置换的k-子集的轨道数
组合数学
2025-08-12 v1 概率论
摘要
令 S n S_n S n n n n g 1 , g 2 ∈ S n g_1, g_2\in S_n g 1 , g 2 ∈ S n g 1 x g 2 = y g_1xg_2=y g 1 x g 2 = y x , y ⊆ S n x, y\subseteq S_n x , y ⊆ S n S n S_n S n k k k T ( n , k ) T(n,k) T ( n , k ) T ( n , 0 ) = T ( n , 1 ) = 1 T(n,0)=T(n, 1)=1 T ( n , 0 ) = T ( n , 1 ) = 1 T ( n , 2 ) = p ( n ) − 1 T(n, 2)=p(n)-1 T ( n , 2 ) = p ( n ) − 1 p ( n ) p(n) p ( n ) n n n 3 ≤ k ≤ n ! − 3 3\leq k\leq n!-3 3 ≤ k ≤ n ! − 3 T ( n , k ) = Λ n ( k ) ( 1 + o n ( 1 ) ) T(n,k) = \Lambda_n(k)(1+o_n(1)) T ( n , k ) = Λ n ( k ) ( 1 + o n ( 1 )) Λ n ( k ) = 1 n ! 2 ( n ! k ) \Lambda_n(k)=\frac{1}{n!^2}\binom{n!}{k} Λ n ( k ) = n ! 2 1 ( k n ! ) 1 Λ n ( n ! / 2 ) T ( n , [ n ! 4 x + n ! 2 ] ) → n → ∞ exp ( − x 2 2 ) , \frac{1}{\Lambda_n(n!/2)}T\!\left(n,\left[\sqrt{\tfrac{n!}{4}}x+\tfrac{n!}{2}\right]\right) ~\xrightarrow{n\to\infty}~ \exp\!\left(-\tfrac{x^2}{2}\right), Λ n ( n ! /2 ) 1 T ( n , [ 4 n ! x + 2 n ! ] ) n → ∞ exp ( − 2 x 2 ) , R \mathbb{R} R
引用
@article{arxiv.2508.07463,
title = {Number of orbits of $k$-subsets of permutations},
author = {Ludovick Bouthat and Raghavendra Tripathi},
journal= {arXiv preprint arXiv:2508.07463},
year = {2025}
}