Benedict Irwin edited section_Another_Result_begin_equation__.tex  almost 8 years ago

Commit id: 262aab4e9d830acc34cf255bea1e814bd21a15bd

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then \begin{equation}  \sum_{q=1}^\infty \sum_{p\in\mathbb{P}} \frac{1}{q}x^{qp}(1-x^{qp}) = \log\left(\prod_{k=1}^\infty 1+x^{p_k} \right)  \end{equation}  which appears to be true by looking at the expansions. If we do something of the opposite to the transform we get \begin{equation}  \sum_{p\in\mathbb{P}} x^p - x^{2p}  \end{equation}