bro what i gave up trying to understand it after the first line my guess is once you decode it, it's probably something simple to prove

not the 17 consecutive right parentheses at the end

$ $

Attachments:

This is like at least 2 MOHS

This is equivalent to the following problem. Quote: Consider the directed graph with vertices $\mathbb{N}$ and edges $m\to km$ and $m\to km+1$, for $m,k\in \mathbb{N}$ and $k\geq 2$. Then, starting from any positive integer $n$, one can reach every sufficiently large integer via a directed path. I was told this is apparently open. So we prove a weaker version of the problem, that we can reach almost every number. Theorem. Fix a positive integer $n$, and let $f(N)$ denote the number of positive integers $m\leq N$ reachable from $n$. Then $f(N)/N\to 1$ as $N\to\infty$. Proof. We can reach any number of the form $kn+1$ from $n$, for $k\geq 2$. In particular, we can reach a prime number by Dirichlet's theorem, so WLOG assume that $n$ is prime. For a positive integer $m=p_1^{a_1}\cdots p_k^{a_k}$, denote by $\Omega(m):=a_1+\cdots+a_k$. We claim that if $\Omega(m)> 2n$, then $m$ is reachable. If $n\mid m$, then $m$ is reachable, so assume that $n\nmid m$. Suppose $\Omega(m)=r>2n$, then there are distinct positive integers $1<q_1\mid q_2\mid\cdots \mid q_r=m$. By pigeonhole and $r>2n$, there exists $i<j<k$ such that $q_i\equiv q_j\equiv q_k\pmod{n}$, so $q_k/q_i\equiv 1\pmod{n}$ and $q_k/q_i>n+1$. Therefore, $q_k/q_i=\frac{q_k/q_i-1}{n}n+1$ is reachable, thus $m$ is reachable since it is a multiple of $q_k/q_i$. Finally, by the Hardy–Ramanujan theorem, almost all numbers $m$ have approximately $\log\log m$ distinct prime factors. In particular, almost all numbers have more than $2n$ prime factors, so by our claim they are reachable. $\square$