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SUR LES PLUS GRANDS FACTEURS PREMIERSD’ENTIERS CONSÉCUTIFS

Published online by Cambridge University Press:  05 April 2018

Zhiwei Wang (Nancy)*
Affiliation:
Institut Élie Cartan de Lorraine, Université de Lorraine, UMR 7502, 54506 Vandœuvre-lès-Nancy, France email zhiwei.wang@univ-lorraine.fr

Abstract

Let $P^{+}(n)$ denote the largest prime factor of the integer $n$ and $P_{y}^{+}(n)$ denote the largest prime factor $p$ of $n$ which satisfies $p\leqslant y$ . In this paper, first we show that the triple consecutive integers with the two patterns $P^{+}(n-1)>P^{+}(n)<P^{+}(n+1)$ and $P^{+}(n-1)<P^{+}(n)>P^{+}(n+1)$ have a positive proportion respectively. More generally, with the same methods we can prove that for any$J\in \mathbb{Z}$ , $J\geqslant 3$ , the $J$ -tuple consecutive integers with the two patterns $P^{+}(n+j_{0})=\min _{0\leqslant j\leqslant J-1}P^{+}(n+j)$ and $P^{+}(n+j_{0})=\max _{0\leqslant j\leqslant J-1}P^{+}(n+j)$ also have a positive proportion, respectively. Second, for $y=x^{\unicode[STIX]{x1D703}}$ with $0<\unicode[STIX]{x1D703}\leqslant 1$ we show that there exists a positive proportion of integers $n$ such that $P_{y}^{+}(n)<P_{y}^{+}(n+1)$ . Specifically, we can prove that the proportion of integers $n$ such that $P^{+}(n)<P^{+}(n+1)$ is larger than 0.1356, which improves the previous result “0.1063” of the author.

MSC classification

Information

Type
Research Article
Copyright
Copyright © University College London 2018 
Figure 0

Figure 1 Graphe de $\unicode[STIX]{x1D6FF}\rightarrow f(\unicode[STIX]{x1D6FF})$ sur $[1/3,1/2]$.