Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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: (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 1:29
-1: uninformative title – Jyotirmoy Bhattacharya Oct 30 '10 at 11:39
@Jyotirmoy, @Jason: I've edited the title. – Jonas Meyer Oct 30 '10 at 15:14
I once tried to solve this problem, but got nowhere... :( – Byron Schmuland Oct 30 '10 at 15:43
up vote 7 down vote accepted

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.

share|cite|improve this answer
I wonder if an attack along these lines might work for saying "there are no counterexamples below n" for sufficiently large n? – J. M. Oct 30 '10 at 10:57
@J.M. That's an interesting thought. – Derek Jennings Oct 30 '10 at 14:59

6561101970383 is a counterexample, and the first such if I computed correctly. See my answer in for more information.

share|cite|improve this answer
Prof. Elkies: I saw this last night on MO. Thanks very much! – Byron Schmuland Aug 26 '11 at 14:33

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

share|cite|improve this answer
Dear Byron Schmuland, Do you think you have found a counterexample? Thank you for your attention. – Paulo Argolo Nov 2 '10 at 14:33
@Paulo: "near miss" - so not really a counterexample. It might however give clues into what a true counterexample looks like. – J. M. Nov 2 '10 at 14:57
See also… – Byron Schmuland Nov 5 '10 at 15:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.