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It is known that the number of digits of a natural number $n > 0$, which represent by $d(n)$ is given by:

$d(n)= 1 + \lfloor\log n\rfloor\qquad (\text{I})$

($\log$ indicates $\log$ base $10$)

Well .. the classical approach to the Stirling factorial natural number $n > 1$ is given by:

$$n! \approx f(n) = [(2n\pi) ^{1/2}] [(n / e) ^ n]$$

The number of digits $n!$, according to equality (I), is:

$d(n!) = 1 + \lfloor\log n!\rfloor$

It seems to me that for all natural $n> 1$, $\log n!$ and $\log [f (n)]$ have the same floor:

$$\lfloor\log(n!)\rfloor = \lfloor\log(f(n))\rfloor$$

Here's my big question!

Therefore, we could write:

$$d (n!) = 1 + \lfloor\log(f(n))\rfloor$$

Hope someone has some little time for the theme.

Cookie
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Paulo Argolo
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  • You've clearly addressed this to Américo Tavares... I do however wonder if an email would be appropiate for this sort of thing. Of course, there's nothing stopping you posting your question in Portuguese, only that the potential audience is very limited. If you can write English, it may be useful to translate. :) – Noldorin Oct 30 '10 at 01:01
  • Let's shorten your question: is $1+\left\lfloor\left(n+\frac12\right)\log_{10}n-n\log_{10}e+\frac12\log_{10}(2\pi)\right\rfloor$ a valid formula for counting the number of digits of $n!$ ? – J. M. ain't a mathematician Oct 30 '10 at 01:01
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    From Mathoverflow: http://mathoverflow.net/questions/19170/how-good-is-kamenetskys-formula-for-the-number-of-digits-in-n-factorial and http://mathoverflow.net/questions/19086/number-of-digits-in-n – Michael Lugo Oct 30 '10 at 01:12
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    Portuguese is not that hard to read, at least if you know French and/or Spanish, which are languages more commonly known to speakers of English. That being said, the meta.mathoverflow discussion on posting in non-English posts: http://tea.mathoverflow.net/discussion/142/nonenglish-posts/. (The consensus there was that posts in all languages should be allowed, but there would probably be few non-English posts.) – Michael Lugo Oct 30 '10 at 01:29
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    -1: uninformative title – Jyotirmoy Bhattacharya Oct 30 '10 at 11:39
  • Please edit the title to be useful. – Jason S Oct 30 '10 at 14:59
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    @Jyotirmoy, @Jason: I've edited the title. – Jonas Meyer Oct 30 '10 at 15:14
  • This is an interesting question that deserves a better title. I was about to edit it, but cannot as the reputation required has been raised now we're out of beta. Also, minor quibble with the translation, please invert "one can" to "can one" as "one can" does not work so well in written language as one cannot hear the intonation at the end of the sentence which turns it into a question. – Derek Jennings Oct 30 '10 at 15:17
  • @Jonas Meyer Your new question title is better than the one I had in mind :-) – Derek Jennings Oct 30 '10 at 15:36
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    I once tried to solve this problem, but got nowhere... :( –  Oct 30 '10 at 15:43
  • As already noted in the comments, your original title was very uninformative. Why did you change it to the original title? (I have rolled it back to the tidied-up version). – J. M. ain't a mathematician Oct 31 '10 at 10:22
  • @Byron Schmuland: (and anyone else who might know). Is there any existing literature on the this question? Thanks. – Derek Jennings Nov 01 '10 at 13:32
  • @Derek: Sorry, I don't know of any literature. In my case, I had written an article on factorials for a high school magazine and the OP's question also occurred to me. I thought that the result is false, and my investigation was just a computer search for a counterexample. My search failed, though there were some "close calls". –  Nov 01 '10 at 14:59
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    @Byron: Interesting... would you mind posting those near-misses as a comment or maybe a CW answer? Maybe they could provide clues to what an actual counterexample might look like. – J. M. ain't a mathematician Nov 02 '10 at 00:14

3 Answers3

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6561101970383 is a counterexample, and the first such if I computed correctly. See my answer in https://mathoverflow.net/questions/19170 for more information.

Community
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Noam D. Elkies
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Here are some thoughts on the conjecture that may lead one to suspect that it is true. This is not a proof that it is true.

We want to know whether

$$ \left\lfloor \log_{10} n! \right\rfloor = \left\lfloor \log_{10} \lfloor f(n) \rfloor \right\rfloor$$

is true for all $n > 1.$

We note that this would NOT be true if the interval $I_n = ( \log_{10} n!,\log_{10} \lfloor f(n) \rfloor)$ contains an integer but is true otherwise.

So let's look at the length of $I_n.$

From the Stirling series we have

$$\frac{1}{\log_{e}10} \left( \frac{1}{12n} - \frac{1}{360n^3} \right) < \log_{10} n! - \log_{10} f(n) < \frac{1}{\log_{e}10} \frac{1}{12n}.$$

And so taking the integer part of $f(n)$ we have

$$\frac{1}{\log_{e}10} \left( \frac{1}{12n} - \frac{1}{360n^3} \right) < \log_{10} n! - \log_{10} \lfloor f(n) \rfloor < \frac{1}{\log_{e}10} \left( \frac{1}{12n} + \frac{1}{ \lfloor f(n) \rfloor } \right),$$

where to achieve the RHS we note that

$$ \log_{10} f(n) - \log_{10} \lfloor f(n) \rfloor < \frac{1}{ \lfloor f(n) \rfloor \log_e 10 }.$$

Hence $$\text{Length}(I_n) < \frac{1}{\log_{e}10} \left( \frac{1}{360n^3} + \frac{1}{ \lfloor f(n) \rfloor } \right).$$

We can verify that the conjecture holds for $n=2,3,\ldots,10,$ so summing up the remaining lengths of the $I_n$ we have

$$\sum_{n=11}^\infty \text{Length}(I_n) < \frac{1}{360 \log_e 10} \left( \sum_{n=11}^\infty \frac{1}{n^3} + \sum_{n=11}^\infty \frac{1}{ \lfloor f(n) \rfloor } \right).$$

Now $ \lfloor f(n) \rfloor \ge (n-1)! $ and so, doing some calculations (replacing the $ \lfloor f(n) \rfloor $ on the RHS by $ (n-1)! $ , we have

$$\sum_{n=11}^\infty \text{Length}(I_n) < \frac{1}{360 \log_e 10} \left( 0.00452492 + 0.00000030 \right) < 5.5 \times 10^{-6}.$$

Hence the probability that an integer falls in any of the intervals is very small.

Derek Jennings
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My numerical work and computing skills are not to be trusted, but the first "near miss" that I recorded was $n=252544447$.