Stirling数-2

Stirling数-1 の続き、基本的公式より少々複雑なものを陳列します。

Stirling数のその他の式

$$\displaystyle\binom{n+1}{m+1}_1=\sum_k\displaystyle\binom{n}{k}_1\displaystyle\binom{k}{m}$$
$$\displaystyle\binom{n+1}{m+1}_2=\sum_k\displaystyle\binom{n}{k}\displaystyle\binom{k}{m}_2$$

$$\displaystyle\binom{n}{m}_1=\sum_k\displaystyle\binom{n+1}{k+1}_1\displaystyle\binom{k}{m}(-1)^{m-k}$$
$$\displaystyle\binom{n}{m}_2=\sum_k\displaystyle\binom{n}{k}\displaystyle\binom{k+1}{m+1}_2(-1)^{n-k}$$

$$m!\displaystyle\binom{n}{m}_2=\sum_k\displaystyle\binom{m}{k}k^n(-1)^{m-k}$$
$$\displaystyle\binom{n+1}{m+1}_1=\sum_{k=0}^n\displaystyle\binom{k}{m}_1n^{\underline{n-k}}=n!\sum_{k=0}^n\frac{\displaystyle\binom{k}{m}_1}{k!}$$
$$\displaystyle\binom{n+1}{m+1}_2=\sum_{k=0}^{n}\displaystyle\binom{k}{m}_2(m+1)^{n-k}$$

$$\displaystyle\binom{m+n+1}{m}_1=\sum_{k=0}^m(n+k)\displaystyle\binom{n+k}{k}_1$$
$$\displaystyle\binom{m+n+1}{m}_2=\sum_{k=0}^mk\displaystyle\binom{n+k}{k}_2$$

$$\displaystyle\binom{n}{m}=\sum_k\displaystyle\binom{n+1}{k+1}_2\displaystyle\binom{k}{m}_1(-1)^{m-k}$$

$$n^{\underline{n-m}}[n\geq m]=\sum_k\displaystyle\binom{n+1}{k+1}_1\displaystyle\binom{k}{m}_2(-1)^{m-k}$$

$$\displaystyle\binom{n}{n-m}_1=\sum_k\displaystyle\binom{m-n}{m+k}\displaystyle\binom{m+n}{n+k}\displaystyle\binom{m+k}{k}_2$$
$$\displaystyle\binom{n}{n-m}_2=\sum_k\displaystyle\binom{m-n}{m+k}\displaystyle\binom{m+n}{n+k}\displaystyle\binom{m+k}{k}_1$$

$$\displaystyle\binom{n}{l+m}_1\displaystyle\binom{l+m}{l}=\sum_k\displaystyle\binom{k}{l}_1\displaystyle\binom{n-k}{m}_1\displaystyle\binom{n}{k}$$
$$\displaystyle\binom{n}{l+m}_2\displaystyle\binom{l+m}{l}=\sum_k\displaystyle\binom{k}{l}_2\displaystyle\binom{n-k}{m}_2\displaystyle\binom{n}{k}$$

Euler数を用いたStirling数の畳み込み式

$\sigma_n(x)=\frac{\displaystyle\binom{x}{x-n}_1}{x(x-1)\cdots(x-n)}$とStirling多項式$\sigma_n(x)$を定義する。

$$rs\displaystyle\sum_{k=0}^{n}\sigma(r+tk)\sigma_{n-k}(s+t(n-k))=(r+s)\sigma_n(r+s+tn)$$
$$s\sum_k^nk\sigma_k(r+tk)\sigma_{n-k}(s+t(n-k))=n\sigma_n(r+s+tn)$$

$$\displaystyle\binom{n}{m}_1=\frac{n!}{(m-1)!}\sigma_{n-m}(n)$$
$$\displaystyle\binom{n}{m}_2=(-1)^{n-m+1}\frac{n!}{(m-1)!}\sigma_{n-m}(-m)$$

$$\displaystyle\left(\frac{ze^z}{e^z-1}\right)^x=x\sum_n\sigma_n(x)z^n$$
$$\left(\frac{1}{z}\ln \frac{1}{1-z}\right)^x=x\sum_n\sigma_n(x+n)z^n$$

一般に、$\mathcal{S}_t(z)$が
$$\ln(1-z\mathcal{S}_t(z))=-z\mathcal{S}_t(z)^t$$を満たすべき級数であるとき、

$$\mathcal{S}_t(z)^x=x\sum_n\sigma_n(x+tn)z^n$$