There is at least one number with $k$ different prime factors in Mersenne sequence, right?

The question(s) of infinitude of Mersenne primes (and semiprimes), that is, is there an infinite number of them in Mersenne sequence are, to the best of my knowledge, still unsolved.

We could ask are there an infinite number of composites in Mersenne sequence that consists of exactly $k\geq 2$ prime factors (either with repetitions allowed or non-allowed) for every $k$, but as the case $k=2$ is unsolved probably all other cases are not much simpler than the $k=2$ case.

But it is not my intention here to discuss about infinitude, I will in this question restrict ourselves with finitude.

We have at least one prime in Mersenne sequence, for example $2^2-1=3$. We have at least one number with two different prime factors in Mersenne sequence, for example $2^4-1=3 \cdot 5$. We have at least one number with three different prime factors in Mersenne sequence, for example $2^8-1=3 \cdot 5 \cdot 17$...

So, I would like to ask:

Is it true that for every integer $k\geq 2$ there exists Mersenne number with exactly $k$ different prime factors?

I believe that this is true and that Mersenne sequence has this kind of "richness" in itself, but I am not sure how to prove this, and, to be honest, I really did not think much on how would/could I prove this, I just "created" the question and decided to ask you.

• See if this helps you: arxiv.org/pdf/1606.08690.pdf – Rohan Dec 29 '17 at 12:29
• @Rohan It could help, although it seems that there this question of mine is not proven, is it? I just went through the abstract. – user480281 Dec 29 '17 at 12:33
• Though I must say it was a very logical and interesting question. – Mohammad Zuhair Khan Dec 29 '17 at 12:34