If $!!$ denotes the double factorial, then $$(2n-1)!!=\color{blue}{\prod_{k=1}^n (2k-1)=\frac{(2n)!}{2^n n!}}$$ for $n\ge 1$. This can be expressed in terms of the gamma function, namely $$\prod_{k=1}^n (2k-1)=\frac{2^n \Gamma \left(n+\frac{1}{2}\right)}{\sqrt{\pi}}.$$
It can be observed that $$\prod_{k=1}^n (3k-1)=\frac{3^n\Gamma \left(n+\frac{2}{3}\right)}{\Gamma \left(\frac{2}{3}\right)}.$$ But can $\prod_{k=1}^n (3k-1)$ be also expressed in terms of the factorial of natural numbers?
Mayble the triplication formula would be useful: $$\Gamma (3z)=\frac{3^{3z-1/2}}{2\pi}\Gamma (z)\Gamma \left(z+\frac{1}{3}\right)\Gamma \left(z+\frac{2}{3}\right)$$ (The duplication formula can be used to arrive at $\prod_{k=1}^n (2k-1)=\frac{(2n)!}{2^n n!}$ from the gamma representation).