To estimate the amount of primes in an interval $\left(2,x\right)$ one might naively sieve by computing $ x \left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)...\left(1-\dfrac{1}{p_i}\right)$ where $p_i$ is the $i$ th prime. (And of course we exclude primes larger than $\sqrt x$).

However by the PNT and the work of Mertens we know we the correct asymptotic result is $ M x \left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)...\left(1-\dfrac{1}{p_i}\right)$.

Now the strong prime twins conjecture basicly works similar , we sieve by computing $ x \left(1-\dfrac{2}{3}\right)\left(1-\dfrac{2}{5}\right)...\left(1-\dfrac{2}{p_i}\right)$.

However here it appears not naive and it actually works pretty well.

So my question is ; what became of the Mertens constant ? Why dont we compute $ M x \left(1-\dfrac{2}{3}\right)\left(1-\dfrac{2}{5}\right)...\left(1-\dfrac{2}{p_i}\right)$ or $ M^2 x \left(1-\dfrac{2}{3}\right)\left(1-\dfrac{2}{5}\right)...\left(1-\dfrac{2}{p_i}\right)$ ?? Or even replace M with another irrational number ?

Although almost nothing has been proved about prime twins it seems I missed something trivial here ?


There is a "twin primes constant" which I think is what you want.

  • $\begingroup$ No that has been taken into account. Basicly the twin constant converts $\prod (1-2/p)$ to its $\dfrac{1}{ln(x)^2}$ asymptotic. $\endgroup$ – mick Feb 20 '13 at 23:07
  • $\begingroup$ No you are right. I was confused. the "log" contains the "M". $\endgroup$ – mick Feb 21 '13 at 23:20

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