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I was doing some research on prime gaps including twin primes and this led me to this finding, which is:

$$ \lim\limits_{n\to \infty} \frac{\pi^2(n)}{n\pi_2(n)} = 0.7550363087870907 \cdots\cdots (1) $$

where the symbols have their usual meanings. Chart shown below. <span class=$\frac{\pi^2(n)}{n\pi_2(n)}$" />

The number $0.755..$ did not make any sense to me until accidentally I inverted it in my Jupyter notebook which resulted in:

$$ \lim\limits_{n\to \infty} \frac{n\pi_2(n)}{\pi^2(n)} = 1.3203236316937392 \cdots\cdots (2) $$

Chart shown below. <span class=$\frac{n\pi_2(n)}{\pi^2(n)}$" />

Now $1.32032... = 2C_2$ where $C_2$ is the twin prime constant.

Question: It appears the graphs are converging but that could be just an illusion. Is the second chart indeed converging to $2C_2$?

Philosophically it would make sense as it gives real meaning to the twin prime constant. This if true will obviously prove the twin prime conjecture. This result is a much stronger result than the Hardy-Littlewood twin prime conjecture. Of course, how I arrived at this is perhaps more interesting but that would take up too much space here.

Finally if we rewrite (2) as: $$ \lim\limits_{n\to \infty} \frac{\frac{\pi_2(n)}{\pi(n)}}{\frac{\pi(n)}{n}} = 1.3203236316937392 \cdots\cdots (3) $$

which is saying the number of twin primes in primes is more than the number of primes in $n$.

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    $\begingroup$ Conjecturally, we do expect the graph to converge to $2 C_2$. You can see this by using the prime number theorem and the conjectured count of twin primes. $\endgroup$ Commented Nov 2, 2023 at 20:00

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By the prime number theorem, we have that $$ \pi(n) \sim \frac{n}{\log n}. $$ The first Hardy-Littlewood Conjecture implies that $$ \pi_2(n) \sim 2 C_2 \frac{n}{\log^2(n)},$$ where $$ C_2 = \prod_{p \geq 3} \left( 1 - \frac{1}{(p-1)^2} \right) \approx 0.6601\ldots$$ is the twin-prime constant. Thus we would expect $$ \frac{\pi^2(n)}{n \pi_2(n)} = \pi^2(n) \frac{1}{n \pi_2(n)} \sim \frac{n^2}{\log^2 n} \frac{1}{n \frac{2 C_2 n}{ \log^2 n}} = \frac{1}{2 C_2},$$ agreeing with your observation.

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  • $\begingroup$ The interesting thing is that although the formulas used for both $\pi(n)$ and $\pi_2(n)$ are approximations, the errors in each of those formulas get offset when we take the ratio resulting in a constant, which is amazing! $\endgroup$ Commented Nov 2, 2023 at 20:24
  • $\begingroup$ @sku , yes but what exactly is the error for the twin prime counting function ?? I asked that here, but it is unanswered. Not even a conjecture. mathoverflow.net/questions/224254/… $\endgroup$ Commented Sep 3, 2024 at 18:43

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