4
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Have you tried inducting on n? If you divide each side by the corresponding terms for n-1 things get much easier. In general $(a)_n/(a)_{n-1} = a-n+1$ $\endgroup$ – Nate Jan 29 '14 at 21:36
5
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ Great answer. Thanks. $\endgroup$ – glebovg Jan 30 '14 at 1:20
3
$\begingroup$

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

$\endgroup$
3
$\begingroup$

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}

$\endgroup$
  • $\begingroup$ Do you have a reference for this formula? $\endgroup$ – glebovg May 6 '14 at 23:31
  • 1
    $\begingroup$ 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. $\endgroup$ – Leucippus May 7 '14 at 2:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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