斯特林数,上升幂,下降幂学习笔记

发布时间 2023-04-16 17:31:15作者: _bzw

斯特林,上升幂,下降幂,普通幂的定义

第二类斯特林数

n \(n\brace 0\) \(n\brace 1\) \(n\brace 2\) \(n\brace 3\) \(n\brace 4\) \(n\brace 5\) \(n\brace 6\) \(n\brace 7\) \(n\brace 8\) \(n\brace9\)
0 \(1\) \(0\) \(0\) \(0\) \(0\) \(0\) \(0\) \(0\) \(0\) \(0\)
1 \(0\) \(1\) \(0\) \(0\) \(0\) \(0\) \(0\) \(0\) \(0\) \(0\)
2 \(0\) \(1\) \(1\) \(0\) \(0\) \(0\) \(0\) \(0\) \(0\) \(0\)
3 \(0\) \(1\) \(3\) \(1\) \(0\) \(0\) \(0\) \(0\) \(0\) \(0\)
4 \(0\) \(1\) \(7\) \(6\) \(1\) \(0\) \(0\) \(0\) \(0\) \(0\)
5 \(0\) \(1\) \(15\) \(25\) \(10\) \(1\) \(0\) \(0\) \(0\) \(0\)
6 \(0\) \(1\) \(31\) \(90\) \(65\) \(15\) \(1\) \(0\) \(0\) \(0\)
7 \(0\) \(1\) \(63\) \(301\) \(350\) \(140\) \(21\) \(1\) \(0\) \(0\)
8 \(0\) \(1\) \(127\) \(966\) \(1701\) \(1050\) \(266\) \(28\) \(1\) \(0\)
9 \(0\) \(1\) \(255\) \(3025\) \(7770\) \(6951\) \(2446\) \(462\) \(36\) \(1\)

定义 \({n\brace m}\) 表示将 \(n\) 件物品分成 \(m\) 个子集(非空)的方案数。

\[{n\brace m} = {n-1\brace m-1} + m{n-1\brace m} \]

第一类斯特林数

n \(n\brack 0\) \(n\brack 1\) \(n\brack 2\) \(n\brack 3\) \(n\brack 4\) \(n\brack 5\) \(n\brack 6\) \(n\brack 7\) \(n\brack 8\) \(n\brace9\)
0 \(1\) \(0\) \(0\) \(0\) \(0\) \(0\) \(0\) \(0\) \(0\) \(0\)
1 \(0\) \(1\) \(0\) \(0\) \(0\) \(0\) \(0\) \(0\) \(0\) \(0\)
2 \(0\) \(1\) \(1\) \(0\) \(0\) \(0\) \(0\) \(0\) \(0\) \(0\)
3 \(0\) \(2\) \(3\) \(1\) \(0\) \(0\) \(0\) \(0\) \(0\) \(0\)
4 \(0\) \(6\) \(11\) \(6\) \(1\) \(0\) \(0\) \(0\) \(0\) \(0\)
5 \(0\) \(24\) \(50\) \(35\) \(10\) \(1\) \(0\) \(0\) \(0\) \(0\)
6 \(0\) \(120\) \(274\) \(225\) \(85\) \(15\) \(1\) \(0\) \(0\) \(0\)
7 \(0\) \(720\) \(1764\) \(1624\) \(735\) \(175\) \(21\) \(1\) \(0\) \(0\)
8 \(0\) \(5040\) \(13068\) \(13132\) \(6769\) \(1960\) \(322\) \(28\) \(1\) \(0\)
9 \(0\) \(40320\) \(109584\) \(11824\) \(67284\) \(22449\) \(4536\) \(546\) \(36\) \(1\)

定义 \(n\brack m\) 表示将 \(n\) 件物品分成 \(m\) 个轮换(非空,圆排列)的方案数。

\[{n\brack m} = (n-1){n-1\brack m}+{n-1\brack m-1} \]

上升幂,下降幂,普通幂

\[\begin{aligned} &x^{\underline{n}} = \frac{x!}{(x-n)!}\\ &x^{\overline{n}} = \frac{(x+n-1)!}{(x-1)!}\\ &x^n = x^n\\ &x^{\overline{n}} = (x+n-1)^{\underline{n}}\\ &x^{\underline{n}} = (x-n+1)^{\overline{n}}\\ \end{aligned} \]

