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Problem 646

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Bounded Divisors

Let $n$ be a natural number and $p_1^{\alpha_1}\cdot p_2^{\alpha_2}\cdots p_k^{\alpha_k}$ its prime factorisation.

Define the Liouville function $\lambda(n)$ as $\lambda(n)=(-1)^{\sum_{i=1}^k\alpha_i}$.
(i.e. $-1$ if the sum of the exponents $\alpha_i$ is odd and $1$ if the sum of the exponents is even. )
Let $S(n,L,H)$ be the sum $\lambda(d)\cdot d$ over all divisors $d$ of $n$ for which $L\le d\le H$.

You are given:
$S(10!,100,1000)=1457$
$S(15!,10^3,10^5)=-107974$
$S(30!,10^8,10^{12})=9766732243224$

Find $S(70!,10^{20},10^{60})$ and give your answer modulo $1\ 000\ 000\ 007$.

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有界因数

考虑自然数$n$及其质因数分解$p_1^{\alpha_1}\cdot p_2^{\alpha_2}\cdots p_k^{\alpha_k}$。

定义刘维尔函数 $\lambda(n)$为$\lambda(n)=(-1)^{\sum_{i=1}^k\alpha_i}$。
（也就是说，若所有指数$\alpha_i$之和为奇数，则该函数取$-1$，若为偶数则取$1$。）
考虑$n$在$L\le d\le H$范围内的所有因数$d$，并记这些$\lambda(d)\cdot d$之和为$S(n,L,H)$。

已知：
$S(10!,100,1000)=1457$
$S(15!,10^3,10^5)=-107974$
$S(30!,10^8,10^{12})=9766732243224$

求$S(70!,10^{20},10^{60})$，并将你的答案对$1\ 000\ 000\ 007$取余。

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