# 斯特灵数

## 第一類史特靈數

### 定義

${\displaystyle (x)_{n}=\sum _{k=0}^{n}s(n,k)x^{k}\,}$

${\displaystyle (x)_{3}=x(x-1)(x-2)\,}$

${\displaystyle (x)_{3}=0\cdot x^{0}+2\cdot x-3\cdot x^{2}+1\cdot x^{3}\,}$

${\displaystyle (x)^{\overline {n}}=\sum _{k=0}^{n}\left[{\begin{matrix}n\\k\end{matrix}}\right]x^{k}\,}$

${\displaystyle s(n,k)=(-1)^{n-k}\left[{\begin{matrix}n\\k\end{matrix}}\right]\,}$

1. (A,B)(C,D)
2. (A,C)(B,D)
3. (A,D)(B,C)
4. (A)(B,C,D)
5. (A)(B,D,C)
6. (B)(A,C,D)
7. (B)(A,D,C)
8. (C)(A,B,D)
9. (C)(A,D,B)
10. (D)(A,B,C)
11. (D)(A,C,B)

s(4,2)=11

### 遞推關係式

${\displaystyle \left[{\begin{matrix}n+1\\k\end{matrix}}\right]=n\left[{\begin{matrix}n\\k\end{matrix}}\right]+\left[{\begin{matrix}n\\k-1\end{matrix}}\right]\,}$

${\displaystyle s(n+1,k)=-ns(n,k)+s(n,k-1)}$

### 第一類史特靈數表

${\displaystyle s(n,k)=(-1)^{n-k}\left[{\begin{matrix}n\\k\end{matrix}}\right]\,}$

k
n
0 1 2 3 4 5 6
0 1
1 0 1
2 0 1 1
3 0 2 3 1
4 0 6 11 6 1
5 0 24 50 35 10 1
6 0 120 274 225 85 15 1

### 簡單性質

${\displaystyle \left[{\begin{matrix}0\\0\end{matrix}}\right]=1\,}$${\displaystyle \left[{\begin{matrix}n\\0\end{matrix}}\right]=0\,}$${\displaystyle n>0\,}$）。

${\displaystyle \left[{\begin{matrix}0\\k\end{matrix}}\right]=0\,}$

${\displaystyle \left[{\begin{matrix}n\\k\end{matrix}}\right]=0\,}$

${\displaystyle \left[{\begin{matrix}n\\1\end{matrix}}\right]=(n-1)!\,}$
${\displaystyle \left[{\begin{matrix}n\\n\end{matrix}}\right]=1\,}$
${\displaystyle \left[{\begin{matrix}n\\n-1\end{matrix}}\right]={n \choose 2}\,}$
${\displaystyle \left[{\begin{matrix}n\\n-2\end{matrix}}\right]={\frac {1}{4}}(3n-1){n \choose 3}\,}$
${\displaystyle \left[{\begin{matrix}n\\n-3\end{matrix}}\right]={n \choose 2}{n \choose 4}\,}$

## 第二類史特靈數

### 定義

${\displaystyle x^{n}=\sum _{k=0}^{n}\left\{{\begin{matrix}n\\k\end{matrix}}\right\}(x)_{k}\,}$

${\displaystyle x^{3}=\left\{{\begin{matrix}3\\0\end{matrix}}\right\}(x)_{0}+\left\{{\begin{matrix}3\\1\end{matrix}}\right\}(x)_{1}+\left\{{\begin{matrix}3\\2\end{matrix}}\right\}(x)_{2}+\left\{{\begin{matrix}3\\3\end{matrix}}\right\}(x)_{3}\,}$

${\displaystyle x^{3}=\left\{{\begin{matrix}3\\0\end{matrix}}\right\}+\left\{{\begin{matrix}3\\1\end{matrix}}\right\}x+\left\{{\begin{matrix}3\\2\end{matrix}}\right\}x(x-1)+\left\{{\begin{matrix}3\\3\end{matrix}}\right\}x(x-1)(x-2)\,}$

${\displaystyle x^{3}=\left\{{\begin{matrix}3\\0\end{matrix}}\right\}+\left(\left\{{\begin{matrix}3\\1\end{matrix}}\right\}-\left\{{\begin{matrix}3\\2\end{matrix}}\right\}+2\left\{{\begin{matrix}3\\3\end{matrix}}\right\}\right)x+\left(\left\{{\begin{matrix}3\\2\end{matrix}}\right\}-3\left\{{\begin{matrix}3\\3\end{matrix}}\right\}\right)x^{2}+\left\{{\begin{matrix}3\\3\end{matrix}}\right\}x^{3}\,}$

