English

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

Number Theory 2015-01-27 v4

Abstract

Let π(n)\pi(n) denote the prime-counting function and let f(n)=lognlogn0.1n/logn1logn1n.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 nn is an integer 60184\ge 60184 and f(n)=0f(n)=0, then π(n)\pi(n) does not divide nn. We also show that if n60184n\ge 60184 and π(n)\pi(n) divides nn, then f(n)=1f(n)=1. In addition, we prove that if n60184n\ge 60184 and n/π(n)n/\pi(n) is an integer, then nn is a multiple of logn1\lfloor\log n-1\rfloor located in the interval [elogn1+1,elogn1+1.1][e^{\lfloor\log n-1\rfloor+1},e^{\lfloor\log n-1\rfloor+1.1}]. This allows us to show that if cc is any fixed integer 12\ge 12, then in the interval [ec,ec+0.1][e^c,e^{c+0.1}] there is always an integer nn such that π(n)\pi(n) divides nn. Let SS denote the sequence of integers generated by the function d(n)=n/π(n)d(n)=n/\pi(n) (where nZn\in\mathbb{Z} and n>1n>1) and let SkS_k denote the kkth term of sequence SS. Here we ask the question whether there are infinitely many positive integers kk such that Sk=Sk+1S_k=S_{k+1}.

Cite

@article{arxiv.1409.2703,
  title  = {When $\pi(n)$ does not divide $n$},
  author = {Germán Paz},
  journal= {arXiv preprint arXiv:1409.2703},
  year   = {2015}
}

Comments

10 pages; some results and a question added

R2 v1 2026-06-22T05:52:21.392Z