Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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 at 21:36

3 Answers 3

up vote 2 down vote accepted

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

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

share|improve this answer
    
Great answer. Thanks. –  glebovg Jan 30 at 1:20

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

share|improve this answer

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}

share|improve this answer
    
Do you have a reference for this formula? –  glebovg May 6 at 23:31
1  
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 at 2:17

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.