Mystical Summation that requires explaination

\[\sum _{ { \sigma }_{ 0 }\left( n \right) \ge 8 }^{ }{ \frac { 1 }{ { n }^{ 2 } } } =A\]

Find \(\displaystyle \left\lfloor 10000A \right\rfloor \).

Inspiration

Bonus: Find the exact value

You may use the following approximations

\(\displaystyle { \pi }^{ 2 }\approx 9.8696044\)

\(\displaystyle P\left( 2 \right) \approx 0.4522474\)

\(\displaystyle P\left( 4 \right) \approx 0.0769931\)

\(\displaystyle P\left( 6 \right) \approx 0.0170700\)

\(\displaystyle P\left( 8 \right) \approx 0.0040614\)

\(\displaystyle P\left( 10 \right) \approx 0.0009936\)

\(\displaystyle P\left( 12 \right) \approx 0.0002460\)

Notation:

\(\displaystyle { \sigma }_{ 0 }\left( n \right) \) is the divisor function, it counts the number of positive factors of a number.

\(\displaystyle P\left( x \right) \) is the prime zeta function or \( \displaystyle P\left( n \right) =\sum _{ p\quad prime }^{ }{ \frac { 1 }{ { p }^{ n } } } \)

\(\displaystyle \left\lfloor x \right\rfloor \) is the floor function.

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