# An identity involving the Pochhammer symbol

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.

• Have you tried inducting on n? If you divide each side by the corresponding terms for n-1 things get much easier. In general $(a)_n/(a)_{n-1} = a-n+1$ – Nate Jan 29 '14 at 21:36

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 $$2^{3n}\cdot\prod_{k=0}^{n-1}(6k+1)(6k+3)(6k+5)=\dfrac{2^{3n}\cdot(6n)!}{\displaystyle\prod_{k=0}^{n-1}(6k+2)(6k+4)(6k+6)}=$$

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

• Great answer. Thanks. – glebovg Jan 30 '14 at 1:20

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

By using the formula \begin{align} (a)_{kn} = k^{kn} \prod_{r=0}^{n-1} \left( \frac{a+r}{k} \right)_{n} \end{align} it is evident that the desired quantity, \begin{align} (1728)^{n} \left( \frac{1}{6} \right)_{n} \left( \frac{3}{6} \right)_{n} \left( \frac{5}{6} \right)_{n}, \end{align} can be seen as \begin{align} 2^{6n} 3^{3n} \left( \frac{1}{6} \right)_{n} \left( \frac{3}{6} \right)_{n} \left( \frac{5}{6} \right)_{n} = \frac{ 6^{6n} \left( \frac{1}{6} \right)_{n} \left( \frac{2}{6} \right)_{n} \left( \frac{3}{6} \right)_{n} \left( \frac{4}{6} \right)_{n} \left( \frac{5}{6} \right)_{n} }{ 3^{3n} \left( \frac{1}{3} \right)_{n} \left( \frac{2}{3} \right)_{n} } = \frac{(1)_{6n}}{(1)_{3n}} = \frac{(6n)!}{(3n)!}. \end{align}

• Do you have a reference for this formula? – glebovg May 6 '14 at 23:31
• The books by E. D. Rainville "Special Functions" or H. M. Srivastava and L. Manocha "A Treatise on Generating Functions" have the generalized Pochammer product formula. – Leucippus May 7 '14 at 2:17