On the number of values with exactly $k$ prime factors of a given polynomial

Question

This is surely be a well studied problem. Let $f(x) \in \mathbb{Z}[x]$. Is there some $k \in \mathbb{N}$ such that there are infinitely many $n \in \mathbb{Z}$ where $f(n)$ has exactly $k$ prime factors?

It is clear that this question is related to many open problems and known results. Obviously we have Dirichlet's theorem, where $f(x) = ax + b$, with $a,b$ coprime. Then the case for $k=1$ holds true. (What about $k > 1$?)

Then we have $f(x) = x^2 -1$, and solving the case for $k=2$ solves the twin prime conjecture.

Then for certain irreducible polynomials, again with the case for $k=1$, this implies the Bunyakovsky conjecture, with the famous unsolved example being $f(x) = x^2 + 1$.

These are all questions with fixed small numbers of prime factors. I'm wondering if, in particular for degree at least $2$, can we even find any number $k$ such that some polynomial gives infinitely many values with exactly $k$ prime factors?

This problem seems to have the same flavor as the results of Zhang and Maynard about fixed (large) gaps between primes occurring infinitely many times. What is already known about this problem?

Answer 1

It is well-known from sieve theory, in fact such a result can be obtained even if one just uses Brun's original combinatorial sieve, that for any irreducible $f \in \mathbb{Z}[x]$ of degree $d$ that one can find infinitely many $n \in \mathbb{Z}$ such that $f(n)$ has at most $d+1$ prime factors. See, for example, Theorem 25.4 in Opera de Cribro by Friedlander and Iwaniec. This is by no means a new result, for example Hooley had already written about it in his book on sieves.

Getting exactly $k$ prime divisors, however, is quite a bit more challenging. Surely it can be achieved for some (especially large, with respect to the degree $d$) values of $k$, but the proof will be somewhat unenlightening: one will have to run an upper bound sieve on the number of integers $n \in [1,X]$ for which $f(n)$ has at most $k-1$ prime factors, and a lower bound result on the number of $n \in [1,X]$ for which $f(n)$ has a most $k$ prime factors, and then take the difference and hope that it's positive. The "right" way to prove results like this, i.e., to get an asymptotic formula, is just as hard for semi-primes as it is for primes since it requires non-trivial Type II information, which we usually do not have.

Answer 2

For $f(x) = ax + b$ with $a, b$ coprime all $k$s are obtained an infinite amount of times - just take some prime of the form $ax + b$, and multiply it by any number of primes of the form $ax + 1$. I expect their density are $\frac{1}{\varphi(a)}$th of all $k$-almost primes, although I didn't prove that. In general, even if $a,b$ aren't coprime you can achieve every $k$ greater than $\Omega(\gcd(a,b))$ infinitely many times.