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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?

  • 0
    [$\Gamma(n+1)=n!$](http://en.wikipedia.org/wiki/Gamma_function)2012-07-21
  • 0
    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.2012-07-21
  • 1
    The expressions involving the gamma function are actually quite clean.2012-07-21
  • 0
    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$2012-07-22

3 Answers 3