Mystical Summation that requires explaination

The numbers $n$ with $\sigma_0(n) \le 7$ are either $1$ , or else prime powers $p^j$ for $1 \le j \le 6$ , or else either of the form $pq$ or of the form $p^2q$ for distinct primes $p$ , $q$ . Thus $\begin{array}{rcl} \displaystyle \zeta(2) - A & = & \displaystyle \sum_{\sigma_0(n) \le 7} \frac{1}{n^2} \; = \; 1 + \sum_{j=1}^6 \sum_{p} \frac{1}{p^{2j}} + \sum_{p \neq q} \frac{1}{p^2q^2} + \sum_{p \neq q} \frac{1}{p^4q^2} \\ & = & \displaystyle 1 + \sum_{j=1}^6 P(2j) + \tfrac12\big[P(2)^2 - P(4)\big] + \big[P(2)P(4) - P(6)\big] \end{array}$ making $A$ equal to $0.0118052...$ , and the answer $\boxed{118}$ .