Ok so here is a question I thought of (while trying to solve some numbers inequalities of course):
Definition: We will call a number $n$ "fair" if for all $p$ prime numbers such that $v_p(n)=\alpha$:
$p^\alpha< \sqrt n$.
$v_p(n)$ is the normal notation for the max power of $p$ that divides $n$.
(That is, there is no majority for one type of primes.)
Show (or disprove): that $\forall k\in \mathbb{N}:$ there exist k consecutive fair numbers.
Remark: I tried using some density arguments but with no luck. I won't write this because it's a long proof but looking only at the density of "fair"s that their primes factorization is $pqr$ for some primes, yields density $0$ but really by a whisker (I was using some basic Stirling stuff).
And also I should mention that when I'm talking about density I'm looking at it on all the consecutive sets: $[1,n],[n+1,2n]\ldots$