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

Abstract

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 $$\sum _{n\ge 2,p(n)\equiv \ell (modk)}\frac{\hat{I}¼(n)}{n}=\frac{-1}{\ddot{I}•(k)},$$ where $\hat{I}¼(n)$ is the Moebius function, $p(n)$ is the smallest prime factor of $n$, and $\ddot{I}•(k)$ 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 $\ddot{I}‰(n)$, 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, $$\sum _{n\ge 2,p(n)\equiv \ell (modk)}\frac{\hat{I}¼(n)\ddot{I}‰(n)}{n}=0.$$ A quantitative version of this result is proved.