Let $p_n$ be the nth prime and $p_L$ be closest to its square root: \begin{equation} p_L^2 \approx p_n \approx x \end{equation}

Let $\sigma \in Z^+$ be a positive integer constant. Define the average slope as \begin{equation} M_{n} = \prod_{\sigma < p_i \leq p_n} \left( \frac{p_i - \sigma}{p_i} \right) \end{equation}

Asymptotically the average slope becomes \begin{equation} M_{n} \sim \frac{K(\sigma)}{e^{\sigma \gamma} (\ln p_n)^\sigma} \quad n \rightarrow \infty \end{equation} where $\gamma$ is the Euler-Mascheroni Constant and \begin{equation} K(\sigma) = \sum_{p_i \leq \sigma} \frac{\sigma}{p_i} \end{equation}

Now define the square weighted slope as \begin{equation} S_{n} = \sum_{p_L < p_i \leq p_n} \frac{p_i^2 - p_{i-1}^2}{p_n^2 - p_L^2} M_{i-1} = \frac{(p_{L+1}^2 - p_{L}^2) M_{L} + \ldots + (p_{n}^2 - p_{n-1}^2) M_{n-1}}{p_n^2 - p_L^2} \end{equation}

Asympototically, \begin{equation} S_{n} \sim \frac{C(\sigma)}{(\ln p_n)^\sigma} \quad n \rightarrow \infty \end{equation} My question is: what is $C(\sigma)$? Is $C(\sigma) = K(\sigma)$? If $C(1) = K(1) = 1$, then I have an exciting new proof of the Prime Number Theorem.

More information can be found at: http://www.ugcs.caltech.edu/~kel/MPP/GammaSquares.pdf Thank you.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.