It is well-known that
$$\frac{\phi(n)}{n}=\sum_{d\mid n}\frac{\mu (d)}{d} \quad n\in \mathbb Z^+$$
Where $\phi $ is Euler's totient function and $\mu$ is the Mobius function. But using the formula for $\phi$ we get
$$\prod_{p\mid n}\left(1-\frac{1}{p}\right)=\sum_{d\mid n}\frac{\mu (d)}{d}$$
And if we pick $n=p_1\dots p_k$ where $p_i$ is the ith prime. We get
$$\prod_{i=1}^k\left(1-\frac{1}{p_i}\right)=\sum_{d\mid n}\frac{\mu (d)}{d}$$
But now the divisors of $n$ are somewhat easier to handle. Let $D_=\{d\in \mathbb N| d\mid n\}$ and $$D_j=\{d\in D\mid d \text{ has exacltly $j$ prime factors.}\}$$
Hence our product becomes $$\prod_{i=1}^k\left(1-\frac{1}{p_i}\right)=\sum_{i=1}^k\sum_{d\in D_i}\frac{\mu(d)}{d}$$
Here I got stuck. I want to get an explicit formula for the RHS but the inner sum is hard to evaluate. How can I evaluate the following sum $\sum_{d\in D_i}\frac{\mu(d)}{d}$. I think binomial coefficients will pop out.
Best Answer
For each $\ell \in [0,k]$, $$\sum_{d \in D_\ell} \frac{\mu(d)}{d} = (-1)^\ell \sum_{i_1 \le \ldots \le i_\ell} \frac{1}{p_{i_1} \cdots p_{i_\ell}}.$$ When you add up these quantities, you just get the developed form of the product $$\prod_{i=1}^k \Big(1-\frac{1}{p_i}\Big).$$ As noted by Tuvasbien, there is no simple formula for the elementary symmetric polynomials of the inverses of the first $\ell$ primes numbers, otherwise it would be a formula giving the list of all prime numbers.