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Benedict Irwin edited untitled.tex
almost 8 years ago
Commit id: ae9e585a46e0cfc34c1c8f47d4a771fb718ec4f5
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Where the right hand side is now the ordinary generating function for primes. Then of course we also have \begin{equation}
\log\left(\frac{\prod_{k=1}^\infty 1+x^{p_k}}{\exp\left(\sum_{q\in\mathbb{P}}^\infty x^q\right)}\right)=\sum_{q\notin\mathbb{P}}^\infty \frac{1}{q}\sum_{p|q} (-1)^{q+1}p^*(q)x^q
\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)
\end{equation}