Are there infinite triples of consecutive integers whose numbers of factors are increasing? We know that there exists infinite numbers of integers $n$ such that $d(n)<d(n+1)$, where $d(n)$ is the number of positive divisors of $n$.
Question: are there infinite numbers of integers $n$ such that $d(n)<d(n+1)<d(n+2)$?
 A: Let $p>3$ be any prime such that $2p-1$ is also prime. Then $2p-1$, being prime, has two factors; and $2p$ has four factors. So the triple $(2p-1,2p,2p+1)$ satisfies your condition as long as $d(2p+1)>4$.
$2p+1$ is divisible by $3$ (because the preceding two integers aren't). So $2p+1=3q$ for some $q$; and this number has at least six factors unless $q$ is prime or is itself a multiple of $3$. If $q$ is a multiple of $3$, then $2p+1=r\cdot 3^k$ for some $k\ge 2$ and some $r$ not divisible by $3$, and $d(2p+1)=(k+1)d(r)$. This is $\ge 5$ unless $r=1$ and $k \le 3$; but these two cases (i.e. $2p+1=9$ or $27$) are ruled out because $8$ is not twice a prime, and $25$ is not prime.
Hence if $p>3$ and


*

*$p$ is prime;

*$2p-1$ is prime;

*$(2p+1)/3$ is composite,


then $(2p-1,2p,2p+1)$ satisfies your conditions.
Such triples are easy to find, but it is beyond my capabilities to prove that there are an infinite number of them. I suspect that it is beyond anyone's capabalities at our current state of knowledge.
