A question about double zeta values

Question

\begin{align*} &\zeta \left( {\bar 2,2} \right) + 2\zeta \left( {\bar 3,1} \right) = \frac{5}{{16}}\zeta \left( 4 \right), \\ &\zeta \left( {\bar 4,2} \right) + 4\zeta \left( {\bar 5,1} \right) = \frac{1}{4}\zeta \left( {4,2} \right) - \frac{{37}}{{64}}\zeta \left( 6 \right) + \frac{1}{2}{\zeta ^2}\left( 3 \right), \\ &\zeta \left( {\bar 6,2} \right) + 6\zeta \left( {\bar 7,1} \right) = - \frac{9}{8}\zeta \left( {6,2} \right) - \frac{{1887}}{{256}}\zeta \left( 8 \right) + 6\zeta \left( 3 \right)\zeta \left( 5 \right). \end{align*} Based on the above three results, I conjecture that \begin{align*} \zeta \left( {\bar 8,2} \right) + 8\zeta \left( {\bar 9,1} \right) =& {a_1}\zeta \left( {8,2} \right) + {a_2}\zeta \left( 2 \right)\zeta \left( {6,2} \right) + {a_3}\zeta \left( {10} \right) + {a_4}{\zeta ^2}\left( 5 \right) \\ &+ {a_5}\zeta \left( 3 \right)\zeta \left( 7 \right) + {a_6}\zeta \left( 2 \right)\zeta \left( 3 \right)\zeta \left( 5 \right) + {a_7}{\zeta ^2}\left( 3 \right)\zeta \left( 4 \right)\quad ?. \end{align*} Here the coefficients $a_i\ (i=1,2,3,4,5,6,7)$ are rational numbers.

In general, the combined alternating mzvs $$\zeta \left( {{\overline {2m}},2} \right) + 2m\zeta \left( {{\overline {2m + 1}},1} \right)=?\quad (m\in \mathbb{N}).$$ It can be expressed in terms of non-alternating mzvs and zeta values?

Here the double zeta values are defined by \begin{align*} \zeta \left( {p,q} \right): = \sum\limits_{{n_1} > {n_2} > 0} {\frac{1}{{n_1^pn_2^q}}} ,\zeta \left( {\bar p,q} \right): = \sum\limits_{{n_1} > {n_2} > 0} {\frac{{{{\left( { - 1} \right)}^{{n_1}}}}}{{n_1^pn_2^q}}} \quad (p,q\in \mathbb{N}). \end{align*}

Answer 1

Your conjecture is quite likely to be true. Let's us take the case $m=4$. We have: \begin{eqnarray} \zeta(\bar{8},2)+8\zeta(\bar{9},1)&=& {\bf H}^{(2)}_8(-1) + 8 {\bf H}^{(1)}_9(-1) -9 Li_{10}(-1) \\ &=& \left( \left\{\zeta(\bar{2},8) - \zeta(\bar{8},2)\right\} - \frac{255}{256} {\bf H}^{(2)}_8(+1) + \frac{237}{32} \zeta(3) \zeta(7)+\frac{15}{4} \zeta(5)^2 - \frac{108201}{5120} \zeta(10)\right) -9 Li_{10}(-1) \\ \end{eqnarray} where in the second line we used my answer to Calculating alternating Euler sums of odd powers . The quantity in curly brackets looks "symmetric" so it is quite likely that it reduces to MZVs at plus unity only. I don't have aproof for that now but by using the MZV-interface http://wayback.cecm.sfu.ca/cgi-bin/EZFace/zetaform.cgi I figured out that : \begin{equation} \left\{\zeta(\bar{2},8) - \zeta(\bar{8},2)\right\}= \frac{1}{2560}\left(-2637 \zeta(10)+1800 \zeta(3) \zeta(7)+840 \zeta(5)^2-270 \zeta(8,2)\right) \end{equation} Bringing this all together we get the following: \begin{eqnarray} \zeta(\bar{8},2)+8\zeta(\bar{9},1)&=&\frac{3}{1024} \left( 1392 \zeta(5)^2 + 2768 \zeta(3) \zeta(7) -4839 \zeta(10) - 376 \zeta(8,2)\right) \end{eqnarray} as expected.

When I look at my list Calculating alternating Euler sums of odd powers of MZVs at minus unity I can indeed see a pattern you conjecture. As a matter of fact there will be many more relationships of this kind one of them being: \begin{eqnarray} &&A_m^{(0)} \cdot\left((2m-1) \zeta(\bar{2 m},2) + \binom{2 m}{2} \zeta(\bar{2m+1},1)+\zeta(\bar{3},2m-1) \right)=\\ && A_m^{(1)} \zeta(2 m,2) + A_m^{(2)} \zeta(2m+2) + \sum\limits_{j=1}^{\lfloor \frac{m}{2} \rfloor} A_m^{(2+j)}\zeta(2 j+1) \zeta(2m-2j+1) \end{eqnarray} where \begin{eqnarray} A_m^{(j)} = \left( \begin{array}{rrrrr} 4&3 &-1&&&\\ 32&0&0&9&&\\ 512&-1260&-6391&5280&0\\ 1024&-3570&-37023&22008&9840&\\ 8192&-36828&-637549&300096&265248 \end{array} \right)_{m=1}^5 \end{eqnarray} where $m$ and $j$ label the rows and the columns of the matrix above, respectively.