I would like to know whether there is a way, thanks to the prime number theorem, to give some kind of an equivalent of the probability that two positive integers $p$ and $q$ less than a given positive integer $n$ are both prime if one knows that $q-p=2r$, where $r$ is a given positive integer. Thank you in advance.

  • 2
    $\begingroup$ Given that the twin primes conjecture is open, what sort of an answer do you expect? $\endgroup$ – Igor Rivin Jul 14 '12 at 10:28
  • $\begingroup$ I expect a heuristics like the one saying that a positive integer around $x$ is prime with probability $1/\log x$, considering the two events "$p$ is prime" and "$q$ is prime" as being independent, except that I add the constraint $q-p=2r$. $\endgroup$ – Sylvain JULIEN Jul 14 '12 at 10:36
  • 2
    $\begingroup$ mathworld.wolfram.com/k-TupleConjecture.html $\endgroup$ – Eric Naslund Jul 14 '12 at 11:07
  • $\begingroup$ Thank you Eric. Can one deduce from this conjecture that the smallest $r$ such that $p+2r$ is a prime number when $p$ is a prime number less than $n$ verifies $r=O(\log^{2}n)$? In a very non rigorous way, one can expect the relation $2r\pi_{2r}(n)\asymp n$ to hold, and thus $r\asymp\dfrac{n}{\pi_{2r}(n)}$, where $\pi_{2r}(n)$ is the number of primes $p'$ less than $n$ such that $p'+2r$ is also prime. In other words, does the k-tuple conjecture make Cramer's conjecture more likely? $\endgroup$ – Sylvain JULIEN Jul 14 '12 at 12:10

A good estimate (in some sense) is $$\frac{\alpha_{2r}C_2}{\ln^2N}$$ where $$C_2 = \prod_{p\ge 3} \frac{p(p-2)}{(p-1)^2} \approx 0.66016.$$ and $$\alpha_{2r}=\prod_{p \mid 2r}\frac{p-1}{p-2}$$ is the product is over all odd primes $p$ which divide $2r.$

Notice that $\alpha_2=\alpha_4=\alpha_8=\dots=1$ and $\alpha_6=\alpha_{12}=\alpha_{18}=\alpha_{24}=\dots=2$

More careful statement: Let $\pi_{2r}(N)$ be the number of primes $p$ up to $N$ with $p+2r$ also prime and $\alpha_{2r}=\lim_{N \to \infty}\frac{\pi_{2r}(N)}{\pi_2(N)}.$ The twin prime conjecture is indeed open: there is no proof that $\lim_{N \to \infty} \pi_2(N)=\infty.$ However there is every reason to expect that

  • $\pi_2(N) \sim C_2\frac{N}{(\ln{N})^2}.$
  • $\alpha_{2r}=\prod_{p \mid 2r}\frac{p-1}{p-2}$ where the product is over all the odd primes which divide $r.$

The great article Heuristic Reasoning in the Theory of Numbers (read it!) discuss the heuristics and some computational (circa 1959) evidence behind this.

  • $\begingroup$ Thank you very much. This article is indeed very interesting and I like the references to physics (I used to study physics several years ago, hence my lack of rigor when I try to do maths)...By the way, what do you think about the possible interplay between this Hardy-Littlewood conjecture and Cramer's conjecture? $\endgroup$ – Sylvain JULIEN Jul 15 '12 at 12:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.