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Under Goldbach's conjecture, I'm trying to find an upper bound for $r_{0}(n):=\inf\{r>0,(n-r,n+r)\in\mathbb{P}^{2}\}$ that would generalize Cramer's conjecture.

Denoting by $k_{0}(n)$ the quantity defined as $\pi(n+r_{0}(n))-\pi(n-r_{0}(n))$, it seems that $2r_{0}(n)\lesssim k_{0}(n)(\log n)^{1+1/k_{0}(n)}$.

Is there a heuristics suggesting this holds or a conditional proof thereof?

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    $\begingroup$ I'm saying since a long time that you need to assume that the $X_{2m+1}=1_{2m+1\ is\ prime}$ are independent random variables with $P(X_{2m+1}=1) = 1/\log(2m+1)$. This gives a probability distribution for $r_0(n)$ from which you get things of the form $Pr(\lim \inf_{n\to \infty}r_0(n)/ \log^2 n= 0) = 1$. Then compare with the truth and search for the minimum assumptions for the results of the random model to hold. Be prepared to throw off $r_0(n)$ and use a better behaved function. $\endgroup$
    – reuns
    Commented Nov 19, 2020 at 17:15
  • $\begingroup$ And I've been saying for almost as long that I'm interested in the primes as they are, a deterministic sequence, and not some random model thereof. $\endgroup$
    – Sylvain JULIEN
    Commented Nov 19, 2020 at 17:18
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    $\begingroup$ .... The random model talks of the primes, not of something which doesn't exist. If you find that the random model doesn't hold (up to minor modifications) then you will get the fields medal. I hope you know that all the prime conjectures are based on it. $\endgroup$
    – reuns
    Commented Nov 19, 2020 at 17:20
  • $\begingroup$ I turned 39 this month so I won't get the Fields medal. And yes, I'm pretty aware that most conjectures are based on the random model but that doesn't mean I have to stick to it. I'm not as good at math as you but I owe you nothing. That is all there is to it. $\endgroup$
    – Sylvain JULIEN
    Commented Nov 19, 2020 at 18:03
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    $\begingroup$ If you ask for "heuristics" but reject the idea of random models giving insight into primes, it's not clear what kind of answer you want. Indeed, our belief in both Goldbach's conjecture and Cramer's conjecture themselves (which you seem to accept as legitimate) is based upon random models of the primes. $\endgroup$
    – Greg Martin
    Commented Nov 19, 2020 at 18:45

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This conjecture is incompatible with Cramer's conjecture. Indeed, Cramer predicts that for arbitrarily large $k$ we have $p_{k+1}-p_k\gg(\log p_k)^2$. Let $n=\frac{p_{k+1}+p_{k-1}}{2}$. Then $r_0(n)=\frac{p_{k+1}-p_{k-1}}{2}\gg(\log n)^2$, while $k_0(n)=\pi(p_{k+1})-\pi(p_{k-1})=2$, so your conjecture would predict $(\log n)^2\ll(\log n)^{3/2}$, which of course fails.

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  • $\begingroup$ Doesn't it depend on the constants that are involved? $\endgroup$
    – Sylvain JULIEN
    Commented Nov 19, 2020 at 16:04
  • $\begingroup$ Regardless of the constants involved, $(\log n)^2\ll(\log n)^{3/2}$ is not true. $\endgroup$
    – Wojowu
    Commented Nov 19, 2020 at 16:52
  • $\begingroup$ Ok, so is Cramer's conjecture compatible with my inequality holding for infinitely many $n$, the set of which being of positive natural density? $\endgroup$
    – Sylvain JULIEN
    Commented Nov 19, 2020 at 17:12
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    $\begingroup$ Presumably the proposed inequality holds for almost all $n$ or a positive proportion of $n$, depending on what you mean by $\lesssim$. Conjecturally $r_0(n)$ is going to be roughly geometrically distributed with parameter $\asymp \log^2 n$; and $k_0(n)$ is going to be approximately $2r_0(n)/\log n$. Therefore your inequality is essentially asking how likely it is that $\log n \lesssim (\log n)^{1+1/k_0(n)} \sim e\log n$. $\endgroup$
    – Greg Martin
    Commented Nov 19, 2020 at 18:43

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