Let $k \in \mathbb{Z}^+$.
Is it possible to prove that, for some given
$m \in \{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23\}$,  

there are only finitely many $k$ such that the closed interval $[k, k+99]$ contains (exactly) $m$ prime numbers?


Backstory: years ago, we wrote an informative article, in Italian, entitled "Quanti numeri primi in 100 interi consecutivi?" (see Matematicamente.it Magazine, Vol. 25(228), 2015 [1]), easily showing the existence of only a finite number of $k$ such that there are more than $23$ primes in $[k, k+99]$ (i.e., if $m \geq 24$, then $2 \leq k \leq 17$) and confirming the existence of infinitely many $k$ such that $m=0$ and $m=1$ (see Lemma 4 from [1]).


Moreover, it seemed reasonable to us to conjecture the existence of infinitely many $k$-tuples such that $m \in [2, 23]$.
**Question:** Is it possible, after more than $7$ years, to prove/disprove (at least) the conjecture above?


Lastly, in [1], we noted that it would be sufficient to prove just a single case of Polignac's conjecture up to a prime gap of $98$ (starting from a prime gap of $2$, then moving on $4$, and so forth, up to $98$) in order to confirm the conjecture, and we also wrote that we were inspired by prof. Tao's article entitled "Polymath8b: Bounded intervals with many primes, after Maynard" (19 Nov. 2013).



Infinitely many $k \in \mathbb{N}$ such that the closed interval $[k, k+99]$ contains from $2$ to $23$ prime numbers
It is compatible with current knowledge that all prime gaps are greater than 100\*, so it's not possible currently to show that for any of m in {2, ..., 23} that there are infinitely many such intervals (though it is surely true). Cases 0 and 1 follow from PNT (or even just Chebyshev's theorem on prime density) and Dirichlet's theorem.


\* The best current bound is 246.


