关于正规约数和函数的Graham问题

On Graham's Problem Concerning the Sum of the Aliquot Parts

设n是大于 1且适合s(n) =[n/2 ]的正整数 ,其中s(n)是n的正规约数和函数 ;ω(n)是n的不同素因数的个数 ,p1,p2 ,… ,pω(n) 是n的适合p1<p2 <… <pω(n) 的素因数。本文证明了 :如果 2 |n ,则必有n =2 ;如果n为奇数且ω(n)≤ 2 ,则必有n =3α,其中α是任意的正整数 ;如果n为奇数且ω(n) =3,则必有p1=3或者p1=5,p2 =7以及 11≤p3≤ 31;如果n为奇数且ω(n) =4 ,则必有p1=3或者p1=5,7≤p2 ≤ 13,11≤p3≤ 17以及 13≤p4 ≤ 2 3,上述结果部分地解决了Graham猜想。

Let n be a positive integer satisfying n>1 and s(n)=[n/2] , where s(n) is the sum of the aliquot parts of n . Further let ω(n) denote the number of distinct prime factors of n and p 1,p 2,…,p ω(n) denote its prime factors with p 1<p 2<…<p ω(n) .In this paper we prove that if 2｜n ,then n=2 , if n is odd and ω(n)≤2 ,then n is a power of 3,if n is odd and ω(n)=3 ,then p 1=3 or p 1=5,p 2=7 and 11≤p 3≤31 , if n is odd and ω(n)=4 ,then p 1=3 or p 1=5,7≤p 2≤13,11≤p 3≤17 and 13≤p 4≤23 . The above-mentioned results partly solve a problem posed by Graham.