# The distribution of fractional parts $\Big\{ \frac{N}{n} \Big\}$

Let $$N$$ be a large integer. What is known about distribution of fractional parts $$\Big\{ \frac{N}{n} \Big\} \in [0,1)$$ after division of $$N$$ by all odd numbers $$n$$ in the range $$3 \leq n < \sqrt{N}$$?

• 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 $k = 1,2, \ldots$. – Vesselin Dimitrov Oct 26 '18 at 16:38
• 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 $\epsilon > 0$. – Vesselin Dimitrov Oct 26 '18 at 16:47
• (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...) – Vesselin Dimitrov Oct 26 '18 at 16:58
• @VesselinDimitrov: why not convert your comments into an answer (expanding them and adding more details)? – Seva Oct 26 '18 at 18:33
• Didn't you just post this question two days ago, Alex? mathoverflow.net/questions/313685/… – Gerry Myerson Oct 26 '18 at 22:32

As $$N \to \infty$$, the set of fractional parts $$\Big\{ \frac{N}{n}\Big\}, \quad 1 \leq n < \sqrt{N}$$ becomes asymptotically equidistributed in the Lebesgue measure of $$[0,1]$$. The same persists with the congruence restriction $$n \equiv 1 \mod{2}$$ asked by the OP, or indeed with $$n$$ running through any fixed arithmetic progression. The same equidistribution answer holds moreover in some other natural variants, such as taking $$n$$ to run through the primes of a fixed primitive arithmetic progression. Regarding this last point I happen to recall de la Vallee Poussin being amused in an 1898 paper by his observation (propriete assez curieuse; un rapprochement tres remarquable) that the mean value $$1-\gamma$$ of fractional parts $$\Big\{\frac{N}{an+b} \Big\}$$, $$an + b \leq N$$ was the same for all arithmetic progressions $$an+b$$, as well as for the prime values $$an+b$$ when $$(a,b) = 1$$. (Of course, $$\sum_{n \leq N} \frac{\Lambda(n)}{n} - N^{-1}\log{N!} = \frac{1}{N}\sum_{n \leq N} \Big\{\frac{N}{n}\Big\} \Lambda(n) \sim 1-\gamma$$ is just the logarithmic form $$\sum_{n \leq N} \frac{\Lambda(n)}{n} = \log{N} - \gamma +o(1)$$ of the prime number theorem, easily equivalent to the usual forms $$\psi(N) \sim N$$ and $$\pi(N) \sim N / \log{N}$$.)
Why equidistribution? For simplicity of notation, let me omit the congruence condition on $$n$$; in practice these are handled straightforwardly as in de la Vallee Poussin's paper, in every situation where we can solve the problem without a congruence condition. It is enough to prove the moment asymptotic $$\frac{1}{\sqrt{N}} \sum_{n < \sqrt{N}} \Big\{ \frac{N}{n} \Big\}^k \sim \frac{1}{k+1} = \int_0^1 t^k \, dt, \quad k = 1,2, \ldots,$$ or what amounts to the same, the estimate $$\sum_{n < \sqrt{N}} B_k\Big(\Big\{ \frac{N}{n} \Big\} \Big) = o(\sqrt{N})$$ for the sum of values of the periodized Bernoulli function $$B_k(\{t\})$$, for each $$k = 1, 2, \ldots$$. This is now a more natural form of posing the problem, for there is a conjecture of S. Chowla and H. Walum ("On the divisor problem," Proc. Symp. Pure Math., vol. VIII, 1965, pp. 138-143) that asserts the much stronger and best-possible ("square root cancellation") bound $$\sum_{n < \sqrt{N}} B_k\Big(\Big\{ \frac{N}{n} \Big\} \Big) \ll_{k,\epsilon} N^{\frac{1}{4} + \epsilon},$$ for any fixed $$k \in \mathbb{N}$$ and $$\epsilon > 0$$. This is one way to generalize the celebrated Dirichlet divisor conjecture, which is the case $$k = 1$$ with $$B_1(\{t\}) = \{t\} - 1/2$$. The Chowla-Walum conjecture thus implies, moreover, a quantitative estimate on the discrepancy (rate of equidistribution) of our fractional parts. It is discussed, for example, in this 1982 paper of R.A. MacLeod.
I have not given you the proof of the equidistribution, but I hope this can convince you of the answer to your question and how you may approach it. I would guess that Voronoi's method for the Dirichlet divisor problem could be adapted to prove the crude $$o(\sqrt{N})$$ bound needed for the qualitative equidistribution. A more precise study of the distribution of fractional parts $$\{N / n\}$$, $$n \leq y$$ is done in these papers of Saffari and Vaughan, but the proved results are very far from the discrepancy prediction that the Chowla-Walum conjecture implies on your question.