English

On the distance between factorials and repunits

Number Theory 2024-11-15 v1

Abstract

We show that if nn0n\ge n_0, b2b\ge 2 are integers, p7p\ge 7 is prime and n!(bp1)/(b1)0n!-(b^p-1)/(b-1)\ge 0, then n!(bp1)/(b1)0.5loglogn/logloglognn!-(b^p-1)/(b-1) \ge 0.5\log\log n/\log\log\log n. Further results are obtained, in particular for the case n!(bp1)/(b1)<0n!-(b^p-1)/(b-1) < 0.

Keywords

Cite

@article{arxiv.2411.09060,
  title  = {On the distance between factorials and repunits},
  author = {Michael Filaseta and Florian Luca},
  journal= {arXiv preprint arXiv:2411.09060},
  year   = {2024}
}
R2 v1 2026-06-28T19:59:14.087Z