查看原题'" class='mdui-btn mdui-btn-raised'>点击加载点击跳转 GCD-Extremetop: 0 设$g(n) = \sum_{i=1}^{n-1} \gcd(i,n)$$$\sum{i=1}^{n-1} \sum{j=i+1}^n \gcd(i,j)\\= \sum_{i=1}^n g(i)$$$$g(n) = \sum_{i=1}^{n-1} \gcd(i,n)\\= \sum{d|n}d \sum{i=1}^{n-1}[gcd(i,n)=d]\\= \sum{d|n}d \sum{i=1}^{\frac nd - 1}[gcd(i,n)=1]\\= \……