# Is $\lfloor \log(n!)\rfloor \alpha$ equidistributed on the unit circle?

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?"

• Could you replace the vertical bars with $\lfloor\dots\rfloor$ floor signs to make the notation more recognizable? – Zsbán Ambrus Mar 5 '14 at 11:22
• Or with $[\dots]$ for those of us a little less hip. – Gerry Myerson Mar 5 '14 at 11:59
• @ZsbánAmbrus: done. – Ian Morris Mar 5 '14 at 12:27
• Would you agree to explain why $\log(n!)\alpha$ equidistributed? – Lior Bary-Soroker Jul 28 '14 at 11:03
• @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. – Trevor Hansen Aug 1 '14 at 12:31