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Brun's constant $B_{2}$ is defined as $B_{2}=1/3+1/5+1/5+1/7+1/11+1/13+...$ where the sum is taken on $p$ such that $p$ is an element of a couple of twin primes. The number of twin primes below is expected to be around $2C_{2}x/\log x$ where $C_{2}=0.66016...$ is the twin prime constant. Whenever $(p,p+2)$ is a couple of twin primes, $6\vert (p+1)$ (except if $p=3$).

Now suppose that the maximal gap between consecutive primes of size $x$ is around $\log^{2} x$ (which is consistant with Cramer's conjecture saying $\limsup \dfrac{p_{n+1}-p_{n}}{\log^2 p_{n}}=1$. As numbers $p+1$ such that $(p,p+2)$ is a couple of twin primes represents at most $1/2$ of all primes (that can be of the form $6n+1$ or $6n+5$) and thus $1/6$ of all integers up to $x$, can one expect the number $\pi_{2}(x)$ of twin primes below $x$ to verify $\dfrac{x}{\pi_{2}(x)}>2(1+o(1))\dfrac{B_{2}}{6}\log^{2}x$?

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The number of twin primes below $x$ is conjectured to be $\sim 2C_2 x/\log^2 x$.

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  • $\begingroup$ Yes, but as far as I know, the best unconditional upper bound is four times the conjectured asymptotic. What I would like to know is whether or not the assumption of Cramer's conjecture can lead to a better upper bound, though not unconditional. $\endgroup$ – Sylvain JULIEN Feb 21 '16 at 11:11
  • $\begingroup$ I see. Then take my answer to be a comment on a typo in your question. $\endgroup$ – Lior Silberman Feb 21 '16 at 11:14

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