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Define for $n \in \mathbb{N}$ the function $$\tau_1(n):=\sum_{\substack{d|n, \\ d+1|n}}1,$$ i.e. the number of consecutive divisors of an integer. The average of $\tau_1(n)$ is $1$ since $$\sum_{n\leq x}\tau_1(n)=\sum_{d<\sqrt{x}}\Big[\frac{x}{d(d+1)}\Big]=x+O(\sqrt{x}).$$ I was wondering whether more is known about $\tau_1(n)$, for example does it have a distribution function, meaning does the following limit exist for all $z \in \mathbb{R}$, $$F(z):=\lim_{x \to +\infty}\frac{1}{x}\#\{n\leq x \ \text{such that} \ \tau_1(n)\geq z\}? $$

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  • $\begingroup$ Of course a suitably modulated question can be posed for $\tau_f(n):=\sum_{[f(d)]|n}1$, where $f:\mathbb{N}\to \mathbb{R}$ is any function, the previous example having $f(d)=d(d+1)$. $\endgroup$ – Captain Darling May 24 '15 at 23:59
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Yes, $F(z)$ exists for all $z$. Let $y\ge 2$, and define $\tau_{1,y}(n)$ as the number of $d\le y$ for which both $d,d+1$ divide n. Let $F_y$ denote the analogue of $F$ with $\tau_{1,y}$ replacing $\tau_1$. It's clear that $F_y(z)$ exists for all $z$, since $\tau_y(n)$ is actually a periodic arithmetic function (with period the lcm of $1, 2, \dots, \lfloor y\rfloor +1$). Moreover, $F_y(z)$ is increasing as a function of $y$, and $F_y(z) \le 1$. So for each $z$, the limit $\lim_{y\to\infty} F_y(z)$ exists.

Claim: The limit definining $F(z)$ coincides with $\lim_{y\to\infty} F_y(z)$.

If $n \le x$ and $\tau_1(n) \ge z$, then either $\tau_{1,y}(n) \ge z$, or $n$ is divisible by some $d(d+1)$ with $d > y$. Hence, $$ \#\{n \le x: \tau_1(n) \ge z\} \le \#\{n \le x: \tau_{1,y}(n) \ge z\} + O(x/y). $$ Divide by $x$, and take the limsup as $x\to\infty$, and then let $y\to\infty$. This shows that the $\limsup$ of the expression in the definition of $F$ is at most $\lim_{y\to\infty} F_y(z)$. On the other hand, since $\tau_{1,y} \le \tau_{1}$, a similar but simpler argument shows that the $\liminf$ in the definition of $F$ is at least $\lim_{y\to\infty} F_y(z)$.

It seems this was known to Erdős and Hall; in their paper, they remark "It is easy to see that the density of integers $n$ for which $\tau_k(n)=r$ exists." Probably they had the above argument in mind.

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  • $\begingroup$ That is rather great, could you please link their paper ? could not find it... $\endgroup$ – Captain Darling May 25 '15 at 21:45
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    $\begingroup$ Here's a PDF: renyi.hu/~p_erdos/1978-26.pdf $\endgroup$ – so-called friend Don May 25 '15 at 22:01
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    $\begingroup$ Ok so the probability distribution result is valid with the same proof even taking any polynomial $f(d)$ with integer coefficients in place of $d(d+1)$ and such that the leading coefficient is positive and the degree of $f$ is greater than $1$. It might be interesting to try and figure explicit formulas for $F(z)$... $\endgroup$ – Captain Darling May 25 '15 at 22:13
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See Katalin Gyarmati, On the density of integers with consecutive divisors, Publ. Math. Debrecen 74 (2009), no. 1-2, 1–17, MR2490419 (2010e:11087).

The first paragraph of the review:

P. Erdős and R. R. Hall [J. Austral. Math. Soc. Ser. A 25 (1978), no. 4, 479–485; MR0506088 (58 #21975)] first studied the number of consecutive divisors. Let $k\ge2$ be a fixed integer; the function $\tau_k(n)$ is defined as the number of positive divisors of the form $m(m+1)\cdots(m+k−1)$ with $m\ge1$. Erdős and Hall proved that the estimate $\tau_k(n)>(\log n)^{e^{1/k}−\epsilon}$ holds for infinitely many $n$. They also proved that $$\sum_{n\le x}\tau_k(n)={x\over(k-1)(k-1)!}+O(x^{1/k})$$

The review goes on to discuss the work of Tenenbaum and other authors, and then gets to Gyamarti's paper, which is concerned with $$A_P(K)=\lim_{N\to\infty}\#\{\,n:1\le n\le N,P(m)\mid n{\rm\ for\ some\ }m\ge K\,\}$$ and finds asymptotics for $P$ a quadratic polynomial with integer coefficients.

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  • $\begingroup$ The reviews indicate that there is work by Tenenbaum and dL Breteche for bounding $\tau_k$ by a power of $\tau$ which is smaller than $1$. The asymptotic estimate is also rather straightforward. But it gives the interesting result that the average number of more than one consecutive divisors is $\int_0^1 \frac{e^x-1}{x}dx$.I am wondering whether it is any easy to prove asymptotics for the higher moments of $\tau_2$. $\endgroup$ – Captain Darling May 25 '15 at 1:42
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This supplements the two previous answers, giving further references that provide information on $F(z)$. Theorem 2 of Konyagin and Soundararajan shows that there exist arbitrarily (square-free) large integers $N$ with $$ \# \{ d: d(d+1)| N \} \ge \exp ((\log N)^{\frac 1{16}}). $$ Therefore for large $z$, there exists an integer $N$ below $\exp( (\log z)^{16})$ with $\# \{ d: d(d+1)|N \} \ge z$. Since all multiples of $N$ have at least $z$ consecutive divisors, it follows that $$ F(z) \ge \exp(- (\log z)^{16}) $$ for large $z$. This improves substantially upon the results of Erdos and Hall, who had a bound like $F(z) \ge \exp(-z^{1/\sqrt{e}+o(1)})$. Harper's paper could probably be adapted slightly to improve the $16$ in the result of Konyagin and Soundararajan to $6$, and further improvements could also be made. One might guess that $F(z)$ (see the heuristic in the paper of Konyagin and Soundararajan) should be something like $\exp(-(\log z)^{2+o(1)})$, but this appears difficult.

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  • $\begingroup$ That is interesting, doesn't the last guess suggest that the distribution of $\tau_1(n)$ should have a heuristic behavior conforming to the gaussian law ? $\endgroup$ – Captain Darling May 26 '15 at 16:35
  • $\begingroup$ @CaptainDarling: Note it's $\exp(-(\log z)^{2+o(1)})$, not like $\exp(-z^2/2)$ (Gaussian). I don't expect it to be anything nice. $\endgroup$ – Lucia May 26 '15 at 16:46
  • $\begingroup$ Indeed you are right, however my implication was that considering $F\left(\exp\left(\frac{t}{\sqrt{2}}\right)\right)$ would lead to a probabilistic interpretation, after all the events $d|n$ and $d+1|n$ are morally independent. $\endgroup$ – Captain Darling May 26 '15 at 17:08

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