2
$\begingroup$

This question was raised in the comment by Todd Trimble at how to proof there is a natural number n, the first four digits of n! Is 2018?. I thought the question may be posted separately, as even partial answers to it may turn out useful and instructive for a number of users (certainly including me).

A trivial remark is that, in view of Stirling's formula, this question can be restated as follows: Is the sequence $((n+1/2)\log n-n\log e\mod1)_{n\in\mathbb N}$ dense in the interval $[0,1]$? It would be surprising to me if the answer here is no or if it depends on the base of the logarithm.

$\endgroup$
  • 1
    $\begingroup$ This should be failry easy to handle by Weyl differencing... $\endgroup$ – Will Sawin Mar 27 '18 at 22:10
  • 1
    $\begingroup$ Indeed: fix $k > 0$, find $n$ large enough with $\operatorname{frac}(\log n) \approx \frac{1}{k}$ and observe that $\log(n!), \log((n+1)!), \ldots, \log((n+k-1)!)$ modulo $1$ are roughly uniformly distributed on $[0,1]$. However, is this sequence equidistributed? $\endgroup$ – Mateusz Kwaśnicki Mar 27 '18 at 22:32
  • 1
    $\begingroup$ For an elementary proof, see John Edward Maxfield's 1970 paper A note on $N!$ in Mathematics Magazine, which I discussed last week in my answer to Short papers for undergraduate course on reading scholarly math. $\endgroup$ – Dave L Renfro Mar 29 '18 at 18:56
  • $\begingroup$ @DaveLRenfro : It appears that the method in that note is similar to the one briefly described in the comment by Mateusz. $\endgroup$ – Iosif Pinelis Mar 30 '18 at 2:13
5
$\begingroup$

Let $a$ be the base of logarithm here. Consider the function $$f(x)=\frac{(x+1/2)\ln x-x}{\ln a}.$$

As noted in the question, it is enough to prove that the sequence $f(n)$ is dense modulo $1$. In fact, this sequence is equidistributed. By Weyl's criterion, it is enough to show that for any nonzero integer $k$ we have

$$\sum_{n \leq N} e^{2\pi ikf(n)}=o(N).$$

To prove this, let us use the van der Corput lemma, which states that if $g \in C^2(I)$, where $I$ is an interval and

$$\lambda\ll |g''(x)| \ll \lambda$$

for all $x \in I$ (constants in Vinogradov symbols are absolute) then

$$\sum_{n \in I\cap \mathbb Z} e^{2\pi ig(n)} \ll |I|\lambda^{1/2}+\lambda^{-1/2}.$$

Taking $g(x)=kf(x)$ and $I=[M,M_1]$ with $M<M_1 \leq 2M$ we get for $M$ large enough

$$\sum_{M\leq n\leq M_1} e^{2\pi i kf(n)} \ll \sqrt{kM},$$

as $f''(x)=\frac{1}{x\ln a}-\frac{1}{2x^2\ln a} \asymp \frac{1}{M}$ for $x \in I$.

Therefore, using the dyadic subdivision of the interval $[1,N]$ we get

$$\sum_{n \leq N} e^{2\pi i kf(n)} \ll \sqrt{kN}+1=o(N),$$

as needed.

$\endgroup$
  • $\begingroup$ Bounds here are of course not uniform in $a$, but this has no effect on anything. Also, this can be fixed by adding something like $\sqrt{\ln a}$ to the rhs if $a \in \mathbb N$. $\endgroup$ – Asymptotiac K Mar 27 '18 at 22:39
  • $\begingroup$ The third equation seems to miss some exponents. $\endgroup$ – Johannes Trost Mar 28 '18 at 7:11
  • $\begingroup$ @JohannesTrost, what do you mean? I don't see any exponents missing.. $\endgroup$ – Asymptotiac K Mar 28 '18 at 8:10
  • $\begingroup$ $|g''(x)|$ being much larger and much smaller than $\lambda$ at the same time does not make sense to me. Or did I miss something ? $\endgroup$ – Johannes Trost Mar 28 '18 at 10:45
  • $\begingroup$ @JohannesTrost, $f\ll g$ means $f=O(g)$, this is Vinoradov's notation $\endgroup$ – Asymptotiac K Mar 28 '18 at 11:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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