Prove $\sum_{d|n} \frac{\Phi(d)}{d} = \prod_{i=1}^r (1 + a_i - \frac{a_i}{p_i})$

Question

I want to prove $\sum_{d|n} \frac{\Phi(d)}{d} = \prod_{i=1}^r (1 + a_i(1 - \frac{1}{p_i}))$, where $\Phi(n)$ is the Euler phi function and given the prime factorisation $n = \prod_{i=1}^r p_i^{a_i} $.

My instinct says to use the Möbius inversion formula, but I am struggling to identify the functions $f(n)$ and $g(n)$ that will allow me to prove this. Any insight would be appreciated.

Answer 1

Assuming that your $\Phi$ is actually the Euler $\phi$ function, use the fact that, since $\phi$ is a multiplicative function, so is $\sum_{d | n}\dfrac{\phi(d)}{d}$.

Then you only need to prove it for $p^a$.

Answer 2

Observe that with your factorization we get for example

$$\sum_{d|n} d = \prod_{q=1}^r (1+p_q+p_q^2+\cdots+p_q^{a_q})$$

Now we have with the product ranging over prime divisors that

$$\frac{\varphi(d)}{d} = \prod_{p|d} \left(1-\frac{1}{p}\right).$$

Using the same scheme again we thus obtain

$$\sum_{d|n} \frac{\varphi(d)}{d} = \prod_{q=1}^r \left(1 + \left(1-\frac{1}{p_q}\right) + \left(1-\frac{1}{p_q}\right) + \cdots + \left(1-\frac{1}{p_q}\right) \right)$$

where the sum contains $a_q$ terms. This yields

$$\prod_{q=1}^r \left(1 + a_q \left(1-\frac{1}{p_q}\right)\right)$$

which is the claim. Here we use the fact that $\varphi(d)/d$ is multiplicative and so is $\alpha\star 1$ with $\alpha$ a multiplicative function as pointed out by M. Cohen.