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There are some easy products that can be written in closed form in terms of factorials:

$ 2 \times 4 \times 6 \times ... 2n = n! \times 2^n$

$ 1 \times 3 \times 5 \times ... (2n-1) = {{(2n)!} \over {n! \times 2^n}}$

$ 3 \times 6 \times 9 \times ... 3n = n! \times 3^n$

But what about these?

$ f_2(n) = 2 \times 5 \times 8 \times ... (3n+2)$

$ f_1(n) = 1 \times 4 \times 7 \times ... (3n+1)$

Wolfram Alpha gives some expressions for partial products in terms of gamma functions, but is there any way to use factorials instead?

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$\Gamma(n+1)=n!$ –  draks ... Jul 21 '12 at 13:15
    
yes, I know. Read the Wolfram Alpha link. The gamma function expressions involve gamma(n+4/3) or gamma(n+5/3) which are not integer factorials. –  Jason S Jul 21 '12 at 13:18
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The expressions involving the gamma function are actually quite clean. –  lhf Jul 21 '12 at 14:36
    
If you aren't fond of gamma functions, you could use Pochhammer symbols instead: $f_1(n)=3^n \left(\dfrac43\right)_n$ and $f_2(n)=2\cdot 3^n \left(\dfrac53\right)_n$ –  J. M. Jul 22 '12 at 9:17
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3 Answers

To supplement lhf's supposition why an expression entirely in terms of factorials of integer argument might not be forthcoming, consider the Gauss multiplication formula for the factorial, suitably specialized:

$$(3n)!=\frac{3^{3n+\frac12}n!\left(n-\frac13\right)!\left(n-\frac23\right)!}{2\pi}$$

If there were some other independent relation between $\left(n-\frac13\right)!$ and $\left(n-\frac23\right)!$, we'd be in business, but since there doesn't seem to be any...

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The sequence $f_1(n)$ is tabulated at http://oeis.org/A007559. No simple closed form is given (because there surely isn't one), but a fairly nice asymptotic result is there. If that interests you, then I'm sure that $f_2(n)$ is tabulated at that site as well.

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In the spirit of the double factorial, where $n!!=n(n-2)(n-4)\ldots $ ending at $1$ or $2$, the same page under multifactorials suggests $n!!!$ and some other notations for what you want. But, like the double factorial, these are just notations. They are less common because the expressions are less common. You can define your third line as $f_3(n)=3 \times 6 \times 9 \times ... 3n = n! \times 3^n$ in which case you get $f_1(n)f_2(n)f_3(n)=(3n)!$ but that doesn't seem very useful.

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At least if you find a nice expression for $f_1$ then you'll find one for $f_2$ and vice-versa. But it seems unlikely. –  lhf Jul 21 '12 at 14:42
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