Five. (20 points) Let $a_{n}$ be the number of subsets of the set $\{1,2, \cdots, n\}$ $(n \geqslant 3)$ that have the following property: each subset contains at least 2 elements, and the difference (absolute value) between any 2 elements in each subset is greater than 1. Find $a_{10}$.
Solution 1: Consider the recurrence relation of $a_{n}$, dividing the subsets that satisfy the condition into two categories:

The first category contains $n$. This type of subset, apart from the $a_{n-2}$ subsets that satisfy the condition in $\{1,2, \cdots, n-2\}$ combined with the element $n$, also includes $n-2$ two-element subsets $\{1, n\},\{2, n\}, \cdots,\{n-2, n\}$, totaling $a_{n-2} + n-2$.

The second category does not contain $n$. These subsets can be given by the subsets that satisfy the condition in $\{1,2, \cdots, n-1\}$, and there are $a_{n-1}$ such subsets.
Thus, we obtain the recurrence relation
$$
a_{n}=a_{n-1}+a_{n-2}+n-2 \quad (n \geqslant 3).
$$

It is easy to see that $a_{3}=1, a_{4}=3$. Therefore,
$$
\begin{array}{l}
a_{5}=7, a_{6}=14, a_{7}=26, \\
\dot{a}_{8}=46, a_{9}=79, a_{10}=133 .
\end{array}
$$

Solution 2: When $n=10$, the elements of the subset can be 2, 3, 4, or 5.

When the subset has only 2 elements, denoted as $a_{1} 、 a_{2}$, and $a_{1}<a_{2}$, then we have
$$
\begin{array}{l}
a_{1} \geqslant 1, \\
a_{2}-a_{1} \geqslant 2 \Rightarrow a_{2}-a_{1}-1 \geqslant 1, \\
10 \geqslant a_{2} \Rightarrow 11-a_{2} \geqslant 1 .
\end{array}
$$

Adding these, we get
$$
a_{1}+\left(a_{2}-a_{1}-1\right)+\left(11-a_{2}\right)=10 .
$$

This indicates that a pair $\left(a_{1}, a_{2}\right)$ corresponds to a positive integer solution of the equation
$$
x_{1}+x_{2}+x_{3}=10
$$

Conversely, a positive integer solution of the equation corresponds to a pair $\left(a_{1}, a_{2}\right)$.

Since the equation has $\mathrm{C}_{9}^{2}$ positive integer solutions, there are $\mathrm{C}_{9}^{2}$ two-element subsets that satisfy the condition.

Similarly, there are $\mathrm{C}_{8}^{3}$ three-element subsets, $\mathrm{C}_{7}^{4}$ four-element subsets, and $\mathrm{C}_{6}^{5}$ five-element subsets.
Therefore, $a_{10}=C_{9}^{2}+C_{8}^{3}+C_{7}^{4}+C_{6}^{5}=133$.