$n$ consecutive composites that have exactly two prime factors

Question

user477343 found an example of three consecutive composites $1041,1042,1043$ with exactly two prime factors.

It came as a surprise that they also found four consecutive composites $445=5 \cdot 89,446=2 \cdot 223,447=3\cdot 149,448=2^6 \cdot 7$ with exactly two prime factors.

Can one find $n$ consecutive composites with exactly two prime factors for some other values of $n \in \mathbb N \setminus \{1\}$? Do they exist for every $n \in \mathbb N \setminus \{1\}$?

Answer 1

The article Consecutive Integers with Equally Many Principal Divisors by Roger B. Eggleton and James A. MacDougall is about this subject. While the exact detailed answer to your question is unknown (at least to those authors at the time of writing), the authors prove that such a sequence of consecutive integers -- they call it a "run in $P_2$", where the 2 stands for 2 distinct prime factors -- has length at most 9. Furthermore, they demonstrate examples of runs of length 8, namely the run from 141 to 148 and the run from 212 to 219. They further conjecture that these are the only runs of length 8.

Answer 2

Partial answer: 54, 55, 56, 57, 58 is a solution for $n=5$. As is 91, 92, 93, 94, 95 or 92, 93, 94, 95, 96.

So we have 91, 92, 93, 94, 95, 96 for $n=6$.

Also 141, 142, 143, 144, 145, 146, 147 is a solution for $n=7$.

And 141,..., 148 is one for $n=8$.