It is probably open can we for every $k \in \mathbb N$ find two composites $a_k$ and $b_k$ such that $a_k$ and $b_k$ have exactly $k$ prime factors and $a_k-b_k=1$.
Smallest examples found so far are:
for $k=1$ $$3^2-2^3=1$$ for $k=2$ $$3 \cdot 5 - 7 \cdot 2=1$$ for $k=3$ $$3\cdot7\cdot11-2\cdot5\cdot23=1$$ for $k=4$ $$5 \cdot 7 \cdot 11 \cdot 19 - 2 \cdot 3 \cdot 23 \cdot 53=1$$ for $k=5$ $$3\cdot7\cdot17\cdot23\cdot31-2^2\cdot5\cdot11\cdot13\cdot89=1$$
It is easy to see that $3$ is a factor of one of numbers that are smallest pair for $k=1,2,3,4,5$ so I will make a pretty dumb conjecture that could be ruled out with a clever computer-check:
If a pair $(a_k,b_k)$ is smallest pair then $3$ is a factor of one of those two numbers.
Until which $k$ is this true?
I am prepared to let this always be true because smallest pairs should tend to have small primes as factors of them and because there is a good chance that one of members is divisible by $3$ because they differ only by one.
I am not sure would I like to see a counterexample, but go for it.
This is true at least for $k=1,2,3,4,5,6,7,8,9,10,11$ by this list .
A same question asked on MSE where there is an answer by Oleg567 proposing smallest examples for $k=12,13,14,15,16$