Kaneko-Yamamoto型多重ゼータ値の和公式

まず,
\begin{align} &\sum_{r+s+2\leq k}x^{k-r-s-2}\sum_{\substack{1\leq k_1,\dots,k_r,l_1,\dots,l_s,2\leq a\\k_1+\cdots+k_r+a+l_1+\cdots+l_s=k}}\zeta\left(\begin{matrix}k_1,\dots,k_r\\l_1,\dots,l_s\end{matrix};a\right)\\ &=\sum_{1\leq k_1,\dots,k_r,l_1,\dots,l_s,2\leq a}x^{k_1-1}\cdots x^{k_r-1}\cdots x^{l_1-1}\cdots x^{l_s-1}\cdot x^{a-2}\sum_{0< n_1<\cdots< n_r< j\geq m_s\geq\cdots\geq m_1>0}\frac 1{n_1^{k_1}\cdots n_r^{k_r}j^am_1^{l_1}\cdots m_s^{l_s}}\\ &=\sum_{0< n_1<\cdots< n_r< j\geq m_s\geq\cdots\geq m_1>0}\frac 1{(n_1-x)\cdots (n_r-x)j(j-x)(m_1-x)\cdots (m_s-x)} \end{align}
が成り立つ. これより,
\begin{align} &\sum_{0\leq r,s}y^rz^s\sum_{r+s+2\leq k}x^{k-r-s-2}\sum_{\substack{1\leq k_1,\dots,k_r,l_1,\dots,l_s,2\leq a\\k_1+\cdots+k_r+a+l_1+\cdots+l_s=k}}\zeta\left(\begin{matrix}k_1,\dots,k_r\\l_1,\dots,l_s\end{matrix};a\right)\\ &=\sum_{0\leq r,s}y^rz^s\sum_{0< n_1<\cdots< n_r< j\geq m_s\geq\cdots\geq m_1>0}\frac 1{(n_1-x)\cdots (n_r-x)j(j-x)(m_1-x)\cdots (m_s-x)}\\ &=\sum_{0\leq j}\frac 1{j(j-x)}\sum_{0\leq r}y^r\sum_{0< n_1<\cdots< n_r< j}\frac 1{(n_1-x)\cdots (n_r-x)}\sum_{0\leq s}z^s\sum_{0< m_1\leq\cdots\leq m_s\leq j}\frac 1{(m_1-x)\cdots (m_s-x)}\\ &=\sum_{0\leq j}\frac 1{j(j-x)}\prod_{0< n< j}\left(1+\frac{y}{n-x}\right)\cdot\prod_{0< m\leq j}\left(1-\frac{z}{m-x}\right)^{-1}\\ &=\sum_{0\leq j}\frac{\prod_{0< n< j}\left(n-x+y\right)}{j\prod_{0< m\leq j}\left(m-x-z\right)} \end{align}
を得る. ここで, $z=0$とすると 多重ゼータ値の和公式 より
\begin{align} &\sum_{0\leq j}\frac{\prod_{0< n< j}\left(n-x+y\right)}{j\prod_{0< m\leq j}\left(m-x\right)}\\ &=\sum_{0\leq r}y^r\sum_{r+2\leq k}x^{k-r-2}\sum_{\substack{1\leq k_1,\dots,k_r,2\leq a\\k_1+\cdots+k_r+a=k}}\zeta\left(k_1,\dots,k_r,a\right)\\ &=\sum_{0\leq r}y^r\sum_{r+2\leq k}x^{k-r-2}\zeta(k)\\ &=\sum_{0\leq r}y^r\sum_{r+2\leq k}x^{k-r-2}\sum_{0< n}\frac 1{n^k}\\ &=\sum_{0< n}\frac 1{(n-x)(n-y)} \end{align}
を得る. ここで, $x,y$を$x+z,y+z$に置き換えて先ほどの等式と合わせると
\begin{align} &\sum_{0\leq r,s}y^rz^s\sum_{r+s+2\leq k}x^{k-r-s-2}\sum_{\substack{1\leq k_1,\dots,k_r,l_1,\dots,l_s,2\leq a\\k_1+\cdots+k_r+a+l_1+\cdots+l_s=k}}\zeta\left(\begin{matrix}k_1,\dots,k_r\\l_1,\dots,l_s\end{matrix};a\right)\\ &=\sum_{0\leq j}\frac{\prod_{0< n< j}\left(n-x+y\right)}{j\prod_{0< m\leq j}\left(m-x-z\right)}\\ &=\sum_{0< n}\frac 1{(n-x-z)(n-y-z)}\\ &=\sum_{0\leq r}y^r\sum_{r+2\leq k}x^{k-r-2}\sum_{0< n}\frac 1{(n-z)^k}\\ &=\sum_{0\leq r,s}y^rz^s\sum_{r+2\leq k}x^{k-r-2}\binom{k+s-1}{s}\sum_{0< n}\frac 1{n^{k+s}}\\ &=\sum_{0\leq r,s}y^rz^s\sum_{r+s+2\leq k}x^{k-r-s-2}\binom{k-1}{s}\sum_{0< n}\frac 1{n^{k}}\qquad(k\mapsto k-s) \end{align}
となる. よって両辺の$x^{k-r-s-2}y^rz^s$の係数を比較して
\begin{align} \sum_{\substack{1\leq k_1,\dots,k_r,l_1,\dots,l_s,2\leq a\\k_1+\cdots+k_r+a+l_1+\cdots+l_s=k}}\zeta\left(\begin{matrix}k_1,\dots,k_r\\l_1,\dots,l_s\end{matrix};a\right)=\binom{k-1}s\zeta(k) \end{align}
を得る.