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Are there any results relevant to the distribution of the sequence $\{\log \log n!\}$ for integers $n$, where $\{x\}$ denotes the fractional part of $x$?

For instance, it is known that for irrational real numbers $\alpha$, the sequence $\{n\alpha\}$ is dense in $[0,1]$ and in fact equidistributed. Does something similar hold for the logarithms of the logarithms of the factorials?

(This curiosity is provoked by this question, and an affirmative answer here would complete the answer to that question.)

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This question popped up today on MSE first page and I just discovered it. Two remarks. First, Benford's law could have been mentioned in relation to {log n} and as a starting point to approach {log log n!}. Second, to accept an answer which has (some interesting remarks but) no definitive argument on the equidistribution (the argument about the density is in a comment), less than 12h after the question was asked, strikes me, once again, as odd. The second point could explain the first one. –  Did Oct 18 '11 at 6:32
@DidierPiau: Given that Andrew's answer along with Gerry Myerson's comment and Andrew's other comment had already answered the question (at least enough for it to be obvious how to complete the proof), and the high number of upvotes which suggested that the community thought the answer has been given, I did not expect that further answers would be forthcoming. Gerry Myerson's comment is what made it clear to me, but I can't accept a comment, unfortunately. :-) Feel free to post a complete answer, and I can mark it as accepted for the benefit of future people who arrive at this question. –  ShreevatsaR Oct 18 '11 at 6:45
(1) There are no results relevant to the distribution of the sequence $\{\log\log n!\}$ for integers $n$ in the answers, except that the sequence is dense. (2) To consider that the reaction of the community is complete and definitive enough after 12h strikes me as odd (and probably excludes de facto a part of said community, if only for time zone reasons). // Having said that, I stress that at least one point of MSE modus operandi is clear to me: the decision to accept an answer is entirely in the hands of the OP. Thus my previous comment was just my two cents and nothing more. –  Did Oct 18 '11 at 7:15
@DidierPiau: (1) From Gerry's comment showing how to prove that $\{\log(n!)\}$ is not equidistributed, and Andrew's comment that "probably the same proof will be valid", we can somewhat easily get a proof that $\{\log\log(n!)\}$ is not equidistributed. Right? If I am mistaken about this, that would be very valuable, please tell me! (2) Yes, it is my mistake in thinking that 12 hours is a very long time without answer.s –  ShreevatsaR Oct 18 '11 at 7:28
Once again, I see no proof that $\{\log\log(n!)\}$ is not equidistributed (do you see one?). And a fortiori nothing about the limiting distribution itself, if it exists. All this despite the fact that your question did not ask only about the sequence being dense but about any results relevant to the distribution. Nice question, by the way... :-) –  Did Oct 18 '11 at 7:56

3 Answers 3

up vote 9 down vote accepted

Sequence $\log n$ is not equidistributed. The reason is $\log[n]$ grows very slowly and so $\{\log n\}$ concentrates to the lower part of $[0,1]$. Since $\log\log n!$ grows approximately as $\log(n\log n)$ probably the same proof will be valid.

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Ah, interesting, thanks… could you suggest a reference for $\{\log n\}$ not being equidistributed? (Of course, even if $\{\log\log n!\}$ it not equidistributed it may still be dense, which would be enough for the other question.) –  ShreevatsaR Jul 1 '11 at 7:10
@Shreevatsa, it's not hard to prove $\log n$ isn't u.d., you just pick any interval, say, $[0,.1]$, and show note that it's in there from $n=e^m$ to $n=e^{m+.1}=e^{.1}e^m$ which is to much for u.d. –  Gerry Myerson Jul 1 '11 at 7:21
It is dense on [0,1] since the sequence $a_n=\log\log n!$ is monotone and difference $a_{n+1}-a_n$ tends to zero as $n\to\infty$. –  Andrew Jul 1 '11 at 10:24
@Andrew: Ah I see. It's all clear now; thanks a lot. –  ShreevatsaR Jul 1 '11 at 17:22
@Gerry: the actual density of $\{\log(n)\}$ is $\frac{e^x}{e-1}$ on $[0,1]$. –  robjohn Oct 18 '11 at 6:40