${\displaystyle {\begin{cases}\left\{{\begin{matrix}3\\0\end{matrix}}\right\}=0\\\left\{{\begin{matrix}3\\1\end{matrix}}\right\}-\left\{{\begin{matrix}3\\2\end{matrix}}\right\}+2\left\{{\begin{matrix}3\\3\end{matrix}}\right\}=0\\\left\{{\begin{matrix}3\\2\end{matrix}}\right\}-3\left\{{\begin{matrix}3\\3\end{matrix}}\right\}=0\\\left\{{\begin{matrix}3\\3\end{matrix}}\right\}=1\end{cases}}\,}$

${\displaystyle \left\{{\begin{matrix}3\\0\end{matrix}}\right\}=0\,}$${\displaystyle \left\{{\begin{matrix}3\\1\end{matrix}}\right\}=1\,}$${\displaystyle \left\{{\begin{matrix}3\\2\end{matrix}}\right\}=3\,}$${\displaystyle \left\{{\begin{matrix}3\\3\end{matrix}}\right\}=1\,}$

${\displaystyle \left\{{\begin{matrix}n\\k\end{matrix}}\right\}={\frac {1}{k!}}\sum _{i=0}^{k}(-1)^{i}\left({\begin{matrix}k\\i\end{matrix}}\right)(k-i)^{n}\,}$

${\displaystyle \left\{{\begin{matrix}3\\2\end{matrix}}\right\}\,}$進行驗算，有

${\displaystyle \left\{{\begin{matrix}3\\2\end{matrix}}\right\}={\frac {1}{2!}}\sum _{i=0}^{2}(-1)^{i}\left({\begin{matrix}2\\i\end{matrix}}\right)(2-i)^{3}\,}$

${\displaystyle \left\{{\begin{matrix}3\\2\end{matrix}}\right\}={\frac {1}{2}}\left(\left({\begin{matrix}2\\0\end{matrix}}\right)\times 2^{3}-\left({\begin{matrix}2\\1\end{matrix}}\right)\times 1^{3}+\left({\begin{matrix}2\\2\end{matrix}}\right)\times 0^{3}\right)\,}$

${\displaystyle \left\{{\begin{matrix}3\\2\end{matrix}}\right\}={\frac {1}{2}}\left(1\times 8-2\times 1+1\times 0\right)=3\,}$

### 拓展示例

${\displaystyle n\,}$個人分成${\displaystyle k\,}$組的分組方法的數目。例如有甲、乙、丙、丁四人，若所有人分成${\displaystyle 1\,}$組，只能所有人在同一組，因此${\displaystyle \left\{{\begin{matrix}4\\1\end{matrix}}\right\}=1\,}$；若所有人分成${\displaystyle 4\,}$組，只能每人獨立一組，因此${\displaystyle \left\{{\begin{matrix}4\\4\end{matrix}}\right\}=1\,}$；若分成${\displaystyle 2\,}$組，可以是甲乙一組、丙丁一組，或甲丙一組、乙丁一組，或甲丁一組、乙丙一組，或其中三人同一組另一人獨立一組，即：

1. {甲, 乙}{丙, 丁}
2. {甲, 丙}{乙,丁}
3. {甲, 丁}{乙, 丙}
4. {甲}{乙, 丙, 丁}
5. {乙}{甲, 丙, 丁}
6. {丙}{甲, 乙, 丁}
7. {丁}{甲, 乙, 丙}

### 遞推關係式

${\displaystyle \left\{{\begin{matrix}n+1\\k\end{matrix}}\right\}=k\left\{{\begin{matrix}n\\k\end{matrix}}\right\}+\left\{{\begin{matrix}n\\k-1\end{matrix}}\right\}\,}$

### 第二類史特靈數表

k
n
0 1 2 3 4 5 6
0 1
1 0 1
2 0 1 1
3 0 1 3 1
4 0 1 7 6 1
5 0 1 15 25 10 1
6 0 1 31 90 65 15 1

### 簡單性質

${\displaystyle \left\{{\begin{matrix}0\\0\end{matrix}}\right\}=1\,}$${\displaystyle \left\{{\begin{matrix}n\\0\end{matrix}}\right\}=0\,}$${\displaystyle n>0\,}$）。

