Relation connecting $(3n)!$, $3^n$ and $n!$

Question

Any idea on the relation for $(3n)!$ in terms of $3^n$ and $n!$ ? I have seen that there exist such relation for $(2n)!$ in terms of double factorials, ie. \begin{equation} (2n-1)!!= \frac{(2n)!}{2^n n!}. \end{equation}

Answer 1

Just duplicate the exact same idea.

$(3n)! = 1\cdot 2 \cdot ....... \dot 3n = $

$(3\cdot 6 \cdot ...... \cdot 3n)\times (4\cdot 7 \cdot ....... \cdot (3n-2)) \times (5\cdot 8\cdot..... \cdot (3n-1))=$

$3^n\times(1\cdot 2 \cdot ..... \cdot n) \times (3n-2)!!! (3n-1)!!!$

$3^n\cdot n!\cdot (3n-2)!!!\cdot (3n-1)!!!$

where $K\underbrace{!!!....!}_{m}$ is defined to be the product of natural numbers up to $K$ that have the same remainder as $K$ does when divided by $m$ (or in math jargon-- the product of all natural numbers less than or equal to $K$ that are in the same equivalence class modulo $m$ as $K$).

So $(3n-1)!!!(3n-2)!!! = \frac {(3n)!}{3^n\cdot n!}$

.....

Take it further:

$(kn)! = k^n\cdot n!\times (k(n-1) + 1)\underbrace{!!!...!}_k\times (k(n-1) + 1)\underbrace{!!!...!}_k.... \times (kn- 1)\underbrace{!!!...!}_k$

or

$(k(n-1) + 1)\underbrace{!!!...!}_k\times (k(n-1) + 1)\underbrace{!!!...!}_k.... \times (kn- 1)\underbrace{!!!...!}_k =\frac {(nk)!}{k^n\cdot n!}$.

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Thomas Andrews points out that we although $!!$ may (or may not?) be standard for $n!! = \prod_{m:m\equiv n\pmod 2;m\le n}m$, that $\underbrace{!!!...!}_k$ meaning $n\underbrace{!!!...!}_k=\prod_{m: m\equiv n\pmod k; n\le n}m$ is probably not.

I could write my final result, as he suggested in the comments, as:

$\frac {(kn)!}{k^n\cdot n!} = \prod_{j=1}^{k-1}(\prod_{m:m\equiv j\pmod k,m\le n} m)$

(Not sure that's a very useful result but....)