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 $n>2$.

  • What's $\gamma$? – Gerry Myerson Mar 9 '14 at 11:41
  • Are you still here, Manjil? Do you have any thoughts on the answer I posted? – Gerry Myerson Mar 11 '14 at 5:40
  • $\gamma$ is the Euler's function. Sorry I was off the loop for a while. – Manjil P. Saikia Mar 17 '14 at 14:05
  • Welcome back. So: any thoughts on the two answers that have been posted? – Gerry Myerson Mar 17 '14 at 22:39
  • This is the way I originally thought, but I don't know how does one get a formula involving $\pi$ and $\gamma$? – Manjil P. Saikia Mar 18 '14 at 10:44
up vote 3 down vote accepted

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 $(P_n+1)/2$.

  • Having worked a lot recently with $(2n)!!$, this looks right (+1) Of course, it would be nice to know what $\gamma$ is. – robjohn Mar 12 '14 at 20:07
  • @rob, I'm guessing it's the Gamma function. But only OP knows, and he/she is maintaining radio silence. – Gerry Myerson Mar 12 '14 at 22:13

Depending on what $\gamma$ is, here is an asymptotic approximation: $$ \begin{align} R\prod_{i=1}^nP_i &=P_n!!\\ &=\frac{(P_n+1)!}{2^{(P_n+1)/2}\left(\frac{P_n+1}{2}\right)!}\\ &\sim\frac1{\sqrt2}\left(\frac{P_n+1}{e}\right)^{\frac{P_i+1}{2}} \end{align} $$

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