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I need the closed form of the sequence $(3n)\times (3n-1) \times \cdots \times 6 \times 5 \times 3 \times 2$?

The title is fairly self-explanatory. I know that a closed form of this exists using the Gamma function but I still have a lot of trouble converting unique sequences into Gamma form. I believe it will be something along the lines of $$\dfrac{3^n\Gamma(\text{something})}{\Gamma(\frac{2}{3})}$$ because I've seen stuff similar to this but could never quite manufacture the solution given the answer.

An explanation of how one goes about solving this would be more beneficial than just the answer, but I will certainly take what I can get.

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Consider $(3n-2) \times \cdots \times 4 \times 1$

This equals:

$3^n \cdot \frac{3n-2}{3} \times \cdots \times \frac{4}{3} \times \frac{1}{3}$

This equals:

$3^n \frac{\Gamma(\frac{3n+1}{3})}{\Gamma(\frac{1}{3})}$

because $\Gamma(t+1) = t \Gamma(t)$.

To explain this step further, using $\Gamma(t+1) = t\Gamma(t)$:

$\Gamma(\frac{4}{3}) = \frac{1}{3}\Gamma(\frac{1}{3})$

and

$\Gamma(\frac{7}{3}) = \frac{4}{3} \times \frac{1}{3}\Gamma(\frac{1}{3})$

and so on, until:

$\Gamma(\frac{3n+1}{3}) = \frac{3n-2}{3} \times \cdots \times \frac{4}{3}\times\frac{1}{3}\Gamma(\frac{1}{3})$.

You see, due to the formula, the left hand side is always one step ahead of the top term in the product. The $\Gamma$ function is a bit funny this way; recall that $\Gamma(n) = (n-1)!$, with the same one step ahead phenomenon.

Thus your original expression is:

$\frac{(3n)!\, \Gamma(\frac{1}{3})}{3^n\, \Gamma(\frac{3n+1}{3})}$

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  • $\begingroup$ Why did the $\frac{3n-2}{3}$ become $\Gamma\frac{3n+1}{3}$? Why did the numerator change?... $\endgroup$
    – Bark Jr.
    Dec 6, 2014 at 3:04
  • $\begingroup$ I've added a section explaining that step in greater detail. $\endgroup$
    – aes
    Dec 6, 2014 at 3:08
  • $\begingroup$ Is there a reason we turn the sequence of integers into fractions (by pulling out the $3^n$) in the first step? Is this necessary to make the Gamma function work correctly? $\endgroup$
    – Bark Jr.
    Dec 6, 2014 at 4:50
  • $\begingroup$ Yes. It's necessary that the distance between the terms be exactly one to make this work. This is due to the $+1$ in the formula $\Gamma(t+1) = t\,\Gamma(t)$. $\endgroup$
    – aes
    Dec 6, 2014 at 4:58
  • $\begingroup$ Because that keeps the originally intended distance between the integers, right? Because $\Gamma(3n+1) = (3n)\Gamma(3n) = (3n)(3n-1)\Gamma(3n-1)\cdots$ is not the intended result: this is simply $(3n+1)!$ Instead (because I want spacing of 3 integers), we use $\Gamma(\frac{3n+1}{3}) = \left(\frac{3n+1}{3}-1\right) \Gamma\left(\frac{3n+1}{3}-1\right) =\left(\frac{3n-2}{3}\right) \Gamma\left(\frac{3n-2}{3}\right) =\left(\frac{3n-2}{3}\right)\left(\frac{3n-5}{3}\right)\Gamma\left(\frac{3n-5}{3}\right) \ldots$ $\endgroup$
    – Bark Jr.
    Dec 6, 2014 at 5:18

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