journal6 ›› 2002, Vol. 23 ›› Issue (1): 1-3.

• Mathematics •     Next Articles

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


  1. (Department of Mathematics,Zhanjiang Normal College,Zhanjiang 524048,Guangdong China)
  • Online:2002-03-15 Published:2013-01-05

Abstract: 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.

Key words: aliquot part, sum of aliquot parts, Graham's problem

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