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 $z \in \mathbb{R}$,
$$F(z):=\lim\_{x \to +\infty}\frac{1}{x}\#\{n\leq x \ \text{such that} \ \tau\_1(n)\geq z\}?
$$



On the number of consecutive divisors of an integer
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.


