Can one show that there exists an $\epsilon>0$ such that there are infinitely many $n$ where $m+\epsilon \omega(m)\leq n$ for all $m<n$?

OPEN

Let $\omega(n)$ count the number of distinct primes dividing $n$. Are there infinitely many $n$ such that, for all $m<n$, we have $m+\omega(m) \leq n$?

Can one show that there exists an $\epsilon>0$ such that there are infinitely many $n$ where $m+\epsilon \omega(m)\leq n$ for all $m<n$?

In [Er79] Erdős calls such an $n$ a 'barrier' for $\omega$. Some other natural number theoretic functions (such as $\phi$ and $\sigma$) have no barriers because they increase too rapidly. Erdős believed that $\omega$ should have infinitely many barriers. In [Er79d] he proves that $F(n)=\prod k_i$, where $n=\prod p_i^{k_i}$, has infinitely many barriers (in fact the set of barriers has positive density in the integers).

Erdős also believed that $\Omega$, the count of the number of prime factors with multiplicity), should have infinitely many barriers. Selfridge found the largest barrier for $\Omega$ which is $<10^5$ is $99840$.

In [ErGr80] this problem is suggested as a way of showing that the iterated behaviour of $n\mapsto n+\omega(n)$ eventually settles into a single sequence, regardless of the starting value of $n$ (see also [412] and [414]).

Erdős and Graham report it could be attacked by sieve methods, but 'at present these methods are not strong enough'.