Duality between prime factors and the Prime Number Theorem for Arithmetic Progressions -- II

In the first paper under this title (1977), the first author utilized a duality identity between the largest and smallest prime factors involving the Moebius function, to establish the following result as a consequence of the Prime Number Theorem for Arithmetic Progressions: If $k$ and $\ell$ are positive integers, with $1\le\ell\le k$ and $(\ell, k)=1$, then $$\sum_{n\ge 2,\, p(n)\equiv\ell(mod\,k)}\frac{\mu(n)}{n}=\frac{-1}{\phi(k)},$$ where $\mu(n)$ is the Moebius function, $p(n)$ is the smallest prime factor of $n$, and $\phi(k)$ is the Euler function. Here we utilize the next level Duality identity between the second largest prime factor and the smallest prime factor, involving the Moebius function and $\omega(n)$, the number of distinct prime factors of $n$, to establish the following result as a consequence of the Prime Number Theorem for Arithmetic Progressions: For all $\ell$ and $k$ as above, $$\sum_{n\ge 2, \, p(n)\equiv\ell(mod\,k)}\frac{\mu(n)\omega(n)}{n}=0.$$ A quantitative version of this result is proved.