$1 \cdot 111$ An increasing integer sequence, if its 1st term is odd, the 2nd term is even, the 3rd term is odd, the 4th term is even, and so on, is called an alternating sequence. The empty set is also considered an alternating sequence. The number of all alternating sequences with each term taken from the set $\{1,2, \cdots, n\}$ is denoted as $A(n)$. Clearly, $A(1)=2, A(2)=3$, find $A(20)$ and explain the reason.
［Solution］When $n=1$, the empty set $\emptyset$ and $\{1\}$ are two alternating sequences, so $A(1)=2$. When $n=2$, in addition to the above two, the alternating sequence $\{1,2\}$ is added, so $A(2)=3$.

Let $\left\{a_{1}, a_{2}, \cdots, a_{m}\right\}$ be a non-empty alternating sequence taken from the set $\{1,2, \cdots, n\}$. By definition, $a_{1}=1$ or $a_{1} \geqslant 3$ and $a_{1}$ is an odd number. When $a_{1}=1$, the sequence $\left\{a_{2}-1, a_{3}-1, \cdots, a_{m}-1\right\}$ is an alternating sequence taken from the set $\{1,2, \cdots, n-1\}$. At this time, define
$$
\left\{1, a_{2}, a_{3}, \cdots, a_{m}\right\} \xrightarrow{f}\left\{a_{2}-1, a_{3}-1, \cdots, a_{m}-1\right\}
$$

Then the mapping $f$ is a bijection from the set of all alternating sequences taken from the set $\{1,2, \cdots, n\}$ with the first term 1 to the set of all alternating sequences taken from the set $\{1,2, \cdots, n-1\}$. Therefore, the number of all alternating sequences taken from the set $\{1,2, \cdots, n\}$ with the first term 1 is $A(n-1)$.

When the odd number $a_{1} \geqslant 3$, the sequence $\left\{a_{1}-2, a_{2}-2, \cdots, a_{m}-2\right\}$ is a non-empty alternating sequence taken from $\{1,2, \cdots, n-2\}$. Define the mapping
$$
\left\{a_{1}, a_{2}, \cdots, a_{m}\right\} \xrightarrow{g}\left\{a_{1}-2, a_{2}-2, \cdots, a_{m}-2\right\},
$$

Then the mapping $g$ is a bijection from the set of all alternating sequences taken from $\{1,2, \cdots, n\}$ with the first term $a_{1} \geqslant 3$ to the set of all non-empty alternating sequences taken from $\{1,2, \cdots, n-2\}$. Therefore, the number of all alternating sequences taken from $\{1,2, \cdots, n\}$ with the first term $a_{1} \geqslant 3$ is $A(n-2) -1$.
In summary, we get the recurrence relation
$$
A(n)=A(n-1)+A(n-2) .
$$

Note that this is precisely the recurrence formula for the Fibonacci sequence.
Since $A(1)=2=F_{2}, A(2)=3=F_{3}$, it follows from (1) that
$$
A(n)=F_{n+1}, n=1,2, \cdots \cdots
$$

In particular, we have $A(20)=F_{21}$. Writing out the first 21 terms of the Fibonacci sequence in sequence
$$
1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584 \text {, }
$$
$4181,6765,10946,17711$.
Thus we have $\square$
$$
A(20)=F_{21}=17711
$$