${\displaystyle \left\{{\begin{matrix}0\\k\end{matrix}}\right\}=0\,}$

${\displaystyle \left\{{\begin{matrix}n\\k\end{matrix}}\right\}=0\,}$

${\displaystyle \left\{{\begin{matrix}n\\1\end{matrix}}\right\}=1\,}$
${\displaystyle \left\{{\begin{matrix}n\\2\end{matrix}}\right\}=2^{n-1}-1\,}$
${\displaystyle \left\{{\begin{matrix}n\\3\end{matrix}}\right\}={\frac {1}{2}}(3^{n-1}+1)-2^{n-1}\,}$
${\displaystyle \left\{{\begin{matrix}n\\n\end{matrix}}\right\}=1\,}$
${\displaystyle \left\{{\begin{matrix}n\\n-1\end{matrix}}\right\}={n \choose 2}\,}$
${\displaystyle \left\{{\begin{matrix}n\\n-2\end{matrix}}\right\}={n \choose 3}+3{n \choose 4}\,}$
${\displaystyle \left\{{\begin{matrix}n\\n-3\end{matrix}}\right\}={n \choose 4}+10{n \choose 5}+15{n \choose 6}\,}$

### 其他性質

${\displaystyle B_{n}=\sum _{k=0}^{n}\left\{{\begin{matrix}n\\k\end{matrix}}\right\}\,}$

## 兩類之間的關係

${\displaystyle \sum _{j\geq 0}s(n,j)S(j,k)=\sum _{j\geq 0}(-1)^{n-j}{\begin{bmatrix}n\\j\end{bmatrix}}{\begin{Bmatrix}j\\k\end{Bmatrix}}=\delta _{nk}}{\displaystyle \sum _{j\geq 0}s(n,j)S(j,k)=\sum _{j\geq 0}(-1)^{n-j}{\begin{bmatrix}n\\j\end{bmatrix}}{\begin{Bmatrix}j\\k\end{Bmatrix}}=\delta _{nk}\,}$

${\displaystyle \sum _{j\geq 0}S(n,j)s(j,k)=\sum _{j\geq 0}(-1)^{j-k}{\begin{Bmatrix}n\\j\end{Bmatrix}}{\begin{bmatrix}j\\k\end{bmatrix}}=\delta _{nk}}{\displaystyle \sum _{j\geq 0}S(n,j)s(j,k)=\sum _{j\geq 0}(-1)^{j-k}{\begin{Bmatrix}n\\j\end{Bmatrix}}{\begin{bmatrix}j\\k\end{bmatrix}}=\delta _{nk}\,}$

## 拉赫數

### 定義

${\displaystyle x^{(n)}=\sum _{k=0}^{n}L(n,k)(x)_{k}\,}$

${\displaystyle (x)_{n}=\sum _{k=0}^{n}(-1)^{n-k}L(n,k)x^{(k)}\,}$

${\displaystyle x^{(3)}=\sum _{k=0}^{3}L(3,k)(x)_{k}\,}$

${\displaystyle x(x+1)(x+2)=L(3,0)\cdot 1+L(3,1)\cdot x+L(3,2)\cdot x(x-1)+L(3,3)\cdot x(x-1)(x-2)\,}$

${\displaystyle 0\cdot x^{0}+2\cdot x+3\cdot x^{2}+1\cdot x^{3}=L(3,0)+[L(3,1)-L(3,2)+2L(3,3)]\cdot x+[L(3,2)-3L(3,3)]\cdot x^{2}+L(3,3)\cdot x^{3}\,}$

${\displaystyle {\begin{cases}{L(3,0)=0}\\{L(3,1)-L(3,2)+2L(3,3)=2}\\{L(3,2)-3L(3,3)=3}\\{L(3,3)=1}\end{cases}}\,}$

${\displaystyle (x)_{3}=\sum _{k=0}^{3}(-1)^{3-k}L(3,k)x^{(3)}\,}$

${\displaystyle x(x-1)(x-2)=L(3,0)\cdot 1+L(3,1)\cdot x+L(3,2)\cdot x(x+1)+L(3,3)\cdot x(x+1)(x+2)\,}$

${\displaystyle 0\cdot x^{0}+2\cdot x-3\cdot x^{2}+1\cdot x^{3}=-L(3,0)+[L(3,1)-L(3,2)+2L(3,3)]\cdot x+[-L(3,2)+3L(3,3)]\cdot x^{2}+L(3,3)\cdot x^{3}\,}$

${\displaystyle {\begin{cases}{L(3,0)=0}\\{L(3,1)-L(3,2)+2L(3,3)=2}\\{-L(3,2)+3L(3,3)=-3}\\{L(3,3)=1}\end{cases}}\,}$

