Three. (50 points) A subset of the set $S_{n}=\{1,2, \cdots, n\}$ is called a "good subset" if it does not contain two consecutive natural numbers. How many good subsets are there in $S_{n}$?
Let $S_{n}$ have $a_{n}$ good subsets.
When $n=1$, $S_{n}$ has subsets $\varnothing, \{1\}$, which are good subsets, $a_{1}=2$.

When $n=2$, $S_{n}$ has 4 subsets: $\varnothing, \{1\}$, $\{2\}, \{1,2\}$, except for $\{1,2\}$, all are good subsets, $a_{2}=3$.
For $n \geqslant 3$, let $M$ be a good subset of $S_{n}$.
If $n \notin M$, then $M$ is also a good subset of $S_{n-1}$, conversely, a good subset $M$ of $S_{n-1}$ is also a good subset of $S_{n}$.

If $n \in M$, then by the definition of a good subset, $n-1 \notin M$, thus, the subset $M^{\prime}$ obtained by removing $n$ from $M$ ($M=M^{\prime} \cup \{n\}$), must be a good subset of $S_{n-2}$, conversely, a good subset $M^{\prime}$ of $S_{n-2}$, when combined with $n$ to form $M=M^{\prime} \cup \{n\}$, is also a good subset of $S_{n}$.
Thus, 
\[
\left\{\begin{array}{l}
a_{n}=a_{n-1}+a_{n-2}, \\
a_{1}=2, a_{2}=3 .
\end{array}\right.
\]
From $a_{n+1}=a_{n}+a_{n-1}$, we have
\[
a_{n+1}=\left(\frac{1+\sqrt{5}}{2}+\frac{1-\sqrt{5}}{2}\right) a_{n}-\left(\frac{1+\sqrt{5}}{2} \times \frac{1-\sqrt{5}}{2}\right) a_{n-1}.
\]
Transforming and recursively, we get
\[
\begin{array}{l}
a_{n+1}-\frac{1-\sqrt{5}}{2} a_{n}=\frac{1+\sqrt{5}}{2}\left(a_{n}-\frac{1-\sqrt{5}}{2} a_{n-1}\right) \\
=\cdots
\end{array}
\]
\[
=\left(\frac{1+\sqrt{5}}{2}\right)^{n-1}\left(a_{2}-\frac{1-\sqrt{5}}{2} a_{1}\right)
\]
\[
=\left(\frac{1+\sqrt{5}}{2}\right)^{n+2}.
\]
Similarly, $a_{n+1}-\frac{1+\sqrt{5}}{2} a_{n}=\left(\frac{1-\sqrt{5}}{2}\right)^{n+2}$.
Eliminating $a_{n+1}$, we get
\[
a_{n}=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n+2}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+2}\right].
\]
(Luo Zengru, Department of Mathematics, Shaanxi Normal University, 710062)