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

${(\binom{n + 1}{m + 1})}_{1} = \sum_{k} {(\binom{n}{k})}_{1} (\binom{k}{m})$
${(\binom{n + 1}{m + 1})}_{2} = \sum_{k} (\binom{n}{k}) {(\binom{k}{m})}_{2}$

${(\binom{n}{m})}_{1} = \sum_{k} {(\binom{n + 1}{k + 1})}_{1} (\binom{k}{m}) (- 1)^{m - k}$
${(\binom{n}{m})}_{2} = \sum_{k} (\binom{n}{k}) {(\binom{k + 1}{m + 1})}_{2} (- 1)^{n - k}$

$m! {(\binom{n}{m})}_{2} = \sum_{k} (\binom{m}{k}) k^{n} (- 1)^{m - k}$
${(\binom{n + 1}{m + 1})}_{1} = \sum_{k = 0}^{n} {(\binom{k}{m})}_{1} n^{\underset{―}{n - k}} = n! \sum_{k = 0}^{n} \frac{{(\binom{k}{m})}_{1}}{k!}$
${(\binom{n + 1}{m + 1})}_{2} = \sum_{k = 0}^{n} {(\binom{k}{m})}_{2} (m + 1)^{n - k}$

${(\binom{m + n + 1}{m})}_{1} = \sum_{k = 0}^{m} (n + k) {(\binom{n + k}{k})}_{1}$
${(\binom{m + n + 1}{m})}_{2} = \sum_{k = 0}^{m} k {(\binom{n + k}{k})}_{2}$

$(\binom{n}{m}) = \sum_{k} {(\binom{n + 1}{k + 1})}_{2} {(\binom{k}{m})}_{1} (- 1)^{m - k}$

$n^{\underset{―}{n - m}} [n \geq m] = \sum_{k} {(\binom{n + 1}{k + 1})}_{1} {(\binom{k}{m})}_{2} (- 1)^{m - k}$

${(\binom{n}{n - m})}_{1} = \sum_{k} (\binom{m - n}{m + k}) (\binom{m + n}{n + k}) {(\binom{m + k}{k})}_{2}$
${(\binom{n}{n - m})}_{2} = \sum_{k} (\binom{m - n}{m + k}) (\binom{m + n}{n + k}) {(\binom{m + k}{k})}_{1}$

${(\binom{n}{l + m})}_{1} (\binom{l + m}{l}) = \sum_{k} {(\binom{k}{l})}_{1} {(\binom{n - k}{m})}_{1} (\binom{n}{k})$
${(\binom{n}{l + m})}_{2} (\binom{l + m}{l}) = \sum_{k} {(\binom{k}{l})}_{2} {(\binom{n - k}{m})}_{2} (\binom{n}{k})$

Euler数を用いたStirling数の畳み込み式

$σ_{n} (x) = \frac{{(\binom{x}{x - n})}_{1}}{x (x - 1) \dots (x - n)}$ とStirling多項式 $σ_{n} (x)$ を定義する。

$r s \sum_{k = 0}^{n} σ (r + t k) σ_{n - k} (s + t (n - k)) = (r + s) σ_{n} (r + s + t n)$
$s \sum_{k}^{n} k σ_{k} (r + t k) σ_{n - k} (s + t (n - k)) = n σ_{n} (r + s + t n)$