I want to calculate the probability that "picking at random" two integers, they have at most $n$ prime factors in common. Reading this nice answer, I found out the probability to have exactly $ k $ prime factors in common, so I just need to sum on $ k $ from $1 $ to $n $
$ K(n)= \frac1{\zeta(2)} (1+ \sum_{k=1}^n \sum_{p_1<\cdots<p_k} \frac1{p_1^2\cdots p_k^2}) $
How can I write that summation as a sum on prime zeta function $ P(n) $? I mean if $ n=2 $ the sum is
$ \frac1{\zeta(2)} (1+ P(2) + \frac{1}{2}(P(2)^2-P(4))) $
but I don't understand clearly how to write it for a generic $n$.
On the other side it seems obvious to me that the leading them for a generic $k$ is of the form $ \frac {P(2)^k}{k!} $, while the other terms are negative, so if I calculate for big values of $n$ this sum
$ L(n) = \frac1{\zeta(2)} (1+ \sum_{k=1}^n \frac{P(2)^k}{k!}) $
I would expect a result $ > 1$ because $L(n)$ should exceed $ K(n)$, which, for bigger $n$, being the probability that "picking at random" two integers they have at most $n$ prime factors in common, should be $1$. But if I run a numerical simulation $L(n)= 0.95556...$, so what's the error in my reasoning? Maybe I should more correctly use the formula
$ K'(n)= \frac1{\zeta(2)} (1+ \sum_{k=1}^n \sum_{p_1<\cdots<p_k} \frac1{(p_1-1)^2\cdots (p_k-1)^2}) $
Which include the multiplicity of the prime factors, while $K(n)$ don't? If this is the case how can I write $K'(n)$ as a sum on prime zeta function $ P(n) $?
