Why is there an unexpected increase in the density of certain types of Goldbach primes?

Question

Note: Posted in MO since it was unanswered in MSE.

I was checking how quickly we can verify Goldbach's conjecture for a given even number $n$ and it was clear that searching backward starting from the largest prime below $n$ was much faster than a forward search starting from the small odd prime $3$. I observed the following.

Consider the set of even numbers $n$ which can be written as the sum of two primes and $p$ be the largest prime less than $n$ such that $n-p$ is also a prime. Let $p_k$ be the $k$-th prime. Then, $p_{\pi(n)-2}$, $p_{\pi(n)-1}$ and $p_{\pi(n)}$ are the three largest primes preceding $n$. Now $p$ can either be the largest prime below $n$, or the prime just before the largest prime or the prime before that and so on. Accordingly, we define:

• $d_1$ is the density of $n$ such that $p = p_{\pi(n)}$
• $d_2$ is the density of $n$ such that $p = p_{\pi(n)-1}$
• $d_3$ is the density of $n$ such that $p = p_{\pi(n)-2}$

Data for $n \le 3 \times 10^{10}$ shows that $d_1$ and $d_2$ decreased as $n$ increased. This was intuitively expected as explained in the answer for the related question "If $p$ is the largest prime less than $2n$, what is the probability that $2n-p$ is a prime?". However, quite counter intuitively, $d_3$ increased with $n$ as shown in the graphs below.

Question 1: Why does $d_3$ increase with $n$ while $d_1$ and $d_2$ decreased?

Question 2: What is the limiting value of $d_3$? Since $d_3 < 1$ and is increasing, it must converge to a positive limiting value.

Note: I have not computed $d_k$ for $k \ge 4$ so I do not know if this increasing trend is observed other values of $k$.

Graphs in normal scale

Answer 1

Disclaimer: this is not a real answer but rather a heuristics that is too long for a comment.

Building upon my comment, denote by $k_{0}(m):=\pi(m+r_{0}(m))-\pi(m-r_{0}(m))$ and say $m$ is $K$-central if $k_{0}(m)=K$. Under some reasonable hypotheses, the number of $K$-central numbers not exceeding $x$ is asymptotically $\frac{\pi(x)}{K}$ for positive $K$ (edit: see this question of mine: Would the following conjectures imply $\lim\inf_{n\to\infty}p_{n+k}-p_{n}=O(k\log k)$?).

Your observation suggests one has $k_{0}(n/2)/2\approx 3$ hence $k_{0}(n/2)\approx 6$. As you count half the primes below $n$ summing to $n$, namely the smallest summands, one can expect a density of $\frac{1}{2}\frac{\pi(n)}{6\pi(n)}=1/12$ so I expect the limit of $d_{3}$, if it exists, to equal $1/12$.

Edit: as for the increasing behavior, I came up with a somewhat hand-wavy idea to explain it, so take it with a grain of salt.

According to the jumping champions conjecture, the most frequent prime gap below some $10^{35}$ is $6$. As $k_{0}(m)$ counts the number of prime gaps in $[m-r_{0}(m),m+r_{0}(m)]$, one can expect $2r_{0}(m)$ to be $\asymp 6k_{0}(m)$ in this range. On the other hand, the Hardy-Littlewood $k$-tuples conjecture predicts the smallest diameter of an admissible $7$-tuple, corresponding to a chain of $7-1=6$ prime gaps to equal $20$, while my formula in the link I gave predicts a value of $6(1+H_{6})\approx 20.7\approx \gamma 6^{2}$ where $\gamma$ is the Euler-Mascheroni constant. As the jumping champions are conjecturally $4$ and the primorials, this suggest $d_{3}$ will eventually decrease after some (huge) threshold and that the value $1/12$ is not its limit but its maximum.

Answer 2

All $d_k$ go to zero, simply because the density of primes goes to zero. Let $\epsilon > 0$ and pick $N$ such that the density of primes in any interval of length at least $N$ is less than $\epsilon$. Since the density of primes goes to zero, almost all even integers lie in a prime gap of size larger than $N$, say between $p_a$ and $p_{a+1}$. The even numbers $n$ between $p_a$ and $p_{a+1}$ 'belong' to $d_k$ if $n-p_{a+1-k}$ is prime. The numbers $n-p_{a+1-k}$ lie in an interval of length at least $N$, so the density of primes here is less than $\epsilon$, showing that $d_k<\epsilon$.

The reason that $d_3$ increases at first, is because al lot of numbers were taken from $d_1$ and $d_2$ before. I suspect that if you plot the density of integers $n$ such that $n-p_{\pi(n)-2}$ is prime, it will start decreasing much sooner.

Answer 3

Here are a few thoughts.

First, $d_i$ is not the density of a certain type of prime, it is the density (as you say) of a certain type of even integer. So let me abuse notation and say that an even integer $n$ is of type $c_i$ if $q=n-p_{\pi(n)+1-i}$ is prime and type $d_i$ if, in addition, its farthest spaced Golbach pair is $n=q+p_{\pi(n)+1-i}.$

That the apparent increase you see in $d_3$ eventually goes away has been explained convincingly by others. I wanted to suggest why it might take a very very long time.

To oversimplify dramatically, $0<d_3(x)<c_3(x)$ and $c_3(x) =O(\frac1{\ln \ln x}).$ This squeezes $d_3$ to zero, but rather gently. Actually, $c_1(x)$ and $c_2(x)$ should all have roughly the same magnitude.

