For a circle with a perimeter of $n\left(n \in \mathbf{N}^{*}\right)$, the smallest positive integer $P_{n}$ that satisfies the following conditions is called the "circle partition number": If there are $P_{n}$ points $A_{1}$, $A_{2}, \cdots, A_{p_{n}}$ on the circumference, for each integer $m$ in $1,2, \cdots, n-1$, there exist two points $A_{i}, A_{j}\left(1 \leqslant i, j \leqslant P_{n}\right)$ such that the arc length between $A_{i}$ and $A_{j}$ equals $m$; the sequence $T_{n}=$ $\left(a_{1}, a_{2}, \cdots, a_{p_{n}}\right)$ formed by the arc lengths between every two adjacent points on the circumference is called the "circle partition sequence". For example: when $n=13$, the circle partition number is $P_{13}=4$, as shown in the figure, the numbers marked in the figure are the arc lengths between adjacent points, and the circle partition sequence is $T_{13}=$ $(1,3,2,7)$ or $(1,2$, $6,4)$. Find $P_{21}$ and $P_{31}$, and provide a corresponding circle partition sequence for each.

Find $P_{21}$ and $P_{31}$, and provide a corresponding circle partition sequence for each.
Solution: Since among $k$ points, every two points can form a major arc and a minor arc, there can be at most $k(k-1)$ arc lengths.
When $k(k-1) \geqslant 20$, $k \geqslant 5$;
and when $k(k-1) \geqslant 30$, $k \geqslant 6$.
On the other hand, when $k=5$, a partition graph can be provided.
Therefore, $P_{21}=5, T_{21}=(1,3,10,2,5)$.
For $n=31$, when $k=6$, a similar partition graph can be provided.

Therefore,
$$
\begin{array}{c}
P_{31}=6, \\
T_{31}=(1,2,7,4,12,5),(1,2,5,4,6,13),(1,3,2,7, \\
8,10),(1,3,6,2,5,14) \text { or }(1,7,3,2,4,14) \text {, etc. }
\end{array}
$$