上升幂,下降幂,普通幂的关系

用下降幂表示普通幂

\[\begin{aligned} \sum_{i=0}^{n}{n\brace i}x^{\underline{i}} &= \sum_{i=0}^{n}(i{n-1\brace i}+{n-1\brace i-1})x^{\underline{i}}\\ &= \sum_{i=0}^{n}i{n-1\brace i}x^{\underline{i}}+\sum_{i=0}^{n}{n-1\brace i-1}x^{\underline{i}}\\ &= \sum_{i=0}^{n}i{n-1\brace i}x^{\underline{i}}+\sum_{i=1}^{n}{n-1\brace i-1}x^{\underline{i}}\\ &= \sum_{i=0}^{n}i{n-1\brace i}x^{\underline{i}}+\sum_{i=1}^{n}{n-1\brace i-1}x^{\underline{(i-1)+1}}\\ &= \sum_{i=0}^{n}i{n-1\brace i}x^{\underline{i}}+\sum_{i=0}^{n-1}{n-1\brace i}x^{\underline{i+1}}\\ &= \sum_{i=0}^{n-1}i{n-1\brace i}x^{\underline{i}}+\sum_{i=0}^{n-1}{n-1\brace i}x^{\underline{i}}(x-i)\\ &= \sum_{i=0}^{n-1}i{n-1\brace i}x^{\underline{i}}+{n-1\brace i}x^{\underline{i}}(x-i)\\ &= x\sum_{i=0}^{n-1}{n-1\brace i}x^{\underline{i}}\\ &= x \times x^{n-1} = x^{n}\\ \end{aligned} \]

用普通幂表示上升幂

\[\begin{aligned} \sum_{i=0}^{n}{n\brack i}x^{i} &= \sum_{i=0}^{n}((n-1){n-1\brack i}+{n-1\brack i-1})x^i\\ &= \sum_{i=0}^{n}(n-1){n-1\brack i}x^i+\sum_{i=0}^{n}{n-1\brack i-1}x^i\\ &= \sum_{i=0}^{n}(n-1){n-1\brack i}x^i+\sum_{i=1}^{n}{n-1\brack i-1}x^{(i-1)+1}\\ &= \sum_{i=0}^{n}(n-1){n-1\brack i}x^i+\sum_{i=0}^{n-1}{n-1\brack i}x^{i+1}\\ &= (n-1+x)\sum_{i=0}^{n}{n-1\brack i}x^i\\ &= (x+n-1)x^{\overline{n-1}}\\ &= x^{\overline{n}}\\ \end{aligned} \]

用普通幂表示下降幂

\[\begin{aligned} x^{\underline{n}} &= (-1)^n(-x)^{\overline{n}}\\ &= (-1)^n\sum_{i=0}^{n}{n\brack i}(-x)^i\\ &= \sum_{i=0}^{n}{n\brack i}(-1)^{n-i}x^i\\ \end{aligned} \]

用上升幂表示普通幂

\[\begin{aligned} x^n &= (-1)^n(-x)^n\\ &= (-1)^n\sum_{i=0}^{n}{n\brace i}(-x)^{\underline{i}}\\ &= (-1)^n\sum_{i=0}^{n}{n\brace i}(-1)^ix^{\underline{i}}\\ &= \sum_{i=0}^{n}{n\brace i}(-1)^{n-i}x^{\underline{i}}\\ \end{aligned} \]

整理一下

\[\begin{aligned} &x^n = \sum_{i=0}^{n}{n\brace i}x^{\underline{i}}\\ &x^n = \sum_{i=0}^{n}{n\brace i}(-1)^{n-i}x^{\overline{i}}\\ &x^{\overline{n}} = \sum_{i=0}^{n}{b\brack i}x^i\\ &x^{\underline{n}} = \sum_{i=0}^{n}{n\brack i}(-1)^{n-i}x^i\\ \end{aligned} \]

表示普通幂时用第二类斯特林数,表示上升幂,下降幂用第一类斯特林数,用大数表示小数时用 \((-1)^{n-i}\)