题解 P6156 【简单题】
\[ans=\sum_{i=1}^n\sum_{j=1}^n(i+j)^kf(gcd(i,j))gcd(i,j)
\]
\[\begin{aligned}
ans
& =\sum_{i=1}^n\sum_{j=1}^n(i+j)^kf(gcd(i,j))gcd(i,j)
\\ & =\sum_{t=1}^nf(t)t^{k+1}\sum_{i=1}^{\left\lfloor\dfrac{n}{t}\right\rfloor}\sum_{j=1}^{\left\lfloor\dfrac{n}{t}\right\rfloor}(i+j)^k[gcd(i,j)=1]
\\ & =\sum_{t=1}^nf(t)t^{k+1}\sum_{i=1}^{\left\lfloor\dfrac{n}{t}\right\rfloor}\sum_{j=1}^{\left\lfloor\dfrac{n}{t}\right\rfloor}(i+j)^k\sum_{d|gcd(i,j)}\mu(d)
\\ & =\sum_{t=1}^nf(t)t^{k+1}\sum_{d=1}^{\left\lfloor\dfrac{n}{t}\right\rfloor}\mu(d)d^k\sum_{i=1}^{\left\lfloor\dfrac{n}{dt}\right\rfloor}\sum_{j=1}^{\left\lfloor\dfrac{n}{dt}\right\rfloor}(i+j)^k
\end{aligned}
\]
我们现在设
\[sum(D)=\sum_{i=1}^{D}\sum_{j=1}^D(i+j)^k
\]
所以
\[\begin{aligned}
ans
& =\sum_{t=1}^nf(t)t^{k+1}\sum_{d=1}^{\left\lfloor\dfrac{n}{t}\right\rfloor}\mu(d)sum(\left\lfloor\dfrac{n}{dt}\right\rfloor)
\\ & =\sum_{D=1}^nsum(\left\lfloor\dfrac{n}{D}\right\rfloor)\sum_{t|D}f(t)t^{k+1}\mu(\frac{D}{t})(\frac{D}{t})^k
\\ & =\sum_{D=1}^nsum(\left\lfloor\dfrac{n}{D}\right\rfloor)D^k\sum_{t|D}f(t)t\mu(\frac{D}{t})
\end{aligned}
\]
我们又设
\[g(D)=\sum_{t|D}f(t)t\mu(\frac{D}{t})
\]
易知,g为积性函数
所以
\[ans=\sum_{D=1}^nsum(\left\lfloor\dfrac{n}{D}\right\rfloor)D^kg(D)
\]
啊，这
我们先来考虑sum该怎么求
\[\begin{aligned}
sum(n) & =\sum_{i=1}^n\sum_{j=1}^n(i+j)^k
\\ & =\sum_{d=2}^{2\cdot n}d^k\times \left\lfloor\dfrac{d}{2}\right\rfloor\times 2
\end{aligned}
\]
这个可以递推？
然后我们看g
\[g(n)=\sum_{t|n}f(t)t\mu(\frac{n}{t})
\]
线筛