To oversimplify somewhat less, the density and pattern of type $d_1$ for even integers the size of $x$ depends on the primes of size up to about $\ln(x).$ These even integers separate into those in the intervals between two primes $p$ and $p+g.$ In such an interval $g$ is on average about $ln(x)$ while the number of type $d_1$ is $\pi(g)-1$ and the pattern is exactly that of the odd primes up to $g.$

Around $10^{10},2\cdot 10^{10}$ and $3 \cdot 10^{10}$ the average gap $g$ between primes takes the values approximately $23,23.7$ and $24.1.$ It is these gaps, specifically the triples which occur, and the specific primes the size of the gaps which control what happens.

Consider four consecutive primes $$p_{m-3},p_{m-2},p_{m-1},p_m=q,q+g_3,q+g_3+g_2,q+g_3+g_2+g_1$$ Then, for the $\frac{g_1}2$ even integers in the range $S=(p_{m-1},p_{m})$, the pattern of which are type $d_1,d_2,d_3$ (or none of these) does not depend on $q$, it depends completely on that triple of gaps $g_3,g_2,g_1$ and the particularities of the odd primes up to $g_1+g_2+g_3.$ More specifically, the primes in these three subintervals:

$I_1=[0,g_1],\ I_2=[g_2,g_2+g_1]$ and $I_3=[g_3+g_2,g_3+g_2+g_1]$ .

Let $\mathcal{P}$ be the odd primes and

• $S_1=\{q+g_3+g_2+i \mid i \in I_1 \cap \mathcal{P}\}$

• $S_2=\{q+g_3+i \mid i \in I_2 \cap \mathcal{P}\}$

• $S_3=\{q+i \mid i \in I_3 \cap \mathcal{P}\}$

The even $n$ ( from $S$ ) of type $d_1$ are exactly those in $S_1.$ Those of type $d_2$ are exactly those in $S_2 \setminus S_1$ and for $d_3,$ $S_3\setminus (S_1 \cup S_2).$

There will always (everyone believes) be triples with these specific gaps $g_3,g_2,g_1,$ but the frequency will decrease to be replaced more often by triples with somewhat larger $g.$ The rate at which this change happens will be exceedingly slow.

Here are some small calculations which are insufficient to do more than suggest that a much larger range would be needed to see meaningful changes in behavior.

• For $101$ consecutive primes starting shortly after $10^{10}$ the difference between the first and last is $2500$ and the frequency of gaps are: $[[6], 15], [[18], 9], [[4, 12], 8], [[26], 7], [[8, 14, 30, 42], 4], [[28, 48], 3], [[2, 10, 16, 20, 32, 36, 40, 46, 52, 70, 120], 2], [[22, 24, 34, 44, 50, 54, 60, 66, 92], 1]$

So the gaps $4$ and $12$ occur $8$ times each.

• For $101$ consecutive primes starting shortly after $2\cdot 10^{10}$ the difference between the first and last is $2232$ and the frequency of gaps are: $[[6], 12], [[10, 12], 9], [[8], 7], [[2, 18], 6], [[22, 24], 5], [[4, 14, 30], 4], [[32], 3], [[20, 26, 36, 38, 40, 42, 50, 52, 90], 2], [[34, 46, 48, 58, 60, 70, 76, 88], 1]$

• For $101$ consecutive primes starting shortly after $3\cdot 10^{10}$ the difference between the first and last is $2566$ and the frequency of gaps are: $ [[12], 11], [[4, 6, 18], 8], [[24], 7], [[20], 5], [[10, 16, 26, 42, 60], 4], [[8, 14, 30], 3], [[2, 28, 32, 34, 36, 48, 70], 2], [[22, 38, 40, 44, 54, 56, 74, 92, 94, 156], 1]$

In the end, I'm not really sure how much this explains. It is merely a suggestion of a place to look. For a stab at the kind of consideration that might arise: The even integers are sectioned off into intervals $[p,p+g]$ where $g$ is the gap from the prime $p$ to the next prime. The local density $d_1$ is $\frac{2(\pi(g)-1)}{g}$ (since there are $\frac{g}2$ even integers and the prime $2$ is not odd.) In those cases where $g$ is $24,26,28$ the local density is $\frac{16}{g}.$ To the extent that $g=28$ occurs more often and $g=24$ less often, the contribution to the growth of $d_1$ is less. This, in turn, gives more chances for $d_3$ to grow. For $g=30$ the local density is $\frac{18}{30}>\frac{16}{28}$ which retards that (tiny) effect ever so slightly. Of course there are many other considerations.

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Two other comments:

I find the question interesting but the stated motivation unconvincing. Given an even $n$ you want to find one representation of it as a sum of two primes. So you suggest checking $n-3$ and, if that is not prime working down through the odd integers $q$ (not congruent to $n \bmod 3$) and , when one is prime, checking if $n-q$ is. It might be faster, for $n \equiv 2 \bmod 6,$ to check $n-3$ and then $n-7,n-13,n-19,n-31,n-37,n-43,n-61$ (though one would need to know when to stop)

You might find it helpful to study $n \equiv 0 \bmod 6$ vs $n \equiv \pm 2 \bmod 6.$ What does the trend of the densities $d_1,d_2,d_3$ look like in each type? I'd expect $n \equiv 0 \bmod 6$ to be perhaps only $1/3$ as likely to be counted by $d_3$ as other even $n:$ Suppose $p\ \equiv 1 \bmod 3$ is a prime. Given only this information, of the even $n$ up to the next prime, those of the form $6m$ could be counted by $d_1,d_2$ or $d_3$ but those of the form $6m + 2$ could possibly be counted by $d_2$ or $d_3$ but not $d_1$.

It might be that these kinds of effects balance out.