# Sorting of prime gaps

Let $$g_i$$ be the $$i^{th}$$ prime gap $$p_{i+1}-p_i.$$

If we rearrange the sequence $$(g_{n,i})_{i=1}^n$$ so that for any finite $$n$$, if the gaps are arranged from smallest to largest, we have a new sequence $$(\hat{g}_{n,i})_{i=1}^n.$$

For example, for $$n = 10,$$

$$(g_{10,i})_{i=1}^{10} = 1, 2, 2, 4, 2, 4, 2, 4, 6, 2;$$ $$(\hat{g}_{10,i})_{i=1}^{10} = 1, 2, 2, 2, 2, 2, 4, 4, 4, 6.$$

We can ask about the number of elements of the original sequence, which are fixed under the sorting operation. For the example above $$g_i = \hat{g}_i$$ for $$i = 1,2,3,5,8.$$

The number of elements fixed by sorting appears to grow in a familiar way. Letting $$f(n)$$ be the number fixed for given $$n$$:

$$\begin{array}{c | c | c | c | c | c | c | c | } n & 500 & 5000 & 50000 & 500000 & 1000000 & 2000000 \\ \hline f(n) & 82 & 614 & 4843 & 39072 & 74195 & 141777 \end{array}$$

For comparison, $$n$$ and $$\pi(n)$$ (number of primes not exceeding $$n$$) are:

$$\begin{array}{c | c | c | c | c | c | c | c |} n & 500 & 5000 & 50000 & 500000 & 1000000 & 2000000 \\ \hline \pi(n) & 95 & 669 & 5133 & 41538 & 78498 & 148933 \end{array}$$

and the ratio $$f(n)/\pi(n)$$ at least for the numbers above appears to approach 1.

So my question is whether this might be an indirect consequence of the prime number theorem?

Edit:

This process seems to approximate the number of primes on intervals generally. For example, if we sort the prime gaps from $$g_{120000}$$ to $$g_{127000}$$, 556 are fixed. And $$\pi(127000) - \pi(120000) = 600.$$ So maybe a better way of putting the question is, why might it be true that

$$f(g_m,g_{m+1},...,g_k) \approx \pi(k) - \pi(m)?$$

Some things I have looked at. The prime number theorem gives us that $$(g_n)$$ is not monotone. I don't have an idea to support the suggestion that $$f(n)/n \sim 1/\log n.$$ If the probability (loosely speaking) that $$g_k = \hat{g_k}$$ is $$1/\log n$$ then maybe this can be explained in terms of the multiplicity, in some average sense, of $$g_k$$ in a particular finite sequence of gaps.

It occurs to me that if the number of gaps less than $$n$$ of size (say) 2 is $$k(n),$$ it is not possible to understand the probabilities involved without knowing whether $$k(n)$$ is bounded as $$n$$ gets large. So maybe the idea can be shown to be incorrect but showing it to be correct might require answering the threshold question about gaps.

Taking the suggestion above as true, assuming infinite gaps of all lengths (a conjecture of Polignac) and letting $$c_k(n)$$ be the number of prime gaps of length $$k$$ for given $$n,$$ if we assume also something akin to a well-distributed property in the distribution of gaps of length $$k$$ (which follows in a qualified way from a conjecture of Hardy-Littlewood$$^1$$) we have a simplification of notation based on geometry:

$$f(n) \approx S = \frac{1}{n}\sum_{k = 1}^{n} c_k(n)^2\approx \frac{n}{\log n}.$$

If nothing else, the expression conveys the idea of an average. While it doesn't seem fortuitous I cannot justify it.

Some numbers:

$$\begin{array}{c | c | c |c | } n & S & \pi(n) & S/ \pi(n) \\ \hline 100 & 21.01 & 25 &0.8404 \\ \hline 1000 & 145.13 & 168 &0.8638\\ \hline 10000 & 1071.45 & 1229 &0.8718\\ \hline 100000 &8562.17& 9592& 0.8926 \\ \hline \end{array}$$

Edit:

Looking carefully at @Sophie's comment below, I get comparable results and so I think it is correct. Can't prove it.

$$^1$$ See Conjectures 3.2, 3.4 of this very readable note by Chris Caldwell (also, this updated 2021 note).

• Your notation looks very sloppy to me. You are using $n$ as the index variable for your sequence $(\hat{g}_n)$, and also as the maximum value that your index variable is allowed to take. You need something like $(\hat{g}_{n,i})_{i=1}^n$. Commented Oct 3, 2014 at 16:59
• A fun fact is that in your first example, the indices of fixed prime gaps are consecutive Fibonacci numbers. Commented Oct 24, 2018 at 6:19
• @SylvainJulien why care about a small example when the number of overlaps is forced.
– user645636
Commented Feb 26, 2019 at 17:42
• @CogitoErgoCogitoSum The OP is asking about the asymptotic behaviour of the fixed points of the gaps after reordering. The twin prime conjecture is equivalent to the limit of the ordered sequence being $(1,2,2,2,2,\dots)$ (with a suitable metric), but that is not what the OP is asking. The asymptotic behaviour of $f(n)$ is not determined by the limit of $\hat{g}_n$
– user765629
Commented Jul 7, 2020 at 1:58
• I suspect $f(n)$ is approximately $n/\log{n}$ for a random permutation of the prime gaps, so that sorting isn't special, but I haven't been able to prove it. Commented Jul 13, 2020 at 3:01