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本页目录
发布时间:2024-09-27
斯特林数
第一类string数
将1-n划分成k个圆排列的方案数,记作
s
[
n
,
k
]
或
[
n
k
]
s[n,k]或[\frac{n}{k}]
s
[
n
,
k
]
或
[
k
n
]
递归求法:
n放到新的圆,
[
n
−
1
k
−
1
]
[\frac{n-1}{k-1}]
[
k
−
1
n
−
1
]
n放到已有的圆,
(
n
−
1
)
[
n
−
1
k
]
(n-1)[\frac{n-1}{k}]
(
n
−
1
)
[
k
n
−
1
]
第一类string数性质
s
(
0
,
0
)
=
1
,
s
(
n
,
0
)
=
0
,
s
(
n
,
n
)
=
1
s(0,0)=1,s(n,0)=0,s(n,n)=1
s
(
0
,
0
)
=
1
,
s
(
n
,
0
)
=
0
,
s
(
n
,
n
)
=
1
s
(
n
,
1
)
=
(
n
−
1
)
!
s(n,1)=(n-1)!
s
(
n
,
1
)
=
(
n
−
1
)!
s
(
n
,
n
−
1
)
=
C
n
2
s(n,n-1)=C_n^2
s
(
n
,
n
−
1
)
=
C
n
2
,只有一个环有2个元素
s
(
n
,
2
)
=
(
n
−
1
)
!
⋅
∑
i
=
1
n
−
1
1
i
s(n,2)=(n-1)! \cdot \sum_{i=1}^{n-1}\frac{1}{i}
s
(
n
,
2
)
=
(
n
−
1
)!
⋅
∑
i
=
1
n
−
1
i
1
s
(
n
,
n
−
2
)
=
2
⋅
C
n
3
+
3
⋅
C
n
4
s(n,n-2)=2 \cdot C_n^3+3 \cdot C_n^4
s
(
n
,
n
−
2
)
=
2
⋅
C
n
3
+
3
⋅
C
n
4
∑
k
=
0
n
s
(
n
,
k
)
=
n
!
\sum_{k=0}^ns(n,k)=n!
∑
k
=
0
n
s
(
n
,
k
)
=
n
!