RT
若
f(k)=∑i=1a∑i=1b[gcd(i,j)=k]\large f(k) = \sum\limits_{i=1}^a\sum\limits_{i=1}^b [\gcd(i,j)=k]f(k)=i=1∑ai=1∑b[gcd(i,j)=k]
则
∑n∣kf(k)=⌊an⌋⌊bn⌋\large\sum\limits_{n|k}f(k)=\lfloor\dfrac{a}{n}\rfloor \lfloor\dfrac{b}{n}\rfloorn∣k∑f(k)=⌊na⌋⌊nb⌋