The definition of factorial gives $$ \log(n!)-\log((n-1)!)=\log(n)\tag{1} $$ Since the derivative of $\log(x)$ is $1/x$, $(1)$ and the Mean Value Theorem yield $$ \begin{align} \log(\log(n!))-\log(\log((n-1)!)) &\in\left(\frac{\log(n)}{\log(n!)},\frac{\log(n)}{\log(n!)-\log(n)}\right)\\ &=\frac{\log(n)}{n\log(n)-n+O(\log(n))}\tag{2} \end{align} $$ The density of $\log(\log(n!))$ is the reciprocal of $(2)$: $n-\frac{n}{\log(n)}+O(1)$ and $$ \log(\log(n!))=\log(n)+\log(\log(n))-\frac{1}{\log(n)}+O\left(\frac{1}{\log(n)^2}\right)\tag{3} $$ As $n\to\infty$, $\log(\log(n!))\sim\log(n)$ and the density $\sim n$. Since the limiting density is determined when $n$ is large, we get that the density of $\{\log(\log(n!))\}$ is $\frac{e^x}{e-1}$ on $[0,1]$.

More explanation about $\frac{e^x}{e-1}$:

Since $\log(\log(n!))=\log(n)+\log(\log(n))-\frac{1}{\log(n)}+O\left(\frac{1}{\log(n)^2}\right)$, $\{\log(\log(n!))\}$ cycles through $[0,1]$ just a bit quicker than $\log(n)$ does; approximately when $n$ goes to $ne\left(1-\frac{1}{\log(n)}\right)$. In each of those cycles, the logarithm of the density, $\log\left(n-\frac{n}{\log(n)}\right)+O\left(\frac{1}{n}\right)$, increases approximately linearly by $1$. Thus, the density of $\{\log(\log(n!))\}$ is proportional to $e^x$, and $\frac{e^x}{e-1}$ is normalized to have total weight $1$.

Density Details:

Let $I_n=\{k\in\mathbb{Z}:n-1<\log(\log(k!))\le n\}$. The density approximated here is the function $\phi:[0,1]\mapsto\mathbb{R}$ so that $$ \int_a^b\phi(x)\;\mathrm{d}x=\lim_{n\to\infty}\left.\left|\{k\in\mathbb{Z}:n-1+a<\log(\log(k!))\le n-1+b\}\right|\middle/\left|I_n\right|\right. $$ Within a given $I_n$ this density is roughly proportional to the reciprocal of the distance between $\log(\log((k-1)!))$ and $\log(\log(k!))$, which is $k-\frac{k}{\log(k)}+O(1)$. For $k\in I_n$, let $x=\log(\log(k!))-n+1=\log(k)+\log(\log(k))-n+1+O\left(\frac{1}{\log(k)}\right)$. Then $$ k=\frac{e^{x+n-1+O(1/(x+n))}}{(x+n-1)^{1-1/(x+n)}}=\frac{e^{n-1+O(1/(x+n))}}{(x+n-1)^{1-1/(x+n)}}e^x $$ Thus, in terms of $x$, the density is $$ k-\frac{k}{\log(k)}+O(1)=\frac{e^{n-1+O(1/(x+n))}}{(x+n-1)^{1-1/(x+n)}}e^x $$ As $n\to\infty$, the coefficient of $e^x$ tends toward constancy. Normalizing this so that the integral over $[0,1]$ is $1$, we get $\phi(x)=\frac{e^x}{e-1}$.

We get the same density if we consider $\displaystyle\lim_{N\to\infty} \bigcup_{n\le N} I_n$. However, if we use a partial $I_N$, the fact that $|I_N|$ is approximately $(e-1)\left|\bigcup_{n<N} I_n\right|$ causes bad behavior.

Thus, the fractional parts of $\log(\log(n!))$ are dense in $[0,1]$, but not uniformly distributed.


On $I_{10}$, the counts in each interval of size $0.01$ are

enter image description here

On $I_{11}$, the counts in each interval of size $0.01$ are

enter image description here

On $I_{12}$, the counts in each interval of size $0.01$ are

enter image description here

On $I_{13}$, the counts in each interval of size $0.01$ are

enter image description here

So as described above, on each $I_n$, the density is $\frac{e^x}{e-1}$. However, since the counts on $I_{n+1}$ are approximately $e$ times the counts on $I_n$, the picture is different if we don't consider complete intervals $I_n$:

enter image description here

The jump has a ratio of about $e\left(1-\frac1{\log(n)}\right)$ since $\{\log(\log(n!))\}$ cycles through $[0,1]$ as $n$ goes to $ne\left(1-\frac1{\log(n)}\right)$, as mentioned above.

