关于正规约数和函数的Graham问题

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

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＝３ａ，其中ａ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。

Let n be a positive integer satisfying n>1 and s(n)=,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 orP 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.