Three. (50 points) On a line $l$, there are $n+1$ points numbered $1, 2, \cdots, n+1$ arranged from left to right. In the plane above $l$, connect these points with $n$ continuous curves, satisfying:
(1) Any curve connects only two different points, and at most one curve connects any two points;
(2) If point $i$ is connected to point $j$, then the connecting line is denoted as $l_{ij}$ $(i<j)$;
(3) If $i<m \leqslant j<n$, $l_{ij}$ and $l_{mn}$ cannot exist simultaneously, i.e., they cannot "cross" or "connect end-to-end".

Let the number of all different connection methods be denoted as $A_{n}(n \geqslant 0, n \in \mathbf{Z})$. Find the expression for $A_{n}$.
Three, first prove a lemma using mathematical induction.
Lemma: Among $n+1$ points, at most $n$ lines can be connected.
Proof: When $n=0,1$, the conclusion is obviously true.
Assume that when $n \leqslant i(i \geqslant 1)$, the conclusion holds.
Suppose when $n=i+1$, at least $i+2$ lines can be connected.
First, explain that there must be a point that is connected by only one line.

If a point (numbered as $a$, hereafter, if not specified, points and numbers are not distinguished) is connected by more than one line, consider all points connected to this point. According to (3), these points are either all to the left of $a$ or all to the right of $a$. Without loss of generality, assume these points are all to the right of $a$. Let the point closest to $a$ among these points be $a_{1}$.

Now consider $a_{1}$, at most $\left|a_{1}-a\right|$ lines can be connected to its left. If $a_{1}$ is connected by more than one line, consider all points connected to this point. According to (3), these points are all to the right of $a$ and to the left of $a_{1}$. Let the point closest to $a_{1}$ among these points be $a_{2}$.

This process can be repeated. Since the total number of points is finite, eventually, a finite sequence $a_{1}, a_{2}, \cdots, a_{k}$ can be obtained. At this point, $a_{k}$ is only connected to $a_{k-1}$.

Remove $a_{k}$ and the line connected to $a_{k}$, leaving at least $i+1$ lines and $i+1$ points, which contradicts the inductive hypothesis. Therefore, the assumption is false.
Thus, when $n=i+1$, the conclusion holds.
Hence, among $n+1$ points, at most $n$ lines can be connected.
The lemma is proved.
Notice that, because the connections of points $1$ and $n+1$ do not affect any other already connected lines, according to (1), points $1$ and $n+1$ must be connected.

Assume that among the points connected to point 1 other than point $n+1$, the point with the largest number is $j$ (specially, if point 1 does not connect to any other points other than point $n+1$, let $j=1$).

At this point, among the $j$ points from point 1 to point $j$, at most $j-1$ lines can be connected. Among the $n-j+1$ points from point $j+1$ to point $n+1$, at most $n-j$ lines can be connected. And since point 1 is connected to point $j$, according to (3), points $2,3, \cdots, j$ and points $j+1, j+2, \cdots, n+1$ are not connected. Therefore, among the $j$ points from point 1 to point $j$ and the $n-j+1$ points from point $j+1$ to point $n+1$, all lines must be connected.
At this point, all possible cases are $A_{j-1} A_{n-j}$.
Then $A_{n}=A_{0} A_{n-1}+A_{1} A_{n-2}+\cdots+A_{n-1} A_{0}(n \geqslant 1)$
$=\sum_{i=0}^{n-1} A_{i} A_{n-1-i}(n \geqslant 1)$.
Notice that, $C_{n}=\frac{1}{n+1} \mathrm{C}_{2 n}^{n}$ satisfies
$$
C_{n}=\sum_{i=0}^{n-1} C_{i} C_{n-1-i}(n \geqslant 1),
$$

and $C_{0}=A_{0}=1, C_{1}=A_{1}=1$.
Therefore, $A_{n}=\frac{1}{n+1} \mathrm{C}_{2 n}^{n}$.