$2 \cdot 116$ Let $S=\{1,2, \cdots, 10\}$. If a subset $T$ of $S$ contains at least 2 elements and the absolute difference between any two elements is greater than 1, then $T$ is said to have property $P$. Find the number of different subsets of $S$ that have property $P$.
[Solution] Let $a_{n}$ be the number of all distinct subsets of the set $\{1,2, \cdots, n\}$ that have property $P$, and consider the relationship between $a_{n+2}$ and $a_{n+1}, a_{n}$.

The subsets of the set $\{1,2, \cdots, n, n+1, n+2\}$ that have property $P$ can be divided into 3 categories:
(1) 2-element subsets $\{j, n+2\}, j=1,2, \cdots, n$, totaling $n$ subsets.
(2) Subsets of the form $T_{n} \cup\{n+2\}$, where $T_{n}$ is a subset of $\{1,2, \cdots, n\}$ that has property $P$, totaling $a_{n}$ subsets.
(3) All subsets of the set $\{1,2, \cdots, n, n+1\}$ that have property $P$, totaling $a_{n+1}$ subsets.
Combining these, we obtain the recurrence relation:
$$
a_{n+2}=a_{n+1}+a_{n}+n .
$$

Clearly, $a_{3}=1, a_{4}=3$. Using the above relation, we can recursively compute $a_{5}=7, a_{6}=14$, $a_{7}=26, a_{8}=46, a_{9}=79, a_{10}=133$, which means the number of all distinct subsets of $S$ that have property $P$ is 133.