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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

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    $\begingroup$ I would (perhaps naively) expect $d_k$ to tend to zero as $n$ tends to infinity for any given $k$, it is just that the scales at which it happens will vary with $k$ since the sum of all the $d_k$ is always $1$. I imagine for smaller $n$ you will see $d_2$ increasing before it starts decreasing, and for large enough $n$, $d_3$ will eventually decrease to $0$. $\endgroup$ Feb 19, 2022 at 14:28
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    $\begingroup$ If true, this would suggest a Cramer type upper bound for $r_{0}(m):=\inf\{r\geq 0,(m-r,m+r)\in\mathbb{P}^{2}\}$. $\endgroup$ Feb 20, 2022 at 14:35
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    $\begingroup$ Maybe I'm repeating what @JoelMoreira already said, but isn't it obvious that for small $n$'s $d_3$ is initially squeezed by $d_1$ and $d_2$? Clearly all the $d_k$'s need to go to zero but only when the initial ones are small enough do the subsequent ones have the asymptotic freedom (so to speak) to almost equal the simple probability of $n-p_{\pi(k)}$ being prime (as the probability of $n-p_{\pi(k)}$ being prime for a smaller $k$ became negligible). $\endgroup$ Feb 20, 2022 at 19:30
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    $\begingroup$ @Yaakov Baruch: don't you mean $n-p_{\pi(n)+1-k}$? $\endgroup$ Feb 20, 2022 at 20:55
  • $\begingroup$ @SylvainJULIEN - yes! Thank you. $\endgroup$ Feb 20, 2022 at 20:56

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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.

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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.

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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.


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.

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