When $\pi(n)$ does not divide $n$

TL;DR

This paper investigates the divisibility of integers by their prime-counting function and introduces conditions under which $ pi(n)$ divides $n$, providing new insights into the distribution of such integers.

Contribution

The paper establishes a threshold for $n$ where the divisibility of $n$ by $ pi(n)$ is characterized by a newly defined function $f(n)$ and explores the distribution of such integers within specific exponential intervals.

Findings

If $f(n)=0$ and $n extgreater=60184$, then $ pi(n)$ does not divide $n$.

If $n extgreater=60184$ and $ pi(n)$ divides $n$, then $f(n)=1$.

Within certain exponential intervals, there always exists an integer $n$ such that $ pi(n)$ divides $n$.

Abstract

Let $\pi(n)$ denote the prime-counting function and let $$f(n)=\left|\left\lfloor\log n-\lfloor\log n\rfloor-0.1\right\rfloor\right|\left\lfloor\frac{\left\lfloor n/\lfloor\log n-1\rfloor\right\rfloor\lfloor\log n-1\rfloor}{n}\right\rfloor\text{.}$$ In this paper we prove that if $n$ is an integer $\ge 60184$ and $f(n)=0$, then $\pi(n)$ does not divide $n$. We also show that if $n\ge 60184$ and $\pi(n)$ divides $n$, then $f(n)=1$. In addition, we prove that if $n\ge 60184$ and $n/\pi(n)$ is an integer, then $n$ is a multiple of $\lfloor\log n-1\rfloor$ located in the interval $[e^{\lfloor\log n-1\rfloor+1},e^{\lfloor\log n-1\rfloor+1.1}]$. This allows us to show that if $c$ is any fixed integer $\ge 12$, then in the interval $[e^c,e^{c+0.1}]$ there is always an integer $n$ such that $\pi(n)$ divides $n$. Let $S$ denote the sequence of integers generated by the function $d(n)=n/\pi(n)$ (where $n\in\mathbb{Z}$ and $n>1$) and let $S_k$ denote the $k$th term of sequence $S$. Here we ask the question whether there are infinitely many positive integers $k$ such that $S_k=S_{k+1}$.