# 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 $\times \ ? \sim \mathbb{N}$)

• do you mean numbers divisible by only one prime from the set of twin primes or by a pair (contiguous)? I can think of a situation where one takes a prime from the set and use it to multiply every single integer by that prime. I can also think of a situation where one takes a pair and multiply the members of that pair and then multiply the result by every single integer. So in my simplistic understanding of the question, as N→∞?, the percentage gets close to 100%. Apr 10, 2015 at 13:35
• Certainly you are not asking for a definitive answer. So what are you looking for? A heuristic argument? Numerical evidence? Upper bound? Apr 10, 2015 at 15:57
• @user25406 By Brun's theorem, the percentage does not converge to 100%. Brun's constant has been calculated to decently-high precision, so one would just need to adjust the computation to $\prod(1-1/p)$ rather than $\sum 1/p$. Apr 10, 2015 at 16:09
• @mjqxxxx Irrelevant. The convergence of the infinite sum implies the convergence of the (appropriate) infinite product, in particular it doesn't diverge to $0$ and its complement doesn't approach $1$. Apr 10, 2015 at 16:12
• You're right... can the actual density being asked for be written in terms of just Brun's constant? It seems like it depends on higher cumulants, too: the sum of the inverse squares of the twin primes, and their inverse cubes, etc. Of course, these are all rapidly convergent and easy to approximate. Apr 10, 2015 at 16:26

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.

• isn't $\prod_{p\in P_2}\left(1-\frac{1}{p}\right)< \exp\left(-\sum_{p\in P_2}\frac{1}{p}\right)?$ Apr 11, 2015 at 10:57
• @martin Yes, $1-e^{-B}$ is actually a lower bound, and the last sentence should be "increase slightly". Actually, I was surprised to learn just now that Brun's constant is not known to any reasonable precision, despite what is reported on MathWorld (maybe one decimal place). So non-trivial lower and upper bounds can be found, but an estimate as sharp as I claimed in the comments is conditional on the Hardy-Littlewood conjecture to estimate the long tail. Apr 11, 2015 at 15:33

As an addendum to mjqxxxx's excellent answer, I present a different approach which offers a minor improvement in accuracy (although a difference of $\approx 2\%$ is sufficiently large to be notable, considering how slowly the product converges at large $N$).

Let $\mathcal {P} (\mathbb{P}_2)$ represent the power set of all twin primes, $\mathcal {P} (\mathbb{P}{_2}(N))$ the power set of the first $N$ twin primes, and $\mathcal {P}_\kappa (\mathbb{P}{_2}(N))$ the set of subsets of cardinality $\kappa.$ Also let $\mathcal {P}_\kappa \small{\left(\prod\frac{1}{p\in \mathbb{P}_2}\right)}$ represent the subset of the products of reciptocals of twin primes with the specified cardinality.

For example, $A=\{3,5,7,11\},\ \mathcal {P}_2 {\left(\prod\frac{1}{A}\right)}$ would represent the set $\left\{\frac{1}{15},\frac{1}{21},\frac{1}{33},\frac{1}{35},\frac{1}{55},\frac{1}{77}\right\}.$

Since

\begin{align} &\quad \prod_{p\in \mathbb{P}_2}^{N}\left(1-\frac{1}{p}\right)&=&\quad 1-\sum^{N} \left (\large\mathcal {P} _ {\text {odd}} \small{\left(\prod\frac{1}{p\in \mathbb{P}_2}\right)} - \large\mathcal {P} _ {\text {even}} \small{\left(\prod\frac{1}{p\in \mathbb{P}_2}\right)} \right) \\ \end{align}

as can be seen easily in the case $N=3:$

\begin{align} &\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}-\frac{1}{15}-\frac{1}{21}-\frac{1}{35}+\frac{1}{105}\right)= 1-\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{7}\right)\\ \end{align}

it follows that

\begin{align} &\quad \prod_{p\in \mathbb{P}_2}\left(1-\frac{1}{p}\right)&= &\quad \quad \sum _{p\in \mathbb{P}_2} \frac{1}{p}\\ &&&-\quad \frac{1}{2} \left(\sum _{p\in \mathbb{P}_2} \frac{1}{p}\right)^2-\frac{1}{2} \sum _{p\in \mathbb{P}_2} \frac{1}{p^2}\\ &&&+\quad \frac{1}{6} \left(\sum _{p\in \mathbb{P}_2} \frac{1}{p}\right)^3-\frac{1}{2} \left(\sum _{p\in \mathbb{P}_2} \frac{1}{p^2}\right) \sum _{p\in \mathbb{P}_2} \frac{1}{p}+\frac{1}{3} \sum _{p\in \mathbb{P}_2} \frac{1}{p^3}\\ &&&-\quad \dots \end{align}

where the coefficients are given in Table 24.2 in 1 (multinomials M2) multiplied by $(-1)^{q},$ where $q$ is the number of elements in the corresponding integer partition.

This representation turns out to be beneficial computationally, since the power sums converge so rapidly. $\infty$ in the sum can be replaced by a reasonably small $N,$ and $\sum _{p\in \mathbb{P}_2} \frac{1}{p}$ replaced by Brun's constant (see note below), to give the slightly improved approximation of $\approx 83.83 \%.$

Note: As Erick Wong notes in the comments below, the current "known" value of Brun's constant is based on a Heuristic argument (Hardy & Littlewood) that $\pi_2(x) \approx 2C_2 \int_2^x \frac{dt}{\left(\log t \right)^2},$ where $C$ is the twin prime constant. Nicely gives here an estimate for $B_2$ of $1.9021605823\pm 8\times10^{-10}.$

1. Milton Abramowitz and Irene A. Stegun (eds.), Handbook of mathematical functions, 9th ed., Dover Publications, New York, 1972.