A continuation of this question :
How does one classify the set of all partitions (into disjoint parts) $\{AP_1, AP_2, AP_3\cdots\}$ of $\mathbb{N}$ into APs such that all the APs have a distinct common difference? One such example was given by @HagenVonEitzen in the linked post: $$\left\{\{(2k+1)2^{i-1} \mid k\in \{0,1,2,\cdots\}\}\ \mid\ i\in\{1,2,3\cdots\}\right \}$$
Naively decomposing numbers into powers of $k$ and a $k$−free part doesn't work because I then end up with APs that have the same common difference.
There must be one AP beginning at $1$, say $AP_1 = \{n: n \equiv 1 \mod m_1\}$. To make things nontrivial, $m_1 \ge 2$. The first natural number not in $AP_1$ is $2$, so it must be in $AP_2 = \{n: n \equiv 2 \mod m_2\}$. In order for this to be disjoint from $AP_1$, we need $\gcd(m_1, m_2) > 1$. We also want $m_2 \ne m_1$. This will also avoid making the process stop after two AP's, which it would if $m_2 = m_1 = 2$.
Proceed inductively: after $k$ steps, we will have $AP_1, \ldots, AP_k$ with differences $m_1, \ldots, m_k$. Let $a_{k+1}$ be the first natural number not included in any of these. It will be in $AP_{k+1} = \{n: n \equiv a_{k+1} \mod m_{k+1}\}$. In order to avoid intersecting previous AP's, we need, for $1 \le j \le k$, $gcd(m_j, m_{k+1})$ not to divide $a_{k+1}-a_j$. Note that $a_{k+1} - a_j$ is not a multiple of $m_j$ (since by assumption $a_{k+1} \notin AP_j$) so we can always take $m_{k+1}$ to be a multiple of $\text{lcm}(m_1, \ldots, m_k)$. Moreover, we want $m_{k+1}$ distinct from $m_1, \ldots, m_k$, and we want to make sure we haven't exhausted all of $\mathbb N$: if that would happen for a particular candidate for $m_{k+1}$, just take a larger candidate.