lines of (\[eq:uxisym\]), and then repeating this for the $2\textup{nd}$, $5\textup{th}$, … or $3\textup{th}$, $6\textup{th}$, … lines, we obtain condition (\[eq:Mmod3\]).

Then it turns out that we can set both $\xi_1$ and $\xi_2$ to be any real numbers, so $\xi_3=u_2-\xi_1-\xi_2$, and $\xi_k$, $k\ge 4$, are given: \_[k+1]{}=\[u\_k-u\_[k-1]{}\]+\[u\_[k-3]{}-u\_[k-4]{}\]+…+\_[(k 3)+1]{}.

Similarly to the asymmetric case, consider the Markov chain $v(t)$ on $\{0,1,2,\dots\}$ and recall the definition of $t_j$ from (\[eqv(t)\]). The following statement is straightforward.

\[prop\_sym1\] For any $k\in\M$ $$v_{k}(t_{j+1})\le v_{k}(t_{j})$$ unless both $v_{k-1}(t_{j})=0$ and $v_{k+1}(t_{j})=0$.

\[prop\_sym2\] For $j$ large enough, in the sequence of $v_k(t_j)$, $k\in\M$, there are no more than two non-zero elements in a row, that is $$\mbox{if }v_k(t_j)>0\mbox{ then either }v_{k-1}(t_j)=0\mbox{ or
 }v_{k+1}(t_j)=0, \mbox{ or both}.$$

Fix some $k\in\M$. Then $v_k(t_j)$ is either $0$ or positive. In the first case, unless both of the neighbours of point $k$ are zeros at time $t_j$, by Proposition \[prop\_sym1\] we have $v_k(t_{j+1})=0$. On the other hand, if $(v_{k-1}(t_j),v_k(t_j),v_{k+1}(t_j))=(0,0,0)$, then at time $t_{j+1}$ either this triple becomes $(0,1,0)$ if both $k-1$ and $k+1$ are chosen, or $v_k(t_{j+1})=0$.

Now suppose that $v_k(t_j)>0$. If both $v_{k-1}(t_{j})=0$ and $v_{k+1}(t_{j})=0$, then from Proposition \[prop\_sym1\] applied to $k-1$ and $k+1$, we conclude $v_{k-1}(t_{j+1})=v_{k+1}(t_{j+1})=0$, hence point $k$ remains 