[ans=sum_{i=1}^nsum_{j=1}^n(i+j)^kf(gcd(i,j))gcd(i,j)
]
[egin{aligned}
ans
& =sum_{i=1}^nsum_{j=1}^n(i+j)^kf(gcd(i,j))gcd(i,j)
\ & =sum_{t=1}^nf(t)t^{k+1}sum_{i=1}^{leftlfloordfrac{n}{t}
ight
floor}sum_{j=1}^{leftlfloordfrac{n}{t}
ight
floor}(i+j)^k[gcd(i,j)=1]
\ & =sum_{t=1}^nf(t)t^{k+1}sum_{i=1}^{leftlfloordfrac{n}{t}
ight
floor}sum_{j=1}^{leftlfloordfrac{n}{t}
ight
floor}(i+j)^ksum_{d|gcd(i,j)}mu(d)
\ & =sum_{t=1}^nf(t)t^{k+1}sum_{d=1}^{leftlfloordfrac{n}{t}
ight
floor}mu(d)d^ksum_{i=1}^{leftlfloordfrac{n}{dt}
ight
floor}sum_{j=1}^{leftlfloordfrac{n}{dt}
ight
floor}(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
[egin{aligned}
ans
& =sum_{t=1}^nf(t)t^{k+1}sum_{d=1}^{leftlfloordfrac{n}{t}
ight
floor}mu(d)sum(leftlfloordfrac{n}{dt}
ight
floor)
\ & =sum_{D=1}^nsum(leftlfloordfrac{n}{D}
ight
floor)sum_{t|D}f(t)t^{k+1}mu(frac{D}{t})(frac{D}{t})^k
\ & =sum_{D=1}^nsum(leftlfloordfrac{n}{D}
ight
floor)D^ksum_{t|D}f(t)tmu(frac{D}{t})
end{aligned}
]
我们又设
[g(D)=sum_{t|D}f(t)tmu(frac{D}{t})
]
易知,g为积性函数
所以
[ans=sum_{D=1}^nsum(leftlfloordfrac{n}{D}
ight
floor)D^kg(D)
]
啊，这
我们先来考虑sum该怎么求
[egin{aligned}
sum(n) & =sum_{i=1}^nsum_{j=1}^n(i+j)^k
\ & =sum_{d=2}^{2cdot n}d^k imes leftlfloordfrac{d}{2}
ight
floor imes 2
end{aligned}
]
这个可以递推？
然后我们看g
[g(n)=sum_{t|n}f(t)tmu(frac{n}{t})
]
线筛