${\displaystyle x^{(n)}=(-1)^{n}\sum _{k=0}^{n}L(n,k)(x)_{k}\,}$

${\displaystyle (x)_{n}=\sum _{k=0}^{n}(-1)^{k}L(n,k)x^{(k)}\,}$

${\displaystyle L(n,k)={n-1 \choose k-1}{\frac {n!}{k!}}\,}$

${\displaystyle L'(n,k)=(-1)^{n}{n-1 \choose k-1}{\frac {n!}{k!}}\,}$

### 遞推關係式

${\displaystyle L(n,k+1)={\frac {n-k}{k(k+1)}}L(n,k)}$

${\displaystyle L(n+1,k)=(n+k)L(n,k)+L(n,k-1)\,}$

### 拉赫數表

k
n
0 1 2 3 4 5 6
0 1
1 0 1
2 0 2 1
3 0 6 6 1
4 0 24 36 12 1
5 0 120 240 120 20 1
6 0 720 1800 1200 300 30 1

### 簡單性質

${\displaystyle L(0,0)=1}$${\displaystyle L(n,0)=0}$${\displaystyle n>0}$

${\displaystyle L(n,1)=n!}$
${\displaystyle L(n,2)={\frac {(n-1)n!}{2}}}$
${\displaystyle L(n,3)={\frac {(n-2)(n-1)n!}{12}}}$
${\displaystyle L(n,n-1)=n(n-1)}$
${\displaystyle L(n,n)=1}$
${\displaystyle \sum _{n\geq k}L(n,k){\frac {x^{n}}{n!}}={\frac {1}{k!}}\left({\frac {x}{1-x}}\right)^{k}}$

### 其他性質

${\displaystyle L(n,k)={n-1 \choose k-1}{\frac {n!}{k!}}={n \choose k}{\frac {(n-1)!}{(k-1)!}}={n \choose k}{n-1 \choose k-1}(n-k)!}$
${\displaystyle L(n,k)={\frac {n!(n-1)!}{k!(k-1)!}}\cdot {\frac {1}{(n-k)!}}=\left({\frac {n!}{k!}}\right)^{2}{\frac {k}{n(n-k)!}}}$

${\displaystyle L(n,k)=\sum _{j=0}^{n}\left[{\begin{matrix}n\\j\end{matrix}}\right]\left\{{\begin{matrix}j\\k\end{matrix}}\right\}\,}$

${\displaystyle L(3,2)\,}$進行驗算，有

${\displaystyle L(3,2)=\sum _{j=0}^{3}\left[{\begin{matrix}3\\j\end{matrix}}\right]\left\{{\begin{matrix}j\\2\end{matrix}}\right\}\,}$

${\displaystyle L(3,2)=\left[{\begin{matrix}3\\0\end{matrix}}\right]\left\{{\begin{matrix}0\\2\end{matrix}}\right\}+\left[{\begin{matrix}3\\1\end{matrix}}\right]\left\{{\begin{matrix}1\\2\end{matrix}}\right\}+\left[{\begin{matrix}3\\2\end{matrix}}\right]\left\{{\begin{matrix}2\\2\end{matrix}}\right\}+\left[{\begin{matrix}3\\3\end{matrix}}\right]\left\{{\begin{matrix}3\\2\end{matrix}}\right\}\,}$

${\displaystyle L(3,2)=\left[{\begin{matrix}3\\2\end{matrix}}\right]\left\{{\begin{matrix}2\\2\end{matrix}}\right\}+\left[{\begin{matrix}3\\3\end{matrix}}\right]\left\{{\begin{matrix}3\\2\end{matrix}}\right\}=3\times 1+1\times 3=6\,}$

${\displaystyle \sum _{j\geq 0}L(n,j)L(j,k)=\delta _{nk}\,}$

## 參考資料

1. ^ Sándor, Jozsef; Crstici, Borislav. Handbook of Number Theory II. Kluwer Academic Publishers. 2004: 464. ISBN 9781402025464.
2. ^ Transformation of Series by a Variant of Stirling's Numbers. Imanuel Marx, The American Mathematical Monthly 69, #6 (June–July 1962). : 530–532,. JSTOR 2311194..
3. ^ Antonio Salmeri (编). Introduzione alla teoria dei coefficienti fattoriali, Giornale di Matematiche di Battaglini 90 (1962). : pp. 44–54.
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7. ^ Lah, Ivo. A new kind of numbers and its application in the actuarial mathematics. Boletim do Instituto dos Actuários Portugueses 9. 1954: 7–15.
8. ^ John Riordan, Introduction to Combinatorial Analysis页面存档备份，存于互联网档案馆）, Princeton University Press (1958, reissue 1980) ISBN 978-0-691-02365-6 (reprinted again in 2002 by Dover Publications).
9. ^ Petkovsek, Marko; Pisanski, Tomaz. Combinatorial Interpretation of Unsigned Stirling and Lah Numbers. Pi Mu Epsilon Journal 12. Fall 2007: 417–424. JSTOR 24340704.
10. ^ Comtet, Louis. Advanced Combinatorics. Dordrecht, Holland: Reidel. 1974: 156.