Let us suppose we have slots $n$ slots $1, \ldots, n$ and $k$ pebbles, each of which is initially placed in some slot. Now the pebbles want to space themselves out as evenly as possible, and so they do the following. At each time step $t$, each pebble moves to the slot closest to the halfway point between its neighboring pebbles; if there is a tie, it chooses the slot to the left. The leftmost and rightmost pebbles apply the same procedure, but imagining that there are slots numbered $0$ and $n+1$ with pebbles in them.
Formally, numbering the pebbles $1, \ldots, k$ from left to right, and letting $x_i(t)$ be the slot of the $i$'th pebble at time t, we have $ x_i(t+1) = \lfloor \frac{x_{i-1}(t)+x_{i+1}(t)}{2} \rfloor, i = 2, \ldots, k-1$ $\lfloor \cdot \rfloor$ rounds down to the closest integer. Similarly, $ x_1(t+1) = \lfloor (1/2) x_2(t) \rfloor, x_k(t+1) = \lfloor \frac{x_{k-1}(t) + (n+1)}{2} \rfloor.$
Now the fixed point of this procedure is the arrangement in which $x_{i+1} - x_i$ and $x_i - x_{i-1}$ differ by $1$. My question: is it true that this fixed point is reached by the above procedure after sufficiently many iterations?
Why I care: no concrete reason really, I am just reading about finite difference methods, and this seemed like a simple problem connected with some of the things which are confusing me.