User kim e lumbard - MathOverflowmost recent 30 from http://mathoverflow.net2013-05-20T07:09:31Zhttp://mathoverflow.net/feeds/user/28125http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/112475/limit-of-sequence-of-unusual-prime-productLimit of Sequence of unusual Prime ProductKim E Lumbard2012-11-15T12:09:52Z2012-11-15T12:09:52Z
<p>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}</p>
<p>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}</p>
<p>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}</p>
<p>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}</p>
<p>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.</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>