For a circle with a circumference 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 $T_{n}=\left(a_{1}, a_{2}, \cdots, a_{p_{n}}\right)$, which consists of the arc lengths between every pair of adjacent points on the circumference, is called the "circle partition sequence". For example, when $n=13$, the circle partition number is $p_{13}=4$, and the numbers marked in Figure 3 represent the arc lengths between adjacent points, with the circle partition sequence being
$$
T_{13}=(1,3,2,7),(1,2,6,4) \text {. }
$$

Find $p_{21}$ and $p_{31}$, and provide a corresponding circle partition sequence for each.
(Tao Pingsheng, Contributed)
Let $p_{n}=k$.
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 $k(k-1) \geqslant 20$, then $k \geqslant 5$;
When $k(k-1) \geqslant 30$, then $k \geqslant 6$.
On the other hand, when $k=5$, a partition graph can be provided (as shown in Figure 7).
Therefore, $p_{21}=5$,
$$
T_{21}=(1,3,10,2,5) \text {. }
$$

For $n=31$, when $k=6$, a similar partition graph can be provided (as shown in Figure 8).
$$
\begin{array}{c}
\text { Therefore, } p_{31}=6, \\
T_{31}=(1,2,7,4,12,5),(1,2,5,4,6,13),(1,3,2, \\
7,8,10),(1,3,6,2,5,14),(1,7,3,2,4,14) \text {, etc. }
\end{array}
$$
(Provided by Tao Pingsheng)