Number of twin primes

Question

I am reading article about pseudorandomness of primes and there is a part that I don't understand. Based on Cramer model(without even numbers) the number of twin primes in $[1,..,N]$ is ~ $2\frac{N}{log^2(N)}$.

Then in general, assume that all numbers n divisible by some prime less than a small threshold w are prime with probability zero, and are prime with a probability of $\Pi_{p<w} (1-\frac{1}{p})^{-1} \frac{1}{logN}$ otherwise.

Then the number of twin primes ~ $2 \Pi_{p<w} (1-\frac{1}{p^2}) \frac{N}{log^2N}$, and if $w \to \infty$ then we got $\Pi_{2} \frac{N}{log^2N}$, where $\Pi_{2}=2 \Pi_{p \geq 3} (1-\frac{1}{(p-1)^2})$.

I wonder how we get $(1-\frac{1}{p^2})$ in this prediction and $(1-\frac{1}{(p-1)^2})$ in twin prime constant? I know this is Euler product formula but I cannot figure out the right calculations. Thanks in advance for help!

Answer 1

The factor $1-\frac1{p^2}$ is wrong. It should be $1-\frac1{(p-1)^2}$ there, too; there's no difference in this regard when you take $w\to\infty$.

The factor $1-\frac1{(p-1)^2}$ arises from a correction for the fact that the estimate $\log^{-2}N$ for the density of twin primes incorrectly assumes that the events of $x$ and $x+2$ not being divisible by a prime $p$ are independent. This assumption yields $\left(1-\frac1p\right)^2$ for the probability of both of these events to occur, whereas for odd primes the correct probability would be $1-\frac2p$. To correct for this, the estimate is multiplied by the ratio

$$ \frac{1-\frac2p}{\left(1-\frac1p\right)^2}=\frac{p^2-2p}{(p-1)^2}=1-\frac1{(p-1)^2} $$

for each odd prime $p$.

See my answer to Heuristics of counting twin primes for more on this.