Are the fractional parts of $\log \log n!$ equidistributed or dense in $[0,1]$?

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.)

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.

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.

Charts

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

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

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

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

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 $I_n$:

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.