If we call a number $k$ $\mathit{minimal}$ with respect to a parameter $Q$ when $k\geq Q$ but there is no single prime divisor we could remove from $k$ and stay greater than $Q$, what bounds can we find on the number of minimal divisors of a square free number $n$?
To make this more clear, here's an example: If $n=210=2\cdot3\cdot5\cdot7$ and $Q=32$, then $d=35=5\cdot7$ is a minimal divisor since $35\geq30$ but if we remove $5$ or $7$ then then it becomes less than $30$. On the otherhand, $d=105=3\cdot5\cdot7$ would not be minimal since $105\geq30$, but if we remove the three we get $35$ which is still greater than $30$.
If we let this number of minimal divisors of a square free $n$ with parameter $Q$ be denoted by $d\left(n,Q\right)$, then what upper bounds can be made on $d\left(n,Q\right)$?
Hopefully, these bounds would be in terms of $n$, $Q$ or $\omega_Q(n)=\#prime\,divisors\,of\, n\,less\,than\,Q$.
My conjecture is that the behavior of $d\left(n,Q\right)$ would have something to do with the "spread" of the prime divisors of $n$, but of course I cannot be sure.
Certainly one way of creating examples with $d(n,Q)$ large is to take $n$ to be the product of primes greater than $Q$. (I think this is probably the best way to do so, though that's a fuzzy statement. I thought about taking primes between $Q^{1/k}$ and $Q^{1/(k-1)}$, but $k=1$ seemed to be the best choice for large $Q$.)
Given $Q$, the number of prime factors of $n$ that exceed $Q$ is certainly at most $(\log n)/(\log Q)$, and the number of prime factors of $n$ in general is at most $(1+o(1))(\log n)/(\log\log n)$. Therefore if we choose $Q$ and $n$ so that $Q\approx\log n$, we can get $d(n,Q) \sim (\log n)/(\log\log n)$ or $d(n,Q) \sim Q/\log Q$.
(Note that here $\omega_Q(n)=0$, so I don't think bounds in terms of this quantity are going to be helpful.)