In this paper, we are going to perform the shuffle products of $Z_-(n) = \sum_{ a+b=m} (-1)^{ b} \zeta(\{ 1\}^{ a},b+2)$ and $Z_+^\star(n) = \sum_{ c+d=n} \zeta^{ \star}(\{ 1\}^{ c},d+2)$ with $m+n = p$. The resulted shuffle relation is a weighted sum formula given by \begin{ equation*} \frac{ (p+1)(p+2)}{ 2} \zeta(p+4) =\sum_{ m+n=p} \sum_{ |\boldsymbol{ \alpha}|=p+3} \zeta(\alpha_{ 0}, \alpha_{ 1}, \ldots, \alpha_{ m}, \alpha_{ m+1}+1) \sum_{ a+b+c=m} \Bigl( W_{ \boldsymbol\alpha}(a,b,c) + W_{ \boldsymbol\alpha}(a,b,c=0) + W_{ \boldsymbol\alpha}(a=0,b,c) + W_{ \boldsymbol\alpha}(a=0,b=m,c=0) \Bigr), \end{ equation*} where $W_{ \boldsymbol\alpha}(a,b,c) = 2^{ \sigma(a+b+1)-\sigma(a)-(b+1)} (1-2^{ 1-\alpha_{ a+b+1}}\ \ )$, with $\sigma(r) = \sum_{ j=0}^{ r} \alpha_{ j}$.
Categories: math.NT