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.
$\frac{\pi^2(n)}{n\pi_2(n)}$" />
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.
$\frac{n\pi_2(n)}{\pi^2(n)}$" />
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 $n$.
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.