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.


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)}. $

  • $\begingroup$ @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, $u(n)=nu(n-1)$. $\endgroup$ May 25 '13 at 17:48

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.