Duality between prime factors and the Prime Number Theorem for Arithmetic Progressions -- II
Fuente:
arXiv
Saved in:
| Main Authors: | , |
|---|---|
| Format: | Preprint |
| Published: |
2024
|
| Subjects: | |
| Online Access: | |
| Tags: |
Add Tag
No Tags, Be the first to tag this record!
|
| _version_ | 1866917813619261440 |
|---|---|
| author | Alladi, Krishnaswami Johnson, Jason |
| author_facet | Alladi, Krishnaswami Johnson, Jason |
| contents | 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. |
| format | Preprint |
| id |
arxiv_https___arxiv_org_abs_2410_18259 |
| institution | arXiv |
| publishDate | 2024 |
| record_format | arxiv |
| spellingShingle | Duality between prime factors and the Prime Number Theorem for Arithmetic Progressions -- II Alladi, Krishnaswami Johnson, Jason Number Theory 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. |
| title | Duality between prime factors and the Prime Number Theorem for Arithmetic Progressions -- II |
| topic | Number Theory |
| url | https://arxiv.org/abs/2410.18259 |