How to prove the following result:
$$\prod_{i=1}^{n}P_i=\frac{2^{(P_n+3)/2}}{\sqrt{\pi}} \gamma (1+P_n/2) \cdot \frac1R$$
where $R$ is the product the odd composite natural numbers less than $P_n$ and $n>2$.
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$\begingroup$ What's $\gamma$? $\endgroup$ – Gerry Myerson Mar 9 '14 at 11:41
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$\begingroup$ Are you still here, Manjil? Do you have any thoughts on the answer I posted? $\endgroup$ – Gerry Myerson Mar 11 '14 at 5:40
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$\begingroup$ $\gamma$ is the Euler's function. Sorry I was off the loop for a while. $\endgroup$ – Manjil P. Saikia Mar 17 '14 at 14:05
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$\begingroup$ Welcome back. So: any thoughts on the two answers that have been posted? $\endgroup$ – Gerry Myerson Mar 17 '14 at 22:39
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$\begingroup$ This is the way I originally thought, but I don't know how does one get a formula involving $\pi$ and $\gamma$? $\endgroup$ – Manjil P. Saikia Mar 18 '14 at 10:44
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Note that $R\prod_1^nP_i$ is just twice the product of all the odd numbers up to $P_n$, and that product of odd numbers is $(P_n+1)!$ divided by the product of the even numbers up to $P_n+1$, and that product in turn is $2^{(P_n+1)/2}$ times the factorial of $(P_n+1)/2$.
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$\begingroup$ Having worked a lot recently with $(2n)!!$, this looks right (+1) Of course, it would be nice to know what $\gamma$ is. $\endgroup$ – robjohn♦ Mar 12 '14 at 20:07
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$\begingroup$ @rob, I'm guessing it's the Gamma function. But only OP knows, and he/she is maintaining radio silence. $\endgroup$ – Gerry Myerson Mar 12 '14 at 22:13
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Depending on what $\gamma$ is, here is an asymptotic approximation: $$ \begin{align} R\prod_{i=1}^nP_i &=P_n!!\\ &=\frac{(P_n+1)!}{2^{(P_n+1)/2}\left(\frac{P_n+1}{2}\right)!}\\ &\sim\frac1{\sqrt2}\left(\frac{P_n+1}{e}\right)^{\frac{P_i+1}{2}} \end{align} $$
