# Is the sequence $\{\{\log(n!)\}\}_n$ dense in $[0,1]$?

I tried to find this problem in Mathematics Stack Exchange and in Math Overflow, but I didn't find it anywhere. Here is my problem:

Is the sequence $\{\{\log(n!)\}\}_n$, the sequence of fractional parts of $\log(n!)$, dense in $[0,1]$?

Note 1: Here log stands for the logarithm with base $10$.

By intuition, I think that the answer is affirmative. I tried to prove that the above sequence is dense in $[0,1]$ by using the density of the sequence $\{\{\log(n)\}\}_n$, but I stucked in the process.

Motivation Problem: Let $a_0,...,a_n$ be natural numbers with $a_n \neq 0,9$. Does there exist a natural number $m$ such that the first digits in the decimal representation of $m!$ are $a_0,...,a_n$?

If $\{\{\log(n!)\}\}_n$ is dense in $[0,1]$ and $l:=a_0+...+a_n10^{n}$, then there is a sufficiently large natural $m$:

$\{\log(l)\}<\{\log(m!)\}<\{\log(l+1)\} \Rightarrow \\\{\log(l)\}+[\log(m!)]<\log(m!)<\{\log(l+1)\}+[\log(m!)]\Rightarrow \\\log(l)+[\log(m!)]-[\log(l)]<\log(m!)<\log(l+1)+[\log(m!)]-[log(l+1)]\ (1)$

Since $a_n \neq0,9$, we have: $10^n\leq l,l+1<10^{n+1}\Rightarrow n\leq \log(l),log(l+1)<n+1 \Rightarrow [log(l)]=[log(l+1)]=n$

Hence $[\log(m!)]-[\log(l)]$ and $[\log(m!)]-[log(l+1)]$ are the same positive integer, say $k$. Then $(1)$ yields that:

$10^kl<m!<10^k(l+1)$ and therefore the answer to the question is "yes".

Note 2: $[x]$ is the integer part of $x$.

Do you have any ideas or hints that will help me prove or disprove the claim of density?

• are tou talking about $\log_{10}$ everywhere ? (anyway the answer is yes) Jan 2, 2017 at 16:23
• @mercio yes that's what I mean everywhere. Can you please tell me why the answer is yes? Jan 2, 2017 at 16:26

Note that $\log(s!)=\sum_{n=1}^s \log n$. To prove that the $\{\sum \log n\}$s are dense in $\mathbb R/\mathbb Z$, consider an arbitrary $x\in[0,1]$ and $k\in\mathbb N_+$, and show that there is some $\{\sum \log n\}$ whose distance from $x$ is at most $1/k$.

This will be the case if we can find $2k$ successive $n$s such that $m+\frac1{2k} < \log n < m+\frac 1k$ for all of them, for some integer $m$.

However, this is easy enough. First, since the derivative of $\log x$ goes to $0$, there will be an $N$ such that each successive increase in $\log n$ from that point onwards is at most $\frac{1}{(2k)^2}$. Additionally $\log n$ itself goes to $+\infty$, so there will be an $M>N$ such that $$\log(M-1) < m+\frac1{2k} < \log M$$ for some $m\in\mathbb N$.

The numbers $M, M+1, M+2, \ldots, M+2k-1$ then have the required property.

• Thanks for the answer. The biggest part is easy, but I am having hard time with understanding why your statement: "This will be...for some integer $m$" implies that for some $n$ we have $|\{\sum \log(n)\} - x|<\frac{1}{k}$? Jan 2, 2017 at 19:00
• @SachpazisStelios: You have $2k$ successive $n$s such that each $\{\log(n!)\}$ is between $\frac1{2k}$ and $\frac1k$ larger than the previous one. This means that the $\{\log(n!)\}$s walk at least one complete cycle around $[0,1]$ from some starting point, in steps no larger than $\frac1k$. Somewhere along the way we must pass $x$. Jan 2, 2017 at 19:12
• What do you mean with "between $\frac{1}{2k}$ and $\frac{1}{k}$ larger than the previous one"? I can't figure out which is the inequality you mean here? Jan 2, 2017 at 19:39
• @SachpazisStelios: I'm unsure what your problem here is. Since the fractional part of $\log n$ is (by assumption) between $\frac1{2k}$ and $\frac1k$, the difference between the fractional parts of $\log((n-1)!)$ and $\log(n!)$ will be between $\frac1{2k}$ and $\frac1k$ too (except when they roll over from $1$ back to $0$, but that's not important). Jan 2, 2017 at 19:51
• Now I get it! Thanks a lot for your time. If I am right, the key idea was that you showed that $2k$ successive terms of the sequence differ from each other more than $\frac{1}{2k}$. Jan 2, 2017 at 20:06

I was searching for an other problem and I encountered this post in Math Overflow. As the O.P. states, in this paper Persi Diaconis proved that $\{\log (n!)\}_n$ is equidistributed (or uniformly distributed) modulo 1. This implies that the sequence $\{\{\log (n!)\}\}_n$ is dense in $[0,1]$.

First, Diaconis gives the definition of a strong Benford sequence. Then he proves that for every strong Benford sequence $\{a_n\}_n$ the sequence $\{\log (a_n)\}_n$ is equidistributed modulo 1 (Theorem 1). Finally he uses this theorem and some theorems from the book Uniform Distribution Sequences of Kuipers and Niederreiter, to prove that $\{n!\}_n$ is a strong Benford sequence (Theorem 3). This finishes the job.