Jie Wu improved Brun's theorem and showed that the number of prime twins up to $n$ satisfies for sufficiently large $n$ :
$$\pi_2(n) < 4.5 \frac{n}{ln(n)^2} $$
However this confused me while trying to get a good estimate of the prime twins.
I was thinking in terms of $1 \mod 30$.
Consider the integer intervals $[30 k + 10,30 k + 20]$.
between $1$ and a large $n$ this means these intervals have together about $n/3$ integers within them.
When we sieve these intervals the usual way ; $C n (1-2/3)(1-2/5)(1-2/7) ...$
we get the usual estimate $ C n / ln(n)^2 $.
$C$ is here the usual value from the Hardy-Littlewood conjecture divided by $3$ because our intervals have a measure of $1/3$.
However I think primes of the form $30 k + 1$ should not be sieved because multiplying integers times $1 (\mod 30)$ does not change anything ($\mod 30$).
In other words prime twins candidates of the form $30 k + 11, 30 k + 13$ and $30 k + 17,30 k + 19$ (these are in the intervals defined above) are not sieved out by number of the form $30 k + 1$ so we should not sieve primes of the form $30 k + 1$.
Assuming that logic is correct my estimate became :
$T(n) = C n (1-2/3)(1-2/5)...$ where where omit terms of the form $(1-2/q)$ where $q$ is a prime of the form $1 \mod 30$.
Since all primes $> 5$ are of the form $1,7,11,13,17,19,23,29$ mod $30$ , $1 \mod 30$ takes $1/8$ of all primes.
Because of the prime number theorem the product over $(1-2/q)$ is then close to the product over $(1-2/(8p))$ where $p$ are the primes.
We simplify $(1-2/8p) = (1-1/4p) $ hence we get a (correcting) factor around $\ln(n)^{1/4}$.
The estimate $T(n)$ is therefore around $C n \ln(n)^{-2+(1/4)}$ or simplified :
$$T(n) = C n \ln(n)^{-7/4}$$
But this contradicts Brun's theorem and Jie Wu's theorem.
Where did I go wrong ?
Im confused.