A mapping from a lattice to itself Consider $\mathbb{Z}^{n}$ for $n = 2^r$ where $r \geq 1$ . Look at the iterates of the following function $T$ from $\mathbb{Z}^n$ to itself.
$T((a_1, a_2, \ldots, a_n)) = (|a_1 - a_n|, |a_2 - a_1|, |a_3 - a_2|, \ldots, |a_n - a_{n-1}|)$.
It has been proved that when $n = 2^{r}$, then for every  $(a_1, a_2, \ldots, a_n) \in \mathbb{Z}^n - \{0\}$, there exists some $i \geq 1$, such that $T^{i}((a_1, a_2, \ldots, a_n)) = 0.$ This does not hold for other values of $n$. Note that, if $T^{i}((a_1, a_2, \ldots, a_n)) = 0$, then $T^{j}((a_1, a_2, \ldots, a_n)) = 0$ for all $j > i.$
Findings so far are the following.
(i)   $T(k(a_1, a_2, \ldots, a_n)) = k T((a_1, a_2, \ldots, a_n))$ for all $k \in \mathbb{Z}$.
(ii)  $T(k + (a_1, a_2, \ldots, a_n)) = T((a_1, a_2, \ldots, a_n))$ for all $k \in \mathbb{Z}$, where $k + (a_1, a_2, \ldots, a_n) = (k + a_1, k + a_2, \ldots, k + a_n).$
(iii) Let, $S_{i} = \{(a_1, a_2, \ldots, a_n) \in Z^{n} : T^{i}((a_1, a_2, \ldots, a_n)) = 0 \text{ and } T^{i-1}((a_1, a_2, \ldots, a_n)) \neq 0\}$ for $i \geq 1$. Note that $S_i$ s are disjoint also their union is equal to $\mathbb{Z}^n$.
The questions that I have are the following.
(i) What's the maximum value of $i$ such that $S_{i}$ is not empty? Putting it in other words, what's the maximum number of times the function T needs to be applied to a vector so that it gets mapped to $0$ vector.
(ii) Since the function $T$ is homogeneous, notions from projective space can be borrowed. How could projective geometry be applied here?
 A: The answer is infinity for $n>2$.
Suppose that there is an $i$ such that $T^i(x)=0$ for all integer vectors $x$. Then the same follows for all rational vectors by homogenuity, and then for all real vectors by approximation.
But there is a real vector that never reaches zero. For example, for $n=4$ consider
$$
 (a^3,a^2,a,1)
$$
where $a>1$ satisfies $a^3=a^2+a+1$. The next iteration is
$$
 (a^3-1,a^3-a^2,a^2-a,a-1)
 = (a-1)\cdot (a^2+a+1, a^2, a, 1) = (a-1)\cdot (a^3, a^2, a, 1) .
$$
Note that it is proportional to the original vector. By induction it follows that the $n$th iteration equals $(a-1)^n$ times the original vector, hence it never becomes zero.
An obvious modification of this example works for every $n>2$.
A: This is not an answer though. I am trying still to solve the problem. I would like to get comments on the approach that I am following. The approach in nutshell is described below.
I am trying to characterize the sets $S_{i}$. Note that, any $a \in S_{1}$ can be derived from the single vector $(1, 1, \ldots, 1)$, by doing either $k + (1, 1, \ldots, 1)$ or  $k (1, 1, \ldots, 1).$
Similarly, any $a \in S_{2}$ can be derived from any of the following vectors by applying  $k + <the vector>$ and  $k.<the vector>$.
The vectors are (i) $(0, 1, 0, 1, \ldots, 0, 1)$, (ii) $(1, 0, 1, 0, \ldots, 1, 0)$ and all the vectors that can be formed from the vector $(1, 0, 1, 2, 1, 0, 1, 2, \ldots, 1, 0, 1, 2)$ by shifting the elements of the vectors by one position in the right.
This can be verified. Can we now iterate in some way this to come to a value of $i$ such that no such vectors can be formed and then we stop. $i-1$ then is the answer.
Is it a viable approach? Or am I missing something?
