TITLE: On the number of consecutive divisors of an integer
QUESTION [14 upvotes]: 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 $z \in \mathbb{R}$,
$$F(z):=\lim_{x \to +\infty}\frac{1}{x}\#\{n\leq x \ \text{such that} \ \tau_1(n)\geq z\}?
$$

REPLY [4 votes]: 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.