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Can it be shown that the sequence $\log(\log((n!)!))$ is uniformly distributed mod 1? I believe it should be, but I'm not certain if Stirling's approximation repeated twice combined with the techniques used to show $\log(n!)$ is uniformly distributed mod 1 are adequate to establish the result in this case.

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  • $\begingroup$ Possibly related is the fact that the sequence $\log\log(n!!)$ — here, $n!!$ means $(n!)!$ — is uniformly distributed mod 1, which is shown in: John Edward Maxfield, A note on $N!$, Mathematics Magazine 43 #2 (March 1970), pp. 64−67. Incidentally, I discussed this paper in my answer to Short papers for undergraduate course on reading scholarly math. $\endgroup$ Jun 30, 2021 at 13:29
  • $\begingroup$ @DaveLRenfro Isn't this exactly an answer to this question? $\endgroup$
    – Wojowu
    Jun 30, 2021 at 13:32
  • $\begingroup$ @Wojowu: I just realized the parentheses are not balanced, which led me to misread this as input $n,$ apply factorial, apply logarithm, apply factorial (gamma function version?), apply logarithm. But what Maxfield did seems to be what is intended here. $\endgroup$ Jun 30, 2021 at 13:45
  • $\begingroup$ @DaveLRenfro Looking at your answer you have linked it seems that what Maxfield has proven is mere density, and Diaconis proved uniform distribution for a single factorial. Is equidistribution known for iterated ones? $\endgroup$
    – Wojowu
    Jun 30, 2021 at 13:47
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    $\begingroup$ The sequence is roughly $n\log n$. There are a couple of things in Kuipers & Niederreiter, Uniform Distribution of Sequences, that might apply. Example 2.8 on page 18 shows that $n\log n$ is u.d. mod $1$. Exercise 2.26 on pages 24-25 says, let $f(x)$ be defined for $x\ge1$ and twice differentiable for sufficiently large $x$ with $f''(x)$ tending monotonically to zero as $x\to\infty$. Suppose also $\lim_{x\to\infty}f'(x)=\pm\infty$ and $$\lim_{x\to\infty}{(f'(x))^2\over x^2|f''(x)|}=0$$ Then $f(n)$, $n=1,2,\dots$, is u.d. mod $1$. $\endgroup$ Jul 1, 2021 at 3:55

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