# Identity involving a recursive product

Here is yet another problem related to plane partitions. Given the recursive formula

\begin{align*} F(0)&=1,\\ F(r)&=\prod_{i=1}^r\frac{i+2r-1}{2i+r-2}F(r-1). \end{align*}

How can we prove

$$F(n)=\prod_{1\leq i\leq j\leq k\leq n}\frac{i+j+k-1}{i+j+k-2}\ ?$$

EDIT: The solution to this problem can be found in the answer section to this question.

First of all, note that, you can write the product as

$$\prod_{i=1}^r\frac{i+2r-1}{2i+r-2}= 2\,{\frac {\Gamma \left( 3\,r \right) \Gamma \left( \frac{r}{2} \right) } {{2}^{r+1}\Gamma \left( \frac{3r}{2} \right) \Gamma \left( 2\,r \right) }},$$

where $\Gamma(x)$ is the gamma function. Second, you have a first order recurrence relation

$$F(r) = g(r)F(r-1)$$

which can be solved using the formula.

Notes:

1) $\prod_{i=1}^r i =r!=\Gamma(r+1).$

2) $\prod_{i=1}^r \frac{f(i)}{g(i)}= \frac{\prod_{i=1}^rf(i)}{\prod_{i=1}^rg(i)}.$

• @CarlNajafi: I recommend to take a simple first order recurrence relation and try to apply the formula to it, so you can see what's going on. It is a good exercise. For instance, $u(n)=nu(n-1)$. May 25 '13 at 17:48