https://www.johndcook.com/blog/2023/06/23/every-factorial-is-a-power/

John D. Cook
Skip to content

  * MATH
      + PROBABILITY
      + SIGNAL PROCESSING
      + NUMERICAL COMPUTING
      + SEE ALL ...
  * STATS
      + EXPERT TESTIMONY
      + FORECASTING
      + RNG TESTING
      + SEE ALL ...
  * PRIVACY
      + HIPAA
      + CRYPTOGRAPHY
      + DIFFERENTIAL PRIVACY
  * WRITING
      + BLOG
      + TWITTER
      + ARTICLES
      + TECH NOTES
      + SUBSCRIBE
      + NEWSLETTER
  * ABOUT
      + CLIENTS
      + ENDORSEMENTS
      + TEAM
      + SERVICES

(832) 422-8646
Contact

Every factorial is a power

Posted on 23 June 2023 by John

The previous post mentioned that 24! [?] 10^24 and 25! [?] 10^25.

For every n, there is some base b such that n! = b^n. For example,
30! [?] 12^30.

It's easy to find b [1]:

b = \exp\left( \frac{\log n!}{n} \right)

What's interesting is that b is very nearly a linear function of n.

[factorial_base2]

In hindsight it's clear that this should be the case--it follows
easily from Stirling's approximation--but I didn't anticipate this
before I plotted it.

Now fix n and find b such that n! = b^n. Since the relationship
between n and b(n) is nearly linear, this suggests

(2n)! \approx (2b)^{2n} = 2^{2n} (n!)^2

which is true. It follows from the multiplication identity for the
gamma function:

[gamma_multiplication_simple]

Let z = n + 1/2 so that the left side is (2n)!. On the right side, G(
z + 1/2) = n! and G(z) is not too different from n!. The rest of the
right side is 2^2n/[?]p.

So our observation that b(n) is nearly linear gave us a hint of
Gauss's multiplication formula.

[1] Numerically you would probably evaluate this function by calling
a routine that computes log G(n + 1) directly without computing G(n +
1) first. This avoids overflow for large n.

This is why mathematical libraries will have not only gamma functions
but also loggamma functions. The latter seems redundant, but it's not
.

Categories : Math
Tags : Special functions
Bookmark the permalink

Post navigation

Previous PostMentally approximating factorials
Next PostNumbers don't typically have many prime factors

2 thoughts on "Every factorial is a power"

 1.
    Michael Albert
    23 June 2023 at 17:37

    The approximation you note could also be seen by observing that
    $C(2n, n) = \frac{(2n)!}{(n!)^2}$ on the one hand, and (as the
    central binomial) is also at least $2^{2n}/n$ (in fact more like
    $2^{2n}/\sqrt{n}$ of course by central limit stuff.)

 2.
    Martin Cohen
    26 June 2023 at 13:17

    The basic approximation I use is
    n! ~ (n/e)^n so ln(n!) ~ n(ln(n)-1).
    The basic inequality is
    (n/e)^n < n! < (n/e)^{n+1}.
    When considering convergence of series using the n-th root test,
    (n!)^{1/n} ~ n/e so
    ((an)!)^{1/n} = (((an)!)^{1/(an)})^a ~(an/e)^a = a^an^a/e^a.

Leave a Reply

Your email address will not be published. Required fields are marked 
*

          [                                             ]
          [                                             ]
          [                                             ]
          [                                             ]
          [                                             ]
          [                                             ]
          [                                             ]
Comment * [                                             ]

Name * [                              ]

Email * [                              ]

Website [                              ]

[Post Comment] 

 [                                             ] 
 [                                             ] 
 [                                             ] 
 [                                             ] 
 [                                             ] 
 [                                             ] 
 [                                             ] 
D[                                             ] 

Search for: [                    ] [Search]
John D. Cook

John D. Cook, PhD

My colleagues and I have decades of consulting experience helping
companies solve complex problems involving data privacy, math,
statistics, and computing.

Let's talk. We look forward to exploring the opportunity to help your
company too.

[                                        ]
[                                        ]
[                                        ]
[                                        ]
[                                        ]
[Send]

 [                                             ] 
 [                                             ] 
 [                                             ] 
 [                                             ] 
 [                                             ] 
 [                                             ] 
 [                                             ] 
D[                                             ] 

John D. Cook

(c) All rights reserved.

Search for: [                    ] [Search]
(832) 422-8646

EMAIL