# How much do these interval collections cover?

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.

• Have you calculated some initial segment of $\cal C$? What does it look like? Nov 3, 2016 at 1:20
• I have only done some approximations to C, which are incomplete. Using the first 50ish primes, I cover all but 205 of the first 1000 and all but 859 of the first 10000 numbers. I wouldn't mind someone posting approximate results. From my limited trials, pq - p seems uncovered for p and q the two largest primes. Gerhard "Which Is Not That Surprising" Paseman, 2016.11.02. Nov 3, 2016 at 1:30
• Also, the title question asks how much. I would be still interested in a lower bound on density results, especially if the answer was something like the $n$th number missed is $\Omega(n^k)$ for some power $k \gt 1$. Gerhard "Assuming Density Should Be One" Paseman, 2016.11.02. Nov 3, 2016 at 1:42
• Non-covered numbers form a good candidate for a sequence in the OEIS. Would you be interested to add it? To test whether $n$ is covered, it's enough to test if $n+k=pq$ for some $k<\sqrt{n}$ (and $q>p>k$). Nov 3, 2016 at 2:10
• @Max If you refer to this post, I don't mind if you enter the sequence. If you want more background, you can send me email and I can give you some motivation for the question. Gerhard "Or You Can Give It" Paseman, 2016.11.03. Nov 3, 2016 at 7:07

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.