chain $u(t)$ are now given by $$\begin{gathered}
 \P\left(u_{i}(t+1)=u_{i}(t)+1_{i\in\{k-1,k,k+1\}}, i=1,\ldots,M|u(t)\right) \\
=\left\{\begin{array}{rcl}
  0, &\mbox{if} & u_k(t)>m(t),\\ \\
  N^{-1}_{\min}(t), &\mbox{if} & u_k(t)=m(t),
 \end{array}\right.
 \end{gathered}$$ for $k\in\M$, where, as before, $N_{\min}(t)\in\{1,2,\dots,M\}$ is the number of $u_i(t)$ equal to $m(t)$.

\[prop:by3\] If $(M\mbox{ mod }3)\ne 0$, then for any $u=(u_1,u_2,\dots,u_M)$ the system \[eq:uxisym\] u\_1&=&\_M+\_[1]{}+\_[2]{}\
u\_2&=&\_1+\_[2]{}+\_[3]{}\
&&\
u\_M&=&\_[M-1]{}+\_[M]{}+\_[1]{} has a unique solution. On the other hand, if $M$ is divisible by $3$, system has a solution if and only if $$\begin{aligned}
 u_1+u_4+\dots+u_{M-2}&= u_2+u_5+\dots+u_{M-1}\nonumber\\
&= u_3+u_6+\dots+u_{M}.\label{eq:Mmod3}
 \end{aligned}$$

If $M$ is not divisible by $3$, then the determinant of the matrix (

[cccccccc]{} 1 & 1 & 0 & 0 &…& 0 & 0 & 1\
1 & 1 & 1 & 0 &…& 0 & 0 & 0\
0 & 1 & 1 & 1 &…& 0 & 0 & 0\
&&&&&&&\
0 & 0 & 0 & 0 &…& 1 & 1 & 1\
1 & 0 & 0 & 0 &…& 0 & 1 & 1\

) corresponding to the equation (\[eq:uxisym\]) is $\pm 3$ (which can be easily proved by induction). Hence the system must have a unique solution.

When $M$ is divisible by $3$, by summing separately the $1^\textup{st}$, $4^\textup{th}$, $5^\textup{th}$, … 