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Oct 26, 2018 at 23:32 history edited Vesselin Dimitrov CC BY-SA 4.0
TeX-ed and mildly edited the wording and the title
Oct 26, 2018 at 23:29 answer added Vesselin Dimitrov timeline score: 3
Oct 26, 2018 at 22:32 comment added Gerry Myerson Didn't you just post this question two days ago, Alex? mathoverflow.net/questions/313685/…
Oct 26, 2018 at 18:33 comment added Seva @VesselinDimitrov: why not convert your comments into an answer (expanding them and adding more details)?
Oct 26, 2018 at 16:58 comment added Vesselin Dimitrov (PS: I missed the word "odd" in your question, but that makes no difference in my comments. Insert the condition $n \equiv 1 \mod{2}$ in the Bernoulli sum above. The answer should be the same with $n$ restricted to any arithmetic progression, or to the primes of an arithmetic progression, or...)
Oct 26, 2018 at 16:47 comment added Vesselin Dimitrov A much stronger conjecture of Chowla and Wallum, generalizing the Dirichlet divisor problem ("On the divisor problem," Proc. Symp. Pure Math., vol. VIII, 1965) predicts this last sum with the periodized Bernoulli function to be $\ll_{k,\epsilon} N^{\frac{1}{4}+\epsilon}$, for every fixed $k$ and all $\epsilon > 0$.
Oct 26, 2018 at 16:38 comment added Vesselin Dimitrov Asymptotically equidistributed in the Lebesgue measure of $[0,1]$. You may try for each $k = 1, 2, \ldots$ to compute the $N \to \infty$ limit of $N^{-1/2} \sum_{n = 1}^{\lfloor \sqrt{N} \rfloor} \{ N/n\}^k$. Proving this $= 1 / (k+1)$ (the moments of the Lebesgue measure) for all $k = 1,2, \ldots$ is enough for equidistribution. It may be more convenient to prove this in the equivalent form $\sum_{n < \sqrt{N}} B_k( \{N/n\} ) = o(\sqrt{N})$ for all $k = 1,2, \ldots$.
Oct 26, 2018 at 16:08 history asked Alex CC BY-SA 4.0