For $A\subseteq \mathbb{N}$, let the upper density of $A$ be defined by $$\mu^+(A) = \lim\sup_{n\to\infty}\frac{|A\cap\{1,\ldots,n\}|}{n}.$$ Let $$A = \{n\in\mathbb{N}: 2n \text{ is the sum of } 2 \text{ primes}\}.$$ Can it be shown that $\mu^+(A) > 0$? What about $\mu^+(A) \geq 1/2$?
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asked Dec 1, 2019 at 7:29
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1 Answer
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The state-of-the art is contained in this paper of János Pintz. Read the introduction, especially (1.9).
answered Dec 1, 2019 at 7:57
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2$\begingroup$ To summarize this answer: we know $\mu^+(A)=1$ and we have good estimates on the size of the complementary set. $\endgroup$– WojowuCommented Dec 1, 2019 at 10:09
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3$\begingroup$ @Wojowu: Yes, and this information is available in the corresponding Wikipedia article: en.wikipedia.org/wiki/Goldbach%27s_conjecture#Rigorous_results $\endgroup$ Commented Dec 1, 2019 at 10:14