設(shè)a所有約數(shù)和為F(a)
首先對互質(zhì)的兩個數(shù)a,b,F(ab)=F(a)*F(b)
令a約數(shù)1=x1,x2.xm
b約數(shù)1=y1,y2,...yn
由a,b互質(zhì),a2->am與b2->bn無公共元素
那么
F(ab)=
sigma(1<=i<=m,1<=j<=n)xiyj
=(sigma(1<=i<=m)xi)*(sigma(1<=j<=n)yj)
=F(a)*F(b)
而
F(pi^qi)=1+pi+pi^2+...+pi^qi=(pi^(qi+1)-1)/(pi-1)
又pi^qi與pj^qj互質(zhì)
所以
F(n)=∏(1<=i<=k)[pi^(qi+1)-1)/(pi-1)]
得證