triple

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令j=j′d,k=k′dj=j'd,k=k'd

∑i=1a∑dμ(d)∑j=1,i⊥j,d∣jb∑k=1,d∣kc1=∑i=1a∑dμ(d)∑j′=1,i⊥j′d[bd]∑k′=1,i⊥k′d[cd]1=∑i=1a∑d⊥iμ(d)∑j′=1,i⊥j′[bd]∑k′=1,i⊥k′[cd]1\begin{aligned} &\sum_{i=1}^a\sum_d\mu(d)\sum_{j=1,i\perp j,d|j}^b\sum_{k=1,d|k}^c1\\ =&\sum_{i=1}^a\sum_d\mu(d)\sum_{j'=1,i\perp j'd}^{[\frac bd]}\sum_{k'=1,i\perp k'd}^{[\frac cd]}1\\ =&\sum_{i=1}^a\sum_{d\perp i}\mu(d)\sum_{j'=1,i\perp j'}^{[\frac bd]}\sum_{k'=1,i\perp k'}^{[\frac cd]}1\\ \end{aligned}

令f(g,x)=∑i=1x[i⊥g]f(g,x)=\sum\limits_{i=1}^x[i\perp g]，也就是≤x\le x与gg互质的数的个数

则

∑i=1a∑d⊥iμ(d)∑j′=1,i⊥j′[bd]∑k′=1,i⊥k′[cd]1=∑i=1a∑d⊥iμ(d)f(i,[bd])f(i,[cd])\begin{aligned} &\sum_{i=1}^a\sum_{d\perp i}\mu(d)\sum_{j'=1,i\perp j'}^{[\frac bd]}\sum_{k'=1,i\perp k'}^{[\frac cd]}1\\ =&\sum_{i=1}^a\sum_{d\perp i}\mu(d)f(i,[\frac bd])f(i,[\frac cd]) \end{aligned}

然后对质因子容斥？