关于除数上函数和式的变换方法咨询

Hey there! Great question—let's break this down in a straightforward way. First, let's simplify the sum you've presented with a quick substitution to make it easier to work with.

Your original sum is:
$$\sum_{d|n} \frac{n/d}{\varphi\left(n/d\right)}$$

If we let $k = n/d$, notice that as $d$ runs over all divisors of $n$, $k$ will also run over all divisors of $n$. So we can rewrite the sum as:
$$\sum_{k|n} \frac{k}{\varphi(k)}$$

That's a cleaner starting point. Now, let's explore two useful ways to re-express this sum:

1. Using the multiplicative property of the function #

First, note that $\frac{k}{\varphi(k)}$ is a multiplicative function (since Euler's totient function $\varphi$ is multiplicative, and the ratio of multiplicative functions is also multiplicative). When dealing with sums over divisors of multiplicative functions, the resulting sum function is also multiplicative. That means we can compute the sum for prime powers first, then multiply the results for each prime power in $n$'s factorization.

Suppose $n = p^m$ where $p$ is prime and $m \geq 1$. Let's compute the sum:
$$\sum_{i=0}^m \frac{p^i}{\varphi(pi)}$$

• For $i=0$: $\frac{1}{\varphi(1)} = 1$ (since $\varphi(1)=1$)
• For $i \geq 1$: $\varphi(p^i) = p^i - p^{i-1} = p^{i-1}(p-1)$, so $\frac{p^i}{\varphi(pi)} = \frac{p}{p-1}$

So the sum for $p^m$ becomes:
$$1 + m \cdot \frac{p}{p-1}$$

For a general $n$ with prime factorization $n = \prod_{p|n} p^{m_p}$, since the sum function is multiplicative, we can write:
$$\sum_{k|n} \frac{k}{\varphi(k)} = \prod_{p|n} \left(1 + m_p \cdot \frac{p}{p-1}\right)$$

This is a nice closed-form expression that's easy to compute once you have $n$'s prime factors.

2. Möbius inversion perspective #

If we let $f(n) = \sum_{k|n} \frac{k}{\varphi(k)}$, we can use Möbius inversion to relate $f(n)$ back to $\frac{n}{\varphi(n)}$. By the Möbius inversion formula, since $f(n)$ is the sum of $\frac{k}{\varphi(k)}$ over divisors $k$ of $n$, we have:
$$\frac{n}{\varphi(n)} = \sum_{k|n} \mu(k) f\left(\frac{n}{k}\right)$$
Where $\mu$ is the Möbius function. While this isn't a direct simplification of your original sum, it gives a useful relationship if you need to work backwards or connect this sum to other number-theoretic functions.

Hope this gives you the insights you were looking for! Feel free to ask if you want to unpack any of these steps further.

备注：内容来源于stack exchange，提问作者AmB