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?