Stepwise irregularity of the divisor function, or does $\limsup_{n\to\infty} |d(n+1)-d(n)| = \infty$?

Question

Defining $d(n):= \sum_{d|n} 1$, I know that $\liminf_{n\to\infty} d(n) = 2$, and Wigert showed that $\limsup_{n\to\infty} \frac{\log(d(n))}{ \log n/\log\log n} = \log 2$ (natural logarithm). This presents a decent picture of how irregular the divisor function is, but I want to know how suddenly it jumps for adjacent values of $n$. The Wikipedia page on the divisor function tells us that Heath-Brown proved in 1984 that $\liminf_{n\to\infty} |d(n+1)-d(n)|= 0$. Are there results about $\limsup_{n\to\infty} |d(n+1)-d(n)|$, or the behavior of the average $\frac 1N \sum_{n=1}^N |d(n+1)-d(n)|$?

Answer 1

Let $\omega(n)$ denote the number of distinct prime factors of $n$. A theorem of Halberstam states that for any fixed irreducible polynomial with integer coefficients $f(x)$ that is not identically a multiple of $x$ satisfies the asymptotic law

$$\sum_{p<x}\omega(f(p))\log p\sim x\log \log x$$

where $p$ runs over primes. Applying this with $f(x)=x+1$ we get that $p+1$ has on average $\log\log p$ distinct prime divisors, and hence on average $d(p+1)\geq 2^{\log \log p}=\log(p)^{\log(2)}$. Thus, we can make the gaps $|d(n+1)-d(n)|$ grow as large as we want, and at least as fast as $\log(n)^{\log(2)}$.