Given a circle with its perimeter equal to $n$( $n \in N^*$), the least positive integer $P_n$ which satisfies the following condition is called the “number of the partitioned circle”: there are $P_n$ points ($A_1,A_2, \ldots ,A_{P_n}$) on the circle; For any integer $m$ ($1\le m\le n-1$), there always exist two points $A_i,A_j$ ($1\le i,j\le P_n$), such that the length of arc $A_iA_j$ is equal to $m$. Furthermore, all arcs between every two adjacent points $A_i,A_{i+1}$ ($1\le i\le P_n$, $A_{p_n+1}=A_1$) form a sequence $T_n=(a_1,a_2,,,a_{p_n})$ called the “sequence of the partitioned circle”. For example when $n=13$, the number of the partitioned circle $P_{13}$=4, the sequence of the partitioned circle $T_{13}=(1,3,2,7)$ or $(1,2,6,4)$. Determine the values of $P_{21}$ and $P_{31}$, and find a possible solution of $T_{21}$ and $T_{31}$ respectively.
Problem
Source: China south east mathematical olympiad 2006 day2 problem 8
Tags: geometry, perimeter, combinatorics unsolved, combinatorics