**Definition:** *Highly composite prime gap*


The three composite numbers between the consecutive primes $643$ and $647$ each have at least three distinct prime factors. This is the first occurrence of prime gap of length $> 1$ where each composite number in the gap has at least $3$ distinct prime factors. We call prime gap between $643$ and $647$ as the highly composite prime gap of order $3$. We have the highly composite prime gaps for of order $k$ for $k = 3,4,5,6$ and $7$ as follows:


* $k = 3; p = 643$
* $k = 4; p = 51427$
* $k = 5; p = 8083633$
* $k = 6; p = 1077940147$
* $k = 7; p = 75582271489$


**Question 1**: Are there infinitely many highly composite prime gaps of order $k \ge 3$?


**Question 2**: Given $k$ is there always a highly composite gap of order $k$?


An ordinary linear regression between $k$ and $\log p$ gives a surprisingly strong fit with $R^2 \approx 0.99915$. Although it is based on only six data points, this suggests a relationship of the form $p \sim ab^k$ forsome fixed $a$ and $b$.


**Definition**: *Maximal highly composite gap*


The maximal highly composite gap is defined as a prime gap which is longer than any previous gap and each composite in the gap has at least $3$ distinct prime factors. The longest such gap I have found is of $75$ consecutive composite between the primes $535473480007$ and $535473480083$.


**Question 3**: Are there arbitrarily long prime gaps in which each composite number in the gap has at least three distinct prime factors?


**Note**: This question was [posted in MSE six months ago](https://math.stackexchange.com/questions/4182509/are-there-arbitrarily-long-prime-gaps-in-which-each-number-has-at-least-three-di); it got many votes but not answer hence posting in MO.



Are there highly composite prime gaps?
Assuming the [prime tuples conjecture](https://en.wikipedia.org/wiki/Dickson%27s_conjecture), all of these questions have affirmative answers. For instance, one can use the Chinese remainder theorem to find $a,b$ such that the tuple
$$ an+b, \frac{an+b+1}{2^2 \times 5}, \frac{an+b+2}{3 \times 7}, \frac{an+b+3}{2 \times 11}, an+b+4$$
(for instance) are an admissible tuple of linear forms (in that they have integer coefficients, and for each prime $p$ there exist a choice of $n$ in which all forms are simultaneously coprime to $p$); concretely, one can take $a=2^2\times 3 \times 5 \times 7 \times 11 = 4620$ and $b=19$. The tuples conjecture then shows that there are infinitely many $n$ in which these forms are all simultaneously prime, giving infinitely many gaps of order $3$ and length $4$. Similarly for other gap orders and lengths.


On the other hand, unconditionally not much can be said. I think at our current level of understanding we cannot even rule out the ridiculous scenario in which every sufficiently large prime gap $(p\_n,p\_{n+1})$ contains a semiprime. (Given that semiprimes are denser than the primes, it is likely that *most* prime gaps contain a semiprime, but it is highly unlikely (and inconsistent with the prime tuples conjecture) that *all* of them do.) The "bounded gaps between primes" technology of Goldston-Yildirim-Pintz, Zhang, Maynard, and Polymath does provide prime gaps of short length, but the method also inevitably produces several almost primes in the vicinity of such gaps, and with our current techniques we cannot prevent one of these almost primes being a semiprime inside the prime gap.


