How much do these interval collections cover?

Question

As usual any related references are appreciated.

Let $p \lt q$ be distinct primes, and for all such pairs, let $m=pq$ and let $\cal{C}$ be the collection $(m-p,m)$ of open intervals. Does (the union of) this collection cover all but finitely many natural numbers?

Suppose we extend the collection to $\cal{D}$ by including $(m-p,m)$ where $m=p^a q^b$, so $m$ ranges over all natural numbers with $\omega(m)=2$ and $p$ is the smallest prime factor of $m$. Does $\cal{D}$ cover all but finitely many natural numbers?

Pick a natural number $t\gt 2$ and consider analogous collections of square free (or not) $m$ with $\omega(m)$ different from 1 and equal to (or at most ) $t$, again with $p$ prime and least dividing $m$, and again just using open intervals of the form $(m-p,m)$. Is there a $t$ such that one of these variations on $\cal{C}$ covers all but finitely many natural numbers?

Gerhard "Yes, It Concerns Jumping Primes" Paseman, 2016.11.02.

Answer 1

Small progress report.

Spurred by Max Alekseyev's comment, I have calculated an initial segment of the complement of $\cal{C}$ in the natural numbers up to over a million, and intend to go further when I make a correct and less stupid algorithm. The complementary sequence starts with (using range notation) 1-4 6-8 10-12 15-18 22-24 26-30 35-36 39-44 46-48 58-60 65-66 69-70 77-80 95-104 and continues on for about almost 450 numbers to 33509-33510. Then nothing until the end of the run. This suggests that $\cal{C}$ (and each variant that contains $\cal{C}$) covers all but finitely many numbers, with finitely many being less than 500.

I find this mildly surprising and would appreciate independent verification. It suggests to me that the sequence of semi-primes (6, 10, 14, 15, ...) have gaps that are smaller than the square root of the semiprime and that the literature might show 33510 is the largest uncovered number.

Gerhard "It Knocks My Socks Off" Paseman, 2016.11.04.

Answer 2

Equivalently we can define a covered number $n$ as such that there exists a prime $p<\sqrt{n}$ such that $p\nmid n$ and $\lceil n/p\rceil$ is also prime. Indeed, taking $q:=\lceil n/p\rceil$ we have that $$pq - p = p(q-1) < n < p\lceil n/p\rceil = pq,$$ that is, $n$ is covered by the interval $(pq-p,pq)$.

This definition implies that as $n$ grows, the heuristic "probability" for it be covered grows as well: there exist about $2\sqrt{n}/\log(n)$ candidate primes $p$, and for each of them the "probability" of $\lceil n/p\rceil$ being prime is at least about $1/\log(n)$. Hence, $n$ would not be covered with the "probability" $$(1-1/\log(n))^{2\sqrt{n}/\log(n)} \approx \exp(-2\sqrt{n}/\log(n)^2).$$ Furthermore, integrating over $n$ suggests that we should expect no more than about 5,000 non-covered numbers. This is well consistent with the practical experiments showing there are likely only 493 of those.