User kim e lumbard - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T07:09:31Z http://mathoverflow.net/feeds/user/28125 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/112475/limit-of-sequence-of-unusual-prime-product Limit of Sequence of unusual Prime Product Kim E Lumbard 2012-11-15T12:09:52Z 2012-11-15T12:09:52Z <p>Let $p_n$ be the nth prime and $p_L$ be closest to its square root: $$p_L^2 \approx p_n \approx x$$</p> <p>Let $\sigma \in Z^+$ be a positive integer constant. Define the average slope as $$M_{n} = \prod_{\sigma &lt; p_i \leq p_n} \left( \frac{p_i - \sigma}{p_i} \right)$$</p> <p>Asymptotically the average slope becomes $$M_{n} \sim \frac{K(\sigma)}{e^{\sigma \gamma} (\ln p_n)^\sigma} \quad n \rightarrow \infty$$ where $\gamma$ is the Euler-Mascheroni Constant and $$K(\sigma) = \sum_{p_i \leq \sigma} \frac{\sigma}{p_i}$$</p> <p>Now define the square weighted slope as $$S_{n} = \sum_{p_L &lt; p_i \leq p_n} \frac{p_i^2 - p_{i-1}^2}{p_n^2 - p_L^2} M_{i-1} = \frac{(p_{L+1}^2 - p_{L}^2) M_{L} + \ldots + (p_{n}^2 - p_{n-1}^2) M_{n-1}}{p_n^2 - p_L^2}$$</p> <p>Asympototically, $$S_{n} \sim \frac{C(\sigma)}{(\ln p_n)^\sigma} \quad n \rightarrow \infty$$ 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.</p> <p>More information can be found at: <a href="http://www.ugcs.caltech.edu/~kel/MPP/GammaSquares.pdf" rel="nofollow">http://www.ugcs.caltech.edu/~kel/MPP/GammaSquares.pdf</a> Thank you.</p>