conjecture concerning a special class of distribution relations.

\[conj:distr\] Let $d$ be a positive integer. Then all the distribution relations in , where $x_j=1$ for all $j$, are consequences of RDS of MPVs of level $d$.

We are able to confirm this conjecture in the special case that $w=2$, $n=1$, and $d$ is a prime.

Write $L(i,j)=L_p(1,1|i,j)$ and $D(i)=L_p(2|i)$. Define for $1\le i,j<p$: $$\begin{aligned}
\ & \text{FDT}:=
   D(0)-p\sum_{j=0}^{p-1}D(j)=0, \qquad
\text{RDS$(i)$}:=
 D(i)+L(i,0)-L(i,-i)=0,\\
\ &\text{FDS$(i,j)$}:=
 D(i+j)+L(i,j)+L(j,i)-L(i,j-i)-L(j,i-j)=0.\end{aligned}$$ Then $$\label{equ:distNull}
{\rm FDT}=\sum_{1\le i<p} {\rm FDS}(i,i)
+2\sum_{1\le j<i<p} {\rm FDS}(i,j)+2\sum_{i=1}^{p-1} {\rm RDS}(i,i).$$

By changing the order of summation we see that $$\begin{aligned}
2\sum_{1\le j<i<p} D(i+j)=& \sum_{i=2}^{p-1}\sum_{j=1}^{i-1} D(i+j) +
\sum_{j=1}^{p-2}\sum_{i=j+1}^{p-1} D(i+j)\\
=&\sum_{i=2}^{p-2}\sum_{i\ne j=1}^{p-1} D(i+j)
+ \sum_{j=1}^{p-2} D(j-1) +\sum_{i=2}^{p-1} D(i+1)\\
=&(p-3)\sum_{j=0}^{p-1} D(j)-\sum_{i=2}^{p-2} D(i)-\sum_{i=1}^{p-1} D(2i)
+ \sum_{j=1}^{p-2} D(j) +\sum_{j=2}^{p-1} D(j)+2D(0)\\
=&(p-1)D(0)+(p-3)\sum_{j=1}^{p-1} D(j)\end{aligned}$$ since $\sum_{j=0}^{p-1} D(i+j)=\sum_{j=1}^{p-1} D(j)$ for all $i$ and $\sum_{i=1}^{p-1} D(2i)=\sum_{i=1}^{p-1} D(i)$. This implies that the dilogarithms on the right hand side of exactly add up to FDT. Thus we only need to show that all the double logarithms on the right hand side of cancel.

First we note that $L(i,0)$ in ${\rm FDS}(i,i)$ and ${\rm RDS}(i,i)$ cancel. Now let us consider the lattice points $(i,j)$ of $\Z^2$ corresponding to $L(i,j)$. The points $(i,j)$ corresponding to $L(i,j)$ 