What percentage of numbers is divisible by the set of twin primes? What percentage of numbers is divisible by the set of twin primes $\{3,5,7,11,13,17,19,29,31\dots\}$ as $N\rightarrow \infty?$
Clarification
Taking the first twin prime and creating a set out of its multiples : $\{3,6,9,12,15\dots\}$ and multiplying by $\dfrac{1}{3}$ gives $\mathbb{N}: \{1,2,3,4,5\dots\}.$ This set then represents $\dfrac{1}{3}$ of $\mathbb{N}.$
Taking the first two: $\{3,5\}$ and creating a set out of its multiples gives: $\{3, 5, 6, 9, 10\dots\}.$ This set represents $\sim \dfrac{7}{15}$ of $\mathbb{N}.$
Taking the first three: $\{3,5,7\}$ and creating a set out of its multiples gives: $\{3, 5, 6, 7, 9, 10, 12, 14\dots\}.$ This set represents $\sim \dfrac{19}{35}$ of $\mathbb{N}.$
What percentage of $\mathbb{N}$ then, does the set consisting of all divisors of all twin primes $\{3, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 17, 18\dots\}$ constitute? (ie This set $\times \ ? \sim \mathbb{N}$)

According to Brun's theorem, the twin primes constitute a small set.  That is, the sum of their reciprocals,
$$
\left(\frac{1}{3} + \frac{1}{5}\right) + \left(\frac{1}{5}+\frac{1}{7}\right) + \left(\frac{1}{11}+\frac{1}{13}\right) + \cdots,
$$
is convergent.  Numerically it is estimated to be about $1.902$, so the sum with the single duplicate $(1/5)$ removed is $B\approx 1.702$.  The probability that a large random number is not divisible by any twin prime approaches
$$
\prod_{p\in P_2}\left(1-\frac{1}{p}\right)=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{7}\right)\cdots < \exp\left(-\sum_{p\in P_2}\frac{1}{p}\right)=e^{-B},
$$
and so the natural density of twin-prime-divisible numbers is at least $1-e^{-B}\approx 0.8177.$  (Estimating the product using the twin primes through two million gives density $> 0.806$.)
As a sanity check on this, the number of twin-prime-divisible numbers from $10^5$ through $2\times 10^5 - 1$ is exactly $81714$, for a density of $0.8171$; and this should increase slightly for larger numbers.
