with positive signs fill in exactly the area inside the square $[1,p-1]\times [1,p-1]$ (boundary inclusive): $L(i,i)$ in ${\rm FDS}(i,i)$ provides the diagonal $y=x$, $\sum_{1\le j<i<p} L(i,j)$ (resp. $\sum_{1\le j<i<p} L(j,i)$) form the lower right (resp. upper left) triangular region.

For the negative terms of the double logs, $L(i,-i)$ in ${\rm RDS}(i)$ provides the diagonal $x+y=p$, $\sum_{1\le j<i<p} L(i,j-i)=\sum_{i=2}^{p-1}\sum_{j=p+1-i}^{p-1}L(i,j)$ form the upper right triangular region. Similarly, by changing the order of summation $\sum_{1\le j<i<p} L(j,i-j)=\sum_{i=1}^{p-2}\sum_{j=i+1}^{p-1}L(i,j-i)
=\sum_{i=2}^{p-1}\sum_{j=1}^{p-1-i}L(i,j)$ fills the lower left region.

Further, numerical evidence up to level $N=49$ supports the following

\[conj:regdistr\] In weight two, all RDT are consequences of RDS and FDT.

Lifted relations from lower weights
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Note that when $N=3$ there are no seeded relations nor (regularized) distribution relations. When we deal with MZVs and alternating Euler sums we expect that all the linear relations come from RDS. Are these enough when $N=3$? Surprisingly, the answer is no.

The first counterexample is in weight four, i.e., $(w,N)=(4,3)$. Easy computation shows that there are 144 MPVs in this case among which there are 239 nontrivial RDS which include 191 FDS. Using these relations we get 127 independent linear relations among the 144 MPVs. But the upper bound of $d(4,3)$ 