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Let $p_n$ be the $n$th prime, $p_n\#\equiv\prod_{k=1}^{n}p_k$ (primorial), and $\sigma(n)=\sum_{d|n}^{}d$ (divisor function).

Does $\text{exp}\bigg(\dfrac{\pi^2}{6 e^{\gamma}}\dfrac{\sigma(p_n\#)}{p_n\#}\bigg)$ bound $p_n$ from above?

(Note: From Daniel Fischer's answer here.)

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Update

... and from below by $\log(p_n\#)$?

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  • $\begingroup$ Is the constant $\frac{\pi^2}{6e^{\gamma}}$ experimental? (just asking). Also, is $\gamma$ supposed to be the Euler-Mascheroni constant? $\endgroup$ – chubakueno May 21 '14 at 15:09
  • $\begingroup$ No - it is from Daniel Fischer's answer here. And yes to the second :) $\endgroup$ – martin May 21 '14 at 15:11
  • $\begingroup$ Since $\sigma$ is multiplicative, $\sigma(p_n\#)/p_n\#$ is just an obscure way of writing $$\prod_{i=1}^n \left(1 + \frac{1}{p_i} \right) $$ $\endgroup$ – Excluded and Offended May 21 '14 at 22:20

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