Benedict Irwin edited section_Another_Result_begin_equation__.tex  almost 8 years ago

Commit id: 8432c5f0779d0dc0b56ae9567d4c207a2ce7dbc1

deletions | additions      

       

\sum_{q=1}^\infty \sum_{p|q} \frac{p}{q}x^q(1-x^q) = \log\left(\prod_{k=1}^\infty 1+x^{p_k} \right)  \end{equation}  then \begin{equation}  \sum_{q=?}^\infty \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.