关于正规约数和函数的Graham问题

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)的素因素.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=3a,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≤23，上述结果部分地解决了Graham猜想.

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

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 ε(n) denote the number of dirts of n.Further let ε(n) denote the number of distinct prime factors of n and p1，p2，…，pω(n) denote its prime factors with p1＜p2＜…＜pω(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 p1=3 or p1=5,p2=7 and 11≤p3≤31；if n is odd and ω(n)=4，then p1=3 or p1=5,7≤p2≤13,11≤p3≤17 and 13≤p4≤23.The above-mentioned results partly solve a problem posed by Graham.