Are there arbitrarily long prime gaps in which each number has at least three distinct prime factors?

Question

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 $k = 3$ 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 $k = 3,4,5$ and $6$ 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.

Update: 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?

Update: Posted in MO since it is unanswered in MSE.

Answer 1

Yes, there are arbitrarily long sequences of integers each of which has at least $3$ prime factors, and the same is true for any $k$ in lieu of $3$.

Given arbitrary positive integers $k$ and $m$, choose integers $n_1,n_2,\ldots,n_m$ with the following properties:

1. Each $n_i$ has at least $k$ distinct prime factors;
2. The $n_i$ are pairwise relatively prime.

It is obvious that such a sequence of integers exists (just pick a batch of distinct primes for each $i$, and multiply them).

Now, using the Chinese Remainder Theorem we can find an integer $N$ such that for each $i$, $N \equiv -i \pmod{n_i}$, or in other words $N+i \equiv 0 \pmod{n_i}$. Therefore, the numbers $N+1, N+2, \ldots, N+m$ all have at least $k$ distinct prime factors.

EDIT: the question seems to further demand that the long stretch of “very composite” integers be bookended by actual primes. This proof does not guarantee that, and in fact I suspect this is very hard to guarantee.