[sum^n_{i=1}lfloorfrac{n}{i}
floor f(i)=sum^n_{i=1}I*f(i)
]
证明:
设$$F(x)=sum^x_{i=1}lfloorfrac{x}{i}
floor f(i)$$
则
[F(x)-F(x-1)=sum^x_{i=1}lfloorfrac{x}{i}
floor f(i)-sum^{x-1}_{i=1}lfloorfrac{x-1}{i}
floor f(i)
]
[=sum^{x-1}_{i=1}(lfloorfrac{x}{i}
floor-lfloorfrac{x-1}{i}
floor)f(i)+f(x)
]
[ ext{又因为}lfloorfrac{x}{i}
floor-lfloorfrac{x-1}{i}
floor=1 ext{当且仅当}i | x
]
所以
[F(x)-F(x-1)=sum_{d|x}f(d)=I*f(x)
]
[=>F(x)=F(1)+sum_{i=2}^xI*f(i)=sum^x_{i=1}I*f(i)
]
即
[sum^n_{i=1}lfloorfrac{n}{i}
floor f(i)=sum^n_{i=1}I*f(i)
]
其中(I)为恒等函数(I(x)=1)
应用举例:
[1.sum^n_{i=1}sum_{j=1}^nmu(i)phi(j)lfloorfrac{n}{ij}
floor=sum_{T=1}^nlfloorfrac{n}{T}
floormu*phi(T)=sum_{i=1}^nphi(i)
]
[2.sum^n_{i=1}lfloorfrac{n}{i}
floorphi(i)=sum_{i=1}^ni=n*(n+1)/2
]
还有很多式子可以用其化简(基本上可以和(I)卷的都可以)
实际上，这就是杜教筛式子特殊化的逆运算
杜教筛式子：$$sum_{i=1}nf(i)sum_{j=1}{lfloorfrac{n}{i} floor}g(i)=sum_{i=1}^nf*g(i)$$
若把(g(i))改为(I(i))则可得上述式子
杜教筛式子在此就不证明，很好证的