It is known that every positive integer can be expressed as $n=s\cdot q $, where $s$ is a powerful number and $q$ a squarefree number, with $(s,q)=1$. It is also known that the set of integers such that $s> q$ is of zero density. Question: is the ratio $\frac{\Omega (n)}{\omega (n)}$ bounded from below with some positive constant $c> 1$ , for integers $n=s\cdot q $ with $s> q$ and $q> 1$? In other words, is there a positive constant $c$ such that inequality $\frac{\Omega (n)}{\omega (n)}\geq c>1$ is valid for all such numbers? Here $\omega (n)$, $\Omega(n)$ denote the number and the total number of prime divisors of $n$.
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$\begingroup$ If $\omega(s)=a\ge1$, $\omega(q)=b\ge1$, then $\frac{\Omega(n)}{\omega(n)}\ge\frac{2a+b}{a+b}\ge1$. Obviously, sharp. Take $q$ with many prime divisors and $s=p^2>q$ for prime $p$ $\endgroup$– te4Commented Dec 7 at 13:24
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$\begingroup$ This is obvious. Sorry, i forgot to add the condition $c> 1$ $\endgroup$– Andrej LeškoCommented Dec 7 at 13:50
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$\begingroup$ I have completed my question with the condition $c> 1$. $\endgroup$– Andrej LeškoCommented Dec 7 at 13:57
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2$\begingroup$ The example of te4 shows that there is no such $c>1$. If $s=p^2>q$, where $p$ is a prime not dividing $q$, then $\Omega(n)/\omega(n)=(2+\omega(q))/(1+\omega(q))$, which is very close to $1$ when $\omega(q)$ is large. $\endgroup$– GH from MOCommented Dec 7 at 14:07
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$\begingroup$ Can we estimate the power of the set $\left\{n\leq x,\frac{\Omega (n)}{\omega (n)}> 1+\varepsilon,\varepsilon > 0 \right\}$ with $n=s\cdot q$,s>q, with fixed $\varepsilon$ $\endgroup$– Andrej LeškoCommented Dec 7 at 16:04
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