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p^*(q)=(-1)^{q/2}\cdot 2,\; p=2\\  p^*(q)=p,\; \mathrm{otherwise}  \end{equation}  We may now manipulate this expression \begin{equation}  \prod_{k=1}^\infty 1+x^{p_k}=\exp\left(\left[\sum_{q\in\mathbb{P}}^\infty \frac{1}{q}\sum_{p|q} (-1)^{q+1}p^*(q)x^q\right] + \left[\sum_{q\notin\mathbb{P}}^\infty \frac{1}{q}\sum_{p|q} (-1)^{q+1}p^*(q)x^q\right]\right) \\  \prod_{k=1}^\infty 1+x^{p_k}=\exp\left(\sum_{q\in\mathbb{P}}^\infty \frac{1}{q}\sum_{p|q} (-1)^{q+1}p^*(q)x^q\right)\exp\left(\sum_{q\notin\mathbb{P}}^\infty \frac{1}{q}\sum_{p|q} (-1)^{q+1}p^*(q)x^q\right)\\  \frac{\prod_{k=1}^\infty 1+x^{p_k}}{\exp\left(\sum_{q\notin\mathbb{P}}^\infty \frac{1}{q}\sum_{p|q} (-1)^{q+1}p^*(q)x^q\right)}\exp\left(\sum_{q\in\mathbb{P}}^\infty \frac{1}{q}\sum_{p|q} (-1)^{q+1}p^*(q)x^q\right)\\  \end{equation}