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In this question $\lfloor a\rfloor$ means the greatest integer not exceeding $a$.

Using van der Corput's inequalities one is able to show that $\log(n!)\alpha$ is equidistributed on the unit circle ($\mathbb{R}\setminus\mathbb{Z}$) for all $\alpha \in \mathbb{R}^{+}$.

I am however having trouble trying to show that $\lfloor\log(n!)\rfloor\alpha$ is equidistributed on the unit circle for $\alpha \in \mathbb{R}\setminus\mathbb{Q}$. The reason I believe this result to hold is due to the research here, showing that $\lfloor\log(n!)\rfloor$ is a good sequence for mean $L^{2}$ convergence. Any ideas of how one could try and prove this would be much appreciated.

I suppose the more general question is: "Are there any results or methods that help in proving equidistribution (or good sequences for mean $L^{2}$ convergence) for functions that are rounded down?"

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    $\begingroup$ Could you replace the vertical bars with $ \lfloor\dots\rfloor $ floor signs to make the notation more recognizable? $\endgroup$
    – Zsbán Ambrus
    Commented Mar 5, 2014 at 11:22
  • $\begingroup$ Or with $[\dots]$ for those of us a little less hip. $\endgroup$
    – Gerry Myerson
    Commented Mar 5, 2014 at 11:59
  • $\begingroup$ @ZsbánAmbrus: done. $\endgroup$
    – Ian Morris
    Commented Mar 5, 2014 at 12:27
  • $\begingroup$ Would you agree to explain why $\log(n!)\alpha$ equidistributed? $\endgroup$
    – Lior Bary-Soroker
    Commented Jul 28, 2014 at 11:03
  • $\begingroup$ @Lior I am afraid in the end I only ended up using $n^{1-\varepsilon}\log(n)$ is equidisitrbuted for all $1>\varepsilon > 0$, which is easy to prove (probably doesn't help you). The reason I believed $\log(n!)$ was equidistributed was due to an exercise in link, I hope the link helps. $\endgroup$
    – Trevor Hansen
    Commented Aug 1, 2014 at 12:31

1 Answer 1

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Yes. This follows from Theorem 3.2 of my paper with Michael Boshernitzan, Gregori Kolesnik and Máté Wierdl, 'Ergodic Averaging Sequences'.

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