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Are there any useful forms for the expression $1!\cdot 2!\cdot 3!\cdot ...\cdot n!$? I'm trying to solve a problem that involves this expression and thought it might help to find a more "workable" form for it, but I didn't get very far.

Thanks

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  • $\begingroup$ What form is useful depends on context. What would make a formula useful for you? $\endgroup$ Dec 10 '14 at 23:04
  • $\begingroup$ See hyperfactorial. $\endgroup$
    – Lucian
    Dec 10 '14 at 23:18
  • $\begingroup$ There is a closed form in terms of $\psi^{(-2)}(x)$, I think, but you probably haven't heard of that yet. $\endgroup$ Dec 11 '14 at 23:55
  • $\begingroup$ (Actually, that might not be true. I'll need to look in to it more.) $\endgroup$ Dec 12 '14 at 0:19
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In the May 2013 Fibonacci Quarterly (Vol. 51, Num. 2) pages 163-173, Michael Hirschhorn proved this result:

Let $P(n) =\prod\limits_{k=0}^n \binom{n}{k} $. Then (this is going to be a pain to enter), as $n \to \infty$,

$$P(n) \sim C^{-1}\dfrac{e^{n(n+2)/2}}{n^{(3n+2))/6}(2\pi)^{(2n+1)/4}} \exp\left(-\sum\limits_{p \ge 1}\dfrac{B_{p+1}+B_{p+2}}{p(p+1)}\dfrac1{n^p}\right) $$ where $$C =\lim\limits_{n \to \infty} \dfrac1{n^{1/12}}\prod\limits_{k=1}^n\left( \dfrac{k!}{\sqrt{2\pi k}(k/e)^k} \right) \approx 1.046335066770503 $$ and the $\{B_p\}$ are the Bernoulli numbers, defined by $$ \sum\limits_{p \ge 0} B_p \dfrac{x^p}{p!} = \dfrac{x}{e^x-1} .$$

I will now try to use this to get an estimate for $f(n) =\prod\limits_{k=0}^{n} n! $.

Since $P(n) =\dfrac{(n!)^{n+1}}{\prod\limits_{k=0}^{n} (n!)^2} $, $f(n) =\prod\limits_{k=0}^{n} n! =\sqrt{\dfrac{(n!)^{n+1}}{P(n)}} $.

I will now try to get an estimate for $f(n)$ from the first term ($p=1$) in the asymptotic formula for $P(n)$..

Since $B_2=\frac16$ and $B_3=0$, $B_2+B_3=\frac16$, so, setting $p=1$ in the formula for $P(n)$,

$\begin{array}\\ P(n) &\sim C^{-1}\dfrac{e^{n(n+2)/2}}{n^{(3n+2))/6}(2\pi)^{(2n+1)/4}} e^{-1/(12n)}\\ &\sim C^{-1}\dfrac{e^{n(n+2)/2}}{n^{(3n+2))/6}(2\pi)^{(2n+1)/4}} \\ \end{array} $

since that $e^{1/(12n)}$ term goes to $1$.

Therefore $\dfrac1{\sqrt{P(n)}} \sim C^{1/2}\dfrac{n^{(3n+2))/12}(2\pi)^{(2n+1)/8}}{e^{n(n+1)/4}} $.

Since $n! \sim \sqrt{2\pi n}\left( \dfrac{n}{e}\right)^n $, $(n!)^{(n+1)/2} \sim (2\pi)^{(n+1)/4} n^{(n+1)/4}\dfrac{n^{n(n+1)/2}}{e^{n(n+1)/2}} = \dfrac{(2\pi)^{(n+1)/4}n^{n(n+1)/2+(n+1)/4}}{e^{n(n+1)/2}} $.

