Consider the sequence $1, 2, 3, \cdots, N$ where $N = p_1^{e_1} \cdot p_2^{e_2} \cdots p_k^{ e_k}$ is the unknown factorization of a given $N$.
Define the set
$$M = \left\{\overbrace{p_1, 2p_1, \cdots, \left(\frac{N}{p_1}\right)}^{\text{multiples of $p_1$}}, \underbrace{p_2, 2p_2, \cdots, \left(\frac{N}{p_2}\right)}_{\text{multiples of $p_2$}}, \cdots, \overbrace{p_k, 2p_k, \cdots, \left(\frac{N}{p_k}\right)}^{\text{multiples of $p_k$}}, N \right\}$$
Suppose we choose a prime $q$ and generate the set of terms in the AP
$$ A_q = \{a_m : a_m = 1 + mq, 0 \le m \le n \} $$
What is the $n$ that we can choose such that $N \in A_q$ and $|A_q \cap M| \ge \beta$ where $\beta \ge 2$ is some chosen bound?
How can we choose $n$ without knowing the size of any factor of $N$? The only fact we can assume is that primes upto $p_{\text{max}}$ have been tested and no prime less than or equal to $p_{\text{max}}$ divides $N$.
Approach so far. Since we want $N \in A_q$ take $a_n = 1 + nq = N$. i.e., $n|(N-1)$. Since no prime less than $p_{\text{max}}$ divides $N$, $N$ is odd. Therefore, $N-1$ is even and $2|(N-1)$. So, $n=2$ will ensure that $N \in A_q$. However, this approach still leaves some open questions.
How do we prove that the cardinality of the intersection $|A_q \cap M| \ge \beta$. Also, it is possible that $q\not|N$ since it is independently chosen. In that case, how do we show that the cardinality constraint is met?