Primes again 2

Well, I would say it's not actually a number theory problem. It's more of an algebra problem.

Let's just replace $S$ by $n$ .

So, we have to find $\displaystyle A = \sum_{\sigma_0(n) = 4}{\frac{1}{n^2}}$ where of course, $\sigma_0(n)$ is the divisor function of order $0$ .

Now, let $n = p_1^{\alpha_1}p_2^{\alpha_2}p_3^{\alpha_3}\cdots p_k^{\alpha_k}$ for primes $p_i$ .

When will $\sigma_0(n) = 4$ ? Well, it's quite easy to find $\sigma_0$ . One can use combinatorics or algebra(for general $\sigma_s(n)$ even) to find (left as an exercise for the reader):

$\displaystyle \sigma_0(n) = (\alpha_1 +1)(\alpha_2 + 1)\cdots (\alpha_k +1) = 4$

Either $(\alpha_1, \alpha_2) = (1,1)$ or $= (3,0)$ [Of course $p$ 's and $\alpha$ 's are inter-convertible]

As a result,

$\displaystyle A = \sum_{p_1 \neq p_2}{\frac{1}{p_1^2 p_2^2}} + \sum_{p_3}{\frac{1}{p_3^6}}$

$\displaystyle A = \frac{1}{2}\left(\sum_{p_1}{\sum_{p_2}{\frac{1}{p_1^2 p_2^2}}} - \sum_{p}{\frac{1}{p^4}}\right) + \sum_{p_3}{\frac{1}{p_3^6}}$ [Using inclusion-exclusion]

$\displaystyle A = \frac{1}{2}\left(P(2)^2 - P(4)\right) + P(6)$

@Julian Poon

It is a nice problem, but when you have provided the values for $P(2), P(4), P(6)$ isn't it easier than level 5?