Therefore,

$\begin{array}\\ f(n) &=\prod\limits_{k=0}^{n} n!\\ &=\sqrt{\dfrac{(n!)^{n+1}}{P(n)}}\\ &\sim \dfrac{(2\pi)^{(n+1)/4}n^{n(n+1)/2+(n+1)/4}}{e^{n(n+1)/2}} C^{1/2}\dfrac{n^{(3n+2))/12}(2\pi)^{(2n+1)/8}}{e^{n(n+2)/4}}\\ &= C^{1/2} \dfrac{(2\pi)^{(n+1)/4+(2n+1)/8}n^{n(n+1)/2+(n+1)/4+(3n+2)/12}}{e^{n(n+1)/2+n(n+2)/4}}\\ &= C^{1/2} \dfrac{(2\pi)^{(4n+3)/8}n^{n(n+1)/2+(6n+5)/12}}{e^{n(2n+2+n+2)/4}}\\ &= C^{1/2} \dfrac{(2\pi)^{(4n+3)/8}n^{n^2/2+(12n+5)/12}}{e^{n(3n+4)/4}}\\ &= C^{1/2} \dfrac{(2\pi)^{(4n+3)/8}n^{n^2/2+(12n+5)/12}}{e^{3n^2/4+n}}\\ &= C^{1/2} \dfrac{(2\pi)^{(4n+3)/8}n^{n^2/2+n+5/12}}{e^{3n^2/4+n}}\\ &= C^{1/2} (2\pi)^{3/8}n^{5/12}(2\pi)^{n/2}(n/e)^n \left(\dfrac{n}{e^{3/2}}\right)^{n^2/2}\\ \end{array} $

(Yep, that definitely was a pain to enter.)

As usual, since the math was done as it was entered, prob(no error) < .5.

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  • $\begingroup$ I get the same $n^{n^2/2}$ as well as the same leading exponent of $e$ (i.e. $e^{-3n^2/4}$), but we differ on the linear terms in the exponents. I could easily have made a mistake as well. $\endgroup$
    – aes
    Dec 11 '14 at 22:24
  • $\begingroup$ Yep. I find it annoyingly easy to make mistakes in these types of computations. I am pretty sure that I transcribed Hirshhorn's results correctly (I really admire his work!), but my work is less likely to be correct. $\endgroup$ Dec 11 '14 at 22:35
  • $\begingroup$ Oops - I see an error in my transcription! I'll fix it up. Oy. $\endgroup$ Dec 11 '14 at 22:37
  • $\begingroup$ Up vote for effort! $\endgroup$
    – D Wiggles
    Dec 11 '14 at 22:58
  • $\begingroup$ Yes, that's a nice result of Hirshhorn! Your formula $\prod_{k=1}^n {n \choose k} \prod_{k=1}^n (k!)^2 = (n!)^{n+1}$ is neat; I used $\prod_{k=1}^n k^k \prod_{k=1}^n k! = (n!)^{n+1}$. Our results are closer now, at least. :) $\endgroup$
    – aes
    Dec 12 '14 at 0:04
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A posibble form is: $$\prod_{i=1}^ni!$$

In other way is $$1^n\cdot2^{n-1}\cdot3^{n-2}\cdot\cdots\cdot(n-1)^2\cdot n^1$$

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It is sometimes referred to as the superfactorial of n (although there is at least one other function that's also called superfactorial), and its values are given by OEIS sequence A000178. I don't believe there are any shortcuts to computing it, other than noting that $sf(n) = \prod_{1 \le i \lt j \le n}\left( j - i \right)$

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Let's get the asymptotic behavior of $\prod_{k=1}^n k! = \prod_{k=1}^n k^{n-k+1}$.

The standard way to estimate such things is by taking the logarithm. But in this case, we can get it directly from Stirling's approximation (which itself can be proved by taking logs).

Stirling's approximation states $n! \sim \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$.

Thus $\prod_{k=1}^n k! \sim (2\pi)^n \sqrt{n!} \frac{\prod_{k=1}^n k^k}{e^{n(n+1)/2}}$

So now we just need to estimate $\prod_{k=1}^n k^k$. We could go do this, but even better is to use:

$\prod_{k=1}^n k! \prod_{k=1}^n k^k = \prod_{k=1}^n k^{n-k+1} \prod_{k=1}^n k^k = \prod_{k=1}^n k^{n+1} = (n!)^{n+1}$

Thus $(\prod_{k=1}^n k!)^2 \sim (2\pi)^n \sqrt{n!} \frac{(n!)^{n+1}}{e^{n(n+1)/2}}$

Then we get, applying Stirling's approximation to $\sqrt{n!}$ and $(n!)^{n+1}$:

$(\prod_{k=1}^n k!)^2 \sim (2\pi)^{2n+3/2}n^{3/4 + n/2}e^{-n(n+1)/2-n/2-n(n+1)} n^{n(n+1)}$

Thus $\prod_{k=1}^n k! \sim (2\pi)^{n+3/4} e^{-3n^2/4 - n} n^{n^2/2 + 3n/4 + 3/8}$

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