目录
模式避免的定义
避免Pattern q 的n-排列计数\(S_n(q)\)
先扔结论,有时间把证明粘过来
q长度是2
\[S_n(12)=S_n(21)=1
\]
q长度是3
All patterns of length three are avoided by the same number of n-permutations.
\[S_n(123)=S_n(132)=S_n(213)=S_n(231)=S_n(312)=S_n(321)\\ S_{n}(132)=C_{n}=\frac{\left(\begin{array}{c} 2 n \\ n \end{array}\right)}{n+1} \]
对一些模式q,做\(S_n(q)\)的阶估计
Backelin, West, and Xin给出的较为一般的Theorem
举例,
\(r=2,k=2,q=34\) it says, \(S_n(1234)=S_n(2134)\)
\(r=2,k=2,q=43\) it says, \(S_n(1243)=S_n(2143)\)
\(r=3,k=1,q=4\) it says, \(S_n(1234)=S_n(3214)\)
q长度是4
本来\(q\)有24种,可以证明最后归结为这三种代表,1234,1342,1324
\[\begin{aligned}
&\text { for } S_{n}(1342) \text { we have } 1,2,6,23,103,512,2740,15485\\
&\text { for } S_{n}(1234) \text { we have } 1,2,6,23,103,513,2761,15767\\
&\text { for } S_{n}(1324) \text { we have } 1,2,6,23,103,513,2762,15793
\end{aligned}
\]
\[\begin{aligned}
S_{n}(1342) &=(-1)^{n-1} \cdot \frac{\left(7 n^{2}-3 n-2\right)}{2} \\
&+3 \sum_{i=2}^{n}(-1)^{n-i} \cdot 2^{i+1} \cdot \frac{(2 i-4) !}{i !(i-2) !} \cdot\left(\begin{array}{c}
n-i+2 \\
2
\end{array}\right)
\end{aligned}
\]
\[S_{n}(1234)=2 \cdot \sum_{k=0}^{n}\left(\begin{array}{c}
2 k \\
k
\end{array}\right)\left(\begin{array}{l}
n \\
k
\end{array}\right)^{2} \frac{3 k^{2}+2 k+1-n-2 n k}{(k+1)^{2}(k+2)(n-k+1)}
\]
\[S_{n}(1234)=\frac{1}{(n+1)^{2}(n+2)} \sum_{k=0}^{n}\left(\begin{array}{c}
2 k \\
k
\end{array}\right)\left(\begin{array}{c}
n+1 \\
k+1
\end{array}\right)\left(\begin{array}{c}
n+2 \\
k+1
\end{array}\right)
\]
证明Stanley-Wilf conjecture
The Stanley-Wilf conjecture
书里给出的思路是先丢个 The Füredi-Hajnal conjecture出来,说这个 The Füredi-Hajnal conjecture可以推导Stanley-Wilf conjecture.
这样我们先来研究The Füredi-Hajnal conjecture
The Füredi-Hajnal conjecture
\[f(n, P) \leq c_{p} n
\]
先空着
资料来自网络
书用的是Combinatorics of permutations by Miklos Bona