})}{2} \Leftrightarrow n \geqslant 8$,
therefore, we can get $F(2)=8$.
Assume $F(1), F(2), \cdots, F(k-1), F(k)$ have all been found, then
$$
f^{(k+1)}(n)>0 \Leftrightarrow f(n) \geqslant F(k) \text {. }
$$

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 $\frac{3+\sqrt{5}}{2} \boldsymbol{F}(k)$, or in other words, $\boldsymbol{F}(k)$ 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 can 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, so $F(k+1)=3 F(k)-\left[\frac{3-\sqrt{5}}{2} F(k)\right]$.
Then from (2) we can get $\left[\frac{3-\sqrt{5}}{2} F(k)\right]=F(k-1)$, thus
$$
F(k+1)=3 F(k)-F(k-1) \text {. }
$$

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

where $\alpha, \beta$ are undetermined constants, by $F(0)=1, F(1)=3$, we can calculate
$$
\alpha=\frac{1}{\sqrt{5}} \cdot \frac{3+\sqrt{5}}{2}, \beta=-\frac{1}{\sqrt{5}} \cdot \frac{3-\sqrt{5}}{2} \text {. }
$$

Thus, $F(k)=\frac{1}{\sqrt{5}}\left(\frac{3+\sqrt{5}}{2}\right)^{k+1}-\frac{1}{\sqrt{5}}\left(\frac{3-\sqrt{5}}{2}\right)^{k+1}, k=0,1,2, \cdots$