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Nice try. Some steps escape me, for example, how you use the comparison of $\log\log(n!)$ and $\log n$, what you mean by the density is $\sim n$, and how you finally conclude that the density exists and is what you write. As a way of calling for more cautious approaches, I mention that the density DOES NOT EXIST, at least in the classical sense of the term. See my own answer for more detailed explanations of these somewhat abrupt comments. –  Did Oct 18 '11 at 17:05
@Didier: I have appended a bit more detail about the density that I was thinking about. –  robjohn Oct 20 '11 at 0:24
to the downvoter: I am not claiming that the distribution has a limit, but only that on the intervals $I_n$, the distribution tends to $\frac{e^x}{e-1}$. –  robjohn Sep 7 at 19:12

A (very weak) version of Stirling's approximation says that $n!=n^{n+o(n)}$, hence $\log\log(n!)=\log(n\log n)+o(1)$. If two sequences $(a_n)$ and $(b_n)$ are such that $a_n\to\infty$, $b_n\to\infty$, and $a_n-b_n=o(1)$, the limiting distributions of $\{a_n\}$ and $\{b_n\}$ are the same (of course, one should prove this) hence, from now on, we look at the asymptotic distribution of $x_n=\{\log(n\log n)\}$.

For every nonnegative $x$, call $n(x)$ the smallest integer such that $n(x)\log n(x)\geqslant\mathrm e^x$. Fix $x$ in $(0,1)$. The set of integers $n$ such that the integer part of $\log(n\log n)$ is $k$ is $n(k+1)-n(k)$. Among these, $n(k+x)-n(k)$ are such that $x_n\leqslant x$. Hence the proportion of integers $n$ in $[n(k),n(k+1)-1]$ such that $x_n\leqslant x$ is $$ F_k(x)=\frac{n(k+x)-n(k)}{n(k+1)-n(k)}. $$ Here is a result: $F_k(x)\to F^0(x)$ when $k\to\infty$, with $$ F^0(x)=\frac{\mathrm e^x-1}{\mathrm e-1}=\int\limits_0^xf^0(y)\mathrm dy,\quad f^0(x)=\frac{\mathrm e^x}{\mathrm e-1}. $$ This (which should also be proved) implies that the proportion of integers $n\leqslant n(k)$ such that $x_n\leqslant x$ converges to $F^0(x)$ when $k\to\infty$. In this sense the sequence of general term $\{x_n\}$ and the sequence of general term $\{\log\log n!\}$ both follow the distribution with density $f^0$ on $(0,1)$.

Note however that this result is not based on the usual version of the density of a subset $A$ of the integers, defined as the limit (when it exists) of the sequence $$ d_n(A)=\frac1n\sum\limits_{k=1}^n\mathbf 1_{k\in A}. $$ Considering the set $A(x)$ of integers $n$ such that $\{x_n\}\leqslant x$, we proved that $d_{n(k)}(A(x))\to F^0(x)$ when $k\to\infty$ but $(d_n(A(x))$ diverges (except for $x=0$ and $x=1$) since for example, $$ d_{n(k+x)}\to\mathrm e^{1-x} F^0(x)>F^0(x). $$ By the way, one can prove that $F^0(x)$ and $F^x(x)=\mathrm e^{1-x} F^0(x)$ are the lower and upper densities of $A(x)$ in the sense that $$ F^0(x)=\liminf d_n(A(x))<\limsup d_n(A(x))=F^x(x). $$ Coming back to our problem, one could very plausibly define the density of $\{x_n\}$ by fixing a real number $\alpha$, considering the proportion of integers $n\leqslant n(k+\alpha)$ such that $x_n\leqslant x$ and the limit $F^\alpha(x)$ of these proportions when $k\to\infty$. It happens that this limit $F^\alpha(x)$ exists for every $x$ and suggests as density the function $f^\alpha$ such that $$ F^\alpha(x)=\int\limits_0^xf^\alpha(y)\mathrm dy. $$ Thus, each $f^\alpha$ is as good a candidate as any to be the asymptotic density of the sequence with general term $\{\log\log(n!)\}$.

Note As said in the comments, Benford's law yields similar oscillatory phenomena. Of course, the procedures developed in the case of Benford's law to escape the conundrum we described above may be applied to the present problem as well.

Other, more advanced, keywords for the interested reader are divergent series or summation methods, see here or the version in French here which seems to mention more references.

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+1, you were the only one to analyze the phenomenon carefully enough. I think "the integer part of $x_n$" should be "the integer part of $\log(n\log n)$"? (The integer part of $x_n$ is $0$.) –  joriki Sep 7 at 6:13
@joriki Thanks for the appreciation and for the correction. Indeed, rereading the whole page after all this time makes for an odd experience. For some reason, at the moment, I obviously failed to convey the message I was trying to convey, namely, that the two other undeleted answers were not addressing the question at all, mathematically speaking (its density part excepted). –  Did Sep 7 at 6:29

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