It is known that every positive integer can be expresed as $n=s\cdot q $, where $s$ is a powerfull 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 $n$.

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 $n$.

It is known that every positive integer can be expresed as $n=s\cdot q $, where $s$ is a powerfull 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 $c$ , for integers $n=s\cdot q $ with $s> q$ and $q> 1$? In other words, is there a positive constant $c$ such that inequality $\frac{\Omega (n)}{\omega (n)}\geq c$ is valid for all such numbers? Here $\omega (n)$, $\Omega(n)$ denote the number and the total number of prime divisors of $n$.

It is known that every positive integer can be expresed as $n=s\cdot q $, where $s$ is a powerfull 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 $n$.