Is it possible to simplify the product $\prod_{i=1}^n (9i^2 - 1)$?

Question

Is is possible to reduce this product to something that only relates (max.) on the calculation of "$\Gamma(x)$ constants (e.g. $\Gamma(\frac{1}{3})$)" and (single) factorials.

I've started with Mathematicas calculated identity: $$\prod_{i=1}^n (9i^2 - 1) = \frac{\Gamma(\frac{2}{3} + n) \Gamma(\frac{4}{3} + n)}{2 π} * 3^{\frac{3}{2} + 2 n}$$

I can simplify $\Gamma(\frac{4}{3} + n) = \Gamma(\frac{1}{3} + (n+1))$ to $$ = \Gamma(\frac{1}{3}) * \frac{(3(n+1) - 2)!!!}{3^{n+1}} = \Gamma(\frac{1}{3}) * \frac{(3n + 1)!!!}{3^{n+1}}$$

Is it now possible to reduce $\Gamma(\frac{1}{3}) * \frac{(3n + 1)!!!}{3^{n+1}}$ and $\Gamma(\frac{2}{3} + n)$ to a 'representation' that is only with n-independent Gamma function-values and single factorials (not third)?

Thanks in advance! :)

Answer 1

Unless I'm being dense, we have \begin{align} \prod_{i=1}^n (9i^2-1) &= \prod_{i=1}^n [(3i-1)(3i+1)] = [2\cdot 4]\cdot[5\cdot 7] \cdots [(3n-1)(3n+1)] \\ &=\frac{(3n+1)!}{3\cdot 6 \cdots 3n} = \frac{(3n+1)!}{3^n n!}. \end{align}

Answer 2

$$\prod_{i=1}^n(3i-1)(3i+1)=\frac{(3n+1)!}{\prod_{i=1}^n 3i}=\frac{(3n+1)!}{n!3^n}.$$