According to Wikpedia, Hardy-Littlewood conjecture says that $$\pi_2(n) \sim 2 C_2 \frac{n}{(\ln n)^2} \sim 2 C_2 \int_2^n {dt \over (\ln t)^2}$$ where $$C_2 = \prod_{p\ge 3} \frac{p(p-2)}{(p-1)^2} \approx 0.66016 18158 46869 57392 78121 10014\dots$$ Wikipedia also says that "This conjecture can be justified (but not proven) by assuming that $1 / \ln t$ describes the density function of the prime distribution, an assumption suggested by the prime number theorem."

Could somebody elaborate a bit on how we can get the above form of the conjecture using probabilistic viewpoint.

Very naively, I could say that prime number theorem says me that a number $p\le x$ is a prime is approximately $\frac1{\ln x}$. If I consider "$p$ is a prime" and "$p+2$ is a prime" as independent events, I would arrive at probability approximately $\frac1{\ln^2 x}$. So this would lead to the conjecture of the form $$\pi_2(x)\sim \frac{x}{\ln^2 x}.$$ Setting aside the assumption that the two events are independent, the above heuristic is rather problematic. We should somehow incorporate the fact that, in order to get a pair of twin primes, $p+2$ cannot be divisible by any smaller primes.

How can we heuristically find out that $C_2$ described above is the "correct" constant for this conjecture? Is there a simple probabilistic approach which would lead to the above formula?

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    $\begingroup$ nice question (+1) $\endgroup$ – tired Nov 8 '15 at 23:05

Wikipedia's description of the heuristic is not complete, for exactly the reason you have noted. The events "$p$ is prime" and "$p+2$" is prime are not independent: if $p$ is prime, then $p+2$ is automatically odd (hence more likely to be prime), and slightly more likely to be divisible by any fixed odd prime than a truly random integer (hence a little less likely to be prime); these effects must be taken into account.

A thorough description of the complete heuristic leading to the conjecture can be found on Terry Tao's blog (compare in particular Model 2 to Model 10).


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