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 \(\begin{aligned} \sum _{n\ge 2,\, p_1(n)\equiv \ell (mod\,k)}\frac{\mu (n)}{n}=\frac{-1}{\phi (k)}, \end{aligned}\) where \(\mu (n)\) is the Moebius function, \(p_1(n)\) 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, \(\begin{aligned} \sum _{n\ge 2, \, p_1(n)\equiv \ell (mod\,k)}\frac{\mu (n)\omega (n)}{n}=0. \end{aligned}\) A quantitative version of this result is proved.
