While trying to find a lower limit for the number of twin primes I noticed the problem of having to compensate for duplicates. Once I overcame this problem the duplicates of the duplicates became a problem and I saw this was leading nowhere. So I came up with a different approach. I noticed this behavior is similar to sieving primes. After further analyzing it even seems likely that twin sieving is actually sieving primes over the results. This is also not easy to calculate so I decided to simplify to an lower limit. I shifted the result of the first sieve together so that they form one continuous sequence of sieved numbers. Then it became much easier to calculate the number because this is $\pi(\pi(n))$. The results for n up to $10^{18}$ are very promising , so my conjecture is that for large enough n: \begin{align*} \pi_2(n) &> \pi(\pi(n)) &f=\frac{\pi_2(n)}{\pi(\pi(n))} > 1\\ \end{align*} The following table shows results for $n$ up to $10^{18}\\$ $\\ \begin{array}{ | l | l | l | l |} \hline n &\pi(\pi(n)) &\pi_2(n) &f\\ \hline 1000 & 39 & 35 & 0,89744\\ \hline 10000 & 201 & 205 & 1,01990\\ \hline 100000 & 1184 & 1224 & 1,03378\\ \hline 1000000 & 7702 & 8169 & 1,06063\\ \hline 10000000 & 53911 & 58980 & 1,09403\\ \hline 100000000 & 397557 & 440312 & 1,10754\\ \hline 1000000000 & 3048955 & 3424506 & 1,12317\\ \hline 10000000000 & 24106415 & 27412679 & 1,13715\\ \hline 100000000000 & 195296943 & 224376048 & 1,14890\\ \hline 1000000000000 & 1613846646 & 1870585220 & 1,15908\\ \hline 10000000000000 & 13556756261 & 15834664872 & 1,16803\\ \hline 100000000000000 & 115465507935 & 135780321665 & 1,17594\\ \hline 1000000000000000 & 995112599484 & 1177209242304 & 1,18299\\ \hline 10000000000000000 & 8663956207026 & 10304195697298 & 1,18932\\ \hline 100000000000000000 & 76105984161825 & 90948839353159 & 1,19503\\ \hline 1000000000000000000 & 673776962356604 & 808675888577436 & 1,20021\\ \hline \end{array}\\$
For small $n$ $f<1$ because for small n the shifting actually causes less sieving. For n=3541 $\pi(\pi(n))$ and $\pi_2(n)$ are both 94, $f=1$)
After that $f>1$ because primes are far enough apart that the shifting reduces the count.
Proof: We determine the number of twin primes by first counting the number of primes that are $6n-1$ then we check the results for $6n+1$ also being a prime number. The first step is obtained by sieving for primes and dividing this by 2 which gives us: $$\frac{\pi(n)}{2}$$ There are more $6n-1$ primes so this slightly lower than the actual count. Then we start eliminating all $6n+1$ that are not prime: We need to remove all $6n+1$ that are 5 multiples: $25+5 \cdot 6x$ next all 7 multiples etc. This is the same as sieving for primes with the exception that 2 and 3 are not used. This proofs sieving for twin primes gives the same result as sieving for primes twice with the exception of 3 for the second pass (2 was necessary because we needed $\frac{\pi(n)}{2}$. Even with a crude compensation for sieving 3 the result is a much better lower limit, even much better than Hardy-Littlewood.
Questions:
$f$ seems to converge, it increases less with every decade. Is this true and is it possible to predict a value for $f \lim_{n \to \infty}\\ \\$
As $f$ only seems to grow for large $n$, this conjecture makes infinitely many twin primes very likely but is it possible to turn it into a proof?