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\end{equation}  Then if we set $x=1$, and change the infinite sums to $n$ (checked that the series are generated at the same rate) we should have have the prime counting function  \begin{equation}  \sum_{q\in\mathbb{P}}^n 1 = \log\left(\frac{\prod_{k=1}^n 2}{\exp\left(\sum_{q\notin\mathbb{P}}^n \frac{1}{q}\sum_{p|q} (-1)^{q+1}p^*(q)\right)}\right)\\  \pi(n) = \log\left(\frac{\prod_{k=1}^n 2}{\exp\left(\sum_{q\notin\mathbb{P}}^n \frac{1}{q}\sum_{p|q}  (-1)^{q+1}p^*(q)\right)}\right) \end{equation}