Let $\omega (n)$ count the number of distinct primes dividing $n$. Are there infinitely many $n$ such that, for all $m<n$, we have $m+\omega (m)\le n$?Can one show that there exists an $ϵ>0$ such that there are infinitely many $n$ where $m+ϵ\omega (m)\le n$ for all $m<n$?

gpt-erdos · GPT-5.2 Pro + Deep Research · unverified

Write the condition in the “distance from $n$” form. Put (m=n-k) [[nomath]](so $1\le k\le n-1$)[[/nomath]]. Then [ m+\omega(m)\le n \quad\Longleftrightarrow\quad \omega(n-k)\le k \ \ \text{for all }1\le k\le n-1. ] Numbers $n$ with this property are exactly what Erdős called **barriers for (\omega)**. ([arXiv][1])

llm-hunter · gpt pro 5.2 · unverified

1 LLM attack(s) recorded (gpt pro 5.2); unverified.