of the Markov chain $u(t)=(u_1(t),\dots,\allowbreak u_{M}(t))$ are given by $$\begin{gathered}
 \P\left(u_{i}(t+1)=u_{i}(t)+1_{i\in\{k-1,k\}}, i=1,\ldots,M| u(t)\right)\\
=\left\{\begin{array}{rcl}
  0, &\mbox{if} & u_k(t)>m(t),\\
  N^{-1}_{\min}(t), &\mbox{if} & u_k(t)=m(t),
 \end{array}\right.\end{gathered}$$ for $k\in\M$, where $N_{\min}(t)\in\{1,2,\dots,M\}$ is the number of $u_i(t)$ equal to $m(t)$.

\[prop:even\] If $M$ is odd, then for any $u=(u_1,u_2,\dots,u_M)$ the system \[eq:uxiasym\] u\_1&=&\_1+\_[2]{}\
u\_2&=&\_2+\_[3]{}\
&&\
u\_M&=&\_M+\_[1]{} has a unique solution. On the other hand, if $M$ is even, system (\[eq:uxiasym\]) has a solution if and only if \[eq:evenodd\] u\_1+u\_3+…+u\_[M-1]{}= u\_2+u\_4+…+u\_[M]{}.

For a fixed $\xi_1$ we can express the remaining $\xi_k$ as \[eq:soluxi\] \_k&=&u\_[k-1]{}-u\_[k-2]{}+u\_[k-3]{}-…+(-1)\^[k-1]{} \_1, for any $k=2$, …, $M$. Now, when $M$ is odd, there is a unique choice of \_1&=&12(u\_M-u\_[M-1]{}+u\_[M-2]{}+…-u\_2+u\_1). When $M$ is even, by summing separately odd and even lines of (\[eq:uxiasym\]) we obtain condition (\[eq:evenodd\]). Then it turns out that we can set $\xi_1$ to be any real number, with $\xi_k$, $k\ge 2$, given by (\[eq:soluxi\]).

The following statement immediately follows from Proposition \[prop:even\].

\[cor\_even\] If $M$ is even, then $$\begin{aligned}
\label{condi}
v_1(t)+v_3(t)+\dots+v_{M-1}(t)=v_2(t)+v_4(t)+\dots+v_{M}(t) \ \ \
\forall t.
 \end{aligned}$$

In the following two Propositions we will show that when either $M$ is odd *or* $M$ is even and condition (\[condi\]) holds, the state ${\bf 0}=(0,\dots,0)$ is recurrent for the Markov chain $v(t)$. 