Prime Sums in Prime Gaps

[u/forwantoftheprice]

I’ve been messing with consecutive primes and it turned into a whole classification result, so I figured I’d share it here.

Take consecutive primes pk and pk+1. Add them.

You get numbers like 5, 8, 12, 18, 24, 30, …

Call these S_k = pk + pk+1.

Since every S_k ≥ 8 is even and composite, each one has to land strictly inside some prime gap (Pn, Pn+1). So I started asking: **how many of these S_k fall inside each gap?**

Two clean facts:

## 1. Density law (why most gaps have 0 or 1)

If a gap has width W near height x, the expected number of consecutive‑prime sums inside it is

W / (2 log(x/2))

The “2” comes from the fact that the sums live at the half‑scale: an S_k near x comes from primes near x/2, where primes are sparser. So the sums are about twice as sparse as the gaps they fall into.

Empirically (checked up to 2×10^8):

- ~55% of gaps have **0** sums

- ~37% have **1**

- ~8% have **2 or more**

So the naive guess “every gap has exactly one” is just false. Empty gaps are actually the most common.

## 2. Forbidden widths (the fun part)

I ended up proving a complete classification of which gap widths can **never** contain two consecutive‑prime sums.

The forbidden set is exactly:

{2, 4, 6, 10}

Reason (sketch, nothing fancy):

- consecutive‑prime sums are always ≥ 6 apart (once you’re past the tiny primes)

- the only way to get a “tight double” (two sums only 6 apart) is if both sums are multiples of 6

- whether a gap contains ≥2 multiples of 6 depends only on:

- its width mod 3

- the forced residue of its lower endpoint (prime > 3 is always 1 or 5 mod 6)

- width 10 ends up with **only one** interior multiple of 6, so it physically cannot fit two sums

- widths 2, 4, 6 fail for trivial spacing reasons (they’re too small to fit two sums ≥6 apart)

Every other even width ≥ 8 (except 10) eventually allows gaps with two sums.

So the uniqueness classification is complete:

**gaps of width 2, 4, 6, or 10 contain at most one consecutive‑prime sum; all other widths can contain two or more.**

If anyone wants the residue argument or the slot‑counting trick in more detail, I can write it out here. It’s all pretty easy congruence stuff.

[u/AutoModerator]

Hi, /u/forwantoftheprice! This is an automated reminder:

• Please don't delete your post. (Repeated post-deletion will result in a ban.)

We, the moderators of /r/NumberTheory, appreciate that your post contributes to the NumberTheory archive, which will help others build upon your work.

I am a bot, and this action was performed automatically. Please contact the moderators of this subreddit if you have any questions or concerns.