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

TL;DR

This paper extends duality identities involving prime factors and the Moebius function, deriving new results related to the Prime Number Theorem for Arithmetic Progressions, including a zero-sum formula involving the number of prime factors.

Contribution

It introduces a new duality identity between the second largest and smallest prime factors, leading to a novel zero-sum result connected to prime number distribution.

Findings

Established a sum involving the Moebius function and prime factors equals zero.

Proved a quantitative version of the zero-sum result.

Extended duality identities to higher prime factors.

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 $$ \sum_{n\ge 2,\, p(n)\equiv\ell(mod\,k)}\frac{\mu(n)}{n}=\frac{-1}{\phi(k)}, $$ 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.