An identity involving the Pochhammer symbol

Question

I need help proving the following identity: $$\frac{(6n)!}{(3n)!} = 1728^n \left(\frac{1}{6}\right)_n \left(\frac{1}{2}\right)_n \left(\frac{5}{6}\right)_n.$$ Here, $$(a)_n = a(a + 1)(a + 2) \cdots (a + n - 1), \quad n > 1, \quad (a)_0 = 1,$$ is the Pochhammer symbol. I do not really know how one converts expressions involving factorials to products of the Pochhammer symbols. Is there a general procedure? Any help would be appreciated.

Answer 1

Pochhammer symbols (sometimes) indicate rising factorials, i.e., $n!=(1)_n$ . This is obviously the case here, since the left hand side is never negative, assuming natural n.

$$\bigg(\frac16\bigg)_n=\prod_{k=0}^{n-1}\bigg(\frac16+k\bigg)=\prod_{k=0}^{n-1}\bigg(\frac{6k+1}6\bigg)=6^{-n}\cdot\prod_{k=0}^{n-1}(6k+1)$$

$$\bigg(\frac12\bigg)_n=\prod_{k=0}^{n-1}\bigg(\frac12+k\bigg)=\prod_{k=0}^{n-1}\bigg(\frac{6k+3}6\bigg)=6^{-n}\cdot\prod_{k=0}^{n-1}(6k+3)$$

$$\bigg(\frac56\bigg)_n=\prod_{k=0}^{n-1}\bigg(\frac56+k\bigg)=\prod_{k=0}^{n-1}\bigg(\frac{6k+5}6\bigg)=6^{-n}\cdot\prod_{k=0}^{n-1}(6k+5)$$

Since $1728=12^3$, our product becomes $$2^{3n}\cdot\prod_{k=0}^{n-1}(6k+1)(6k+3)(6k+5)=\dfrac{2^{3n}\cdot(6n)!}{\displaystyle\prod_{k=0}^{n-1}(6k+2)(6k+4)(6k+6)}=$$

$$=\dfrac{2^{3n}\cdot(6n)!}{2^{3n}\cdot\displaystyle\prod_{k=0}^{n-1}(3k+1)(3k+2)(3k+3)}=\dfrac{(6n)!}{(3n)!}$$

Answer 2

$$1728^n \left(\frac{1}{6}\right)_n \left(\frac{1}{2}\right)_n \left(\frac{5}{6}\right)_n=24^n\prod_{k=0}^{n-1}{(6k+1)(2k+1)(6k+5)}=\\=\frac{4^n}{3^n n!}\prod_{k=0}^{n-1}{(6k+1)(6k+3)(6k+5)(6k+6)}=\frac{\prod_{k=0}^{n-1}{(6k+1)(6k+2)(6k+3)(6k+4)(6k+5)(6k+6)}}{\prod_{k=0}^{n-1}{(3k+1)(3k+2)(3k+3)}}=\frac{(6n)!}{(3n)!}$$