# 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? – Gerry Myerson Nov 3 '16 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. – Gerhard Paseman Nov 3 '16 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. – Gerhard Paseman Nov 3 '16 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 $q>p>k$). – Max Alekseyev Nov 3 '16 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. – Gerhard Paseman Nov 3 '16 at 7:07

## 1 Answer

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.

• My improved algorithm seems to run slow, suggesting no uncovered numbers in (33510,99999997). I am looking for research on semiprime gaps, but have not found good unconditional results yet. There are results that suggest density 1 for covered numbers, but I haven't done the verification yet. Gerhard "Putting My Socks Back On" Paseman, 2016.11.06. – Gerhard Paseman Nov 7 '16 at 3:36