So I've been trying to work out if it is possible to write:

$\large \Pi_{i=1}^n (3i-1)$ as an expression involving the quotient or product of two factorials, or really any expression involving factorials that isn't something like $\large (\Pi_{i=1}^n (3i-1))\frac{n!}{n!}$.

I actually started with another expression, namely $\large \Pi_{i=1}^n (2i-1)$

I managed to work with this expression by noticing we can consider $(2i-1)!$ and then divide by $(2^{n-1})(n-1)!$ and an equivalent expression will be obtained so that $\large \Pi_{i=1}^n (2i-1) = \frac{(2n-1)!}{2^{n-1}(n-1)!}$

When attempting to reason similarly with $\large \Pi_{i=1}^n (3i-1)$ I did not seem to get anywhere, which made me beg the question, is it never possible to do this for expressions of the form $\large \Pi_{i=1}^n (ki-1)$ where $k$ is an integer larger than $2$ ?

  • $\begingroup$ One clarification about your claimed identity. Take $n=6$. Mathematica gives $\prod_{i=1}^n (2i)-1=46079$, while $\frac{(2n-1)!}{2(n-1)!}=166320$... $\endgroup$ – Pierpaolo Vivo Jan 9 '16 at 11:11
  • $\begingroup$ I think he means it to be the product of $(2i-1)$ $\endgroup$ – πr8 Jan 9 '16 at 11:12
  • $\begingroup$ It doesn't work either, does it? $\prod_{i=1}^n (2i-1)=10395$, for $n=6$... $\endgroup$ – Pierpaolo Vivo Jan 9 '16 at 11:15
  • $\begingroup$ You are absolutely correct. I will make an edit to my post to amend this. $\endgroup$ – porridgemathematics Jan 9 '16 at 11:30

Short answer is no, there's nothing quite so concise involving factorials of integers. There are some other tools which allow you to write these expressions in shorthand - for example, your product would be $3^n(2/3)_n$ in the Pochhammer notation.

There are some combinations of the type of product that you mention which can be written in the factorial form. Consider:





and how you could "fill in the gaps" to get an expression involving pure factorials. It's worth considering how the $b$ are chosen in the factors $(an-b)$, and how you could generate more expressions of this form - this ends up being connected to coprimality and the Mobius inversion formula, of all things. Fun to play with.

For reference: if we define

$$U_n(x):=\prod_{0<a<n, (a,n)=1}(nx-a)$$


$$\prod_{x=1}^kU_n(x)=\prod_{d\vert n}[(dk)!(\frac{n}{d})^{dk}]^{\mu(\frac{n}{d})}$$

Furthermore, if you have any interest in the asymptotics of your product, we can compare it to $3^nn!$ quite easily, viz:


If you write the terms of the remaining product as $(1-\frac{1}{3i})=[(1-\frac{1}{3i})e^{1/{3i}}]e^{-1/3i}$, we see that the product is equal to:


We now note the following:

  • The term in the exponential outside the product is the $n^{th}$ Harmonic number $H_n$, which satisfies $H_n=\log(n)+\gamma+o(1)$
    • The term inside the product is $(1-\frac{1}{18i^2}+o(\frac{1}{i^2}))$, which means that the product will converge to a nonzero constant (compare with convergence of $\sum\frac{1}{i^2}$ to convince yourself of this). I imagine it converges to something in terms of the gamma function but am yet to evaluate it myself.
    • The leading factorial can be asymptotically approximated by Stirling's approximation, $n!=\sqrt{2\pi}n^{n+1/2}e^{-n}(1+o(1))$

So, the product can be asymptotically approximated as:



where $C$ is a constant.

  • $\begingroup$ Great answer, very detailed! $\endgroup$ – porridgemathematics Jan 9 '16 at 11:35
  • $\begingroup$ You inspired me ! Thanks for the fun. $\endgroup$ – Claude Leibovici Jan 9 '16 at 15:22

This is not an answer but it is too long for a comment.

Inspired by πr8's answer, I looked at the more general problem of $$P_n=\prod_{i=1}^n(a\,i+b)=a^n \left(\frac{a+b}{a}\right)_n$$ using Pochhammer notation. Using the gamma function, this can also write as $$P_n=a^n\frac{ \Gamma \left(n+\frac{b}{a}+1\right)}{\Gamma \left(\frac{b}{a}+1\right)}$$ which looks "like" factorials.

Taking logarithms and using Stirling approximation for the gamma function, the asymptotics write $$\log(P_n)=n \big(\log (n)+\log (a)-1\big)+\left(\frac{b}{a}+\frac{1}{2}\right) \log (n)+\log \left(\frac{\sqrt{2 \pi }}{\Gamma \left(\frac{a+b}{a}\right)}\right)+$$ $$\left(\frac{a^2+6 a b+6 b^2}{12 a^2 }\right)\frac 1n+O\left(\frac{1}{n^{3/2}}\right)$$ which, at least to me, looks quite nice.


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