论文标题
一些代数身份及其与第二类的斯特林数字的关系
Some algebraic identity and its relations to Stirling numbers of the second kind
论文作者
论文摘要
在简短的说明中，我们提供了一些代数身份，并提供了利用其概率解释的证据。我们展示了身份的几个后果，特别是我们获得了第二类的新表示形式， $ s（n，d）= {1 \ fover d！} \ sum_ {1 \ leq j_1 <j_2 <j_2 <\ ldots <j_ {d-1} <n} <n} <n} 1 \ cdot2^{j__ {d-1} {d-1} - 对于整数$ n \ geq d $。将此与其他已知公式（n，d）$相关 $$ \ sum_ {1 \ leq j_1 \ leq j_2 \ leq \ cdots \ leq j_ {n-d} \ leq d} j_1j_2 \ ldots，j_ {n-d} = d！ \ sum_ {1 \ leq j_1 <j_2 <\ ldots <j_ {d-1} <n} 1 \ cdot2^{j_ {d-1} -j__ {d-2}} \ cdots d^{j_1}。 作为副作用，我们有新的证据证明已知结果表明，对于任何整数$ d \ $$ \ sum_ {r = 0}^d（-1）^r {d \ select r}（x-r）^d = d！$$保持。这是提出身份的特殊情况。
In this short note we provide some algebraic identity with a proof exploiting its probabilistic interpretation. We show several consequences of the identity, in particular we obtain a new representation of a Stirling number of second kind, $$ S(n,d)={1\over d!} \sum_{1\leq j_1<j_2<\ldots<j_{d-1}< n} 1\cdot2^{j_{d-1}-j_{d-2}}\cdots d^{j_1}$$ for integers $n\geq d$. Relating this to other known formula for $S(n,d)$ we also obtain $$ \sum_{1\leq j_1\leq j_2\leq \cdots\leq j_{n-d}\leq d} j_1j_2\ldots,j_{n-d} =d! \sum_{1\leq j_1<j_2<\ldots<j_{d-1}< n} 1\cdot2^{j_{d-1}-j_{d-2}}\cdots d^{j_1}.$$ As a side effect, we have new proof of a known result stating that for any integer $d\in\mathbb{N}$ and any $x\in\mathbb{R}$ equality $$\sum_{r=0}^d (-1)^r{d\choose r}(x-r)^d=d!$$ holds. This is a special case of the presented identity.
