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 $$ \sum_{n\ge 2,\, p(n)\equiv\ell(mod\,k)}\frac{μ(n)}{n}=\frac{-1}{ϕ(k)}, $$ where $μ(n)$ is the Moebius function, $p(n)$ is the smallest prime factor of $n$, and $ϕ(k)$ 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 $ω(n)$, 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, $$ \sum_{n\ge 2, \, p(n)\equiv\ell(mod\,k)}\frac{μ(n)ω(n)}{n}=0. $$ A quantitative version of this result is proved.