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

Abstract

Let $\pi (n)$ denote the prime-counting function and let $$f(n)=\left|\lfloor \log n-\lfloor \log n\rfloor -0.1\rfloor \right|\lfloor \frac{\lfloor n/\lfloor \log n-1\rfloor \rfloor \lfloor \log n-1\rfloor }{n}\rfloor \text{.}$$ 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 $S_k=S_{k+1}$.

10 pages; some results and a question added

References in corpus (2)

• Estimates of Some Functions Over Primes without R.H.
• An explicit formula for the prime counting function