About the Hardy-Littlewood conjecture by Terence Tao:

Conjecture 2 (Prime tuples conjecture, quantitative form) Let ${k_0 \geq 1}$ be a fixed natural number, and let ${{\mathcal H}}$ be a fixed admissible ${k_0}$-tuple. Then the number of natural numbers ${n < x}$ such that ${n+{\mathcal H}}$ consists entirely of primes is ${({\mathfrak G} + o(1)) \frac{x}{\log^{k_0} x}}$.

It was written there: From the methods of sieve theory, one can obtain an upper bound of ${(C_{k_0} {\mathfrak G} + o(1)) \frac{x}{\log^{k_0} x}}$ for the number of ${n < x}$ with ${n + {\mathcal H}}$ all prime, where ${C_{k_0}}$ depends only on ${k_0}$. Sieve theory can also give analogues of Conjecture 2 if the primes are replaced by a suitable notion of almost prime (or more precisely, by a weight function concentrated on almost primes).


1) Where can I find these results? Could you give a reference?

2) How does ${C_{k_0}}$ depend on ${k_0}$?

Thank you!

  • 3
    $\begingroup$ This is a consequence of the fundamental lemma of sieve theory, and can be proved in many ways - these days most people would prove it using Selberg's upper bound sieve. Opera de Cribro is a good reference. $\endgroup$
    – zeb
    Commented Oct 30, 2015 at 4:31
  • 2
    $\begingroup$ Also, IIRC the current best bound for $C_k$ is $k!2^k$, using Selberg's upper bound sieve. $\endgroup$
    – zeb
    Commented Oct 30, 2015 at 4:43
  • $\begingroup$ For those who don't want to click on the link: gothic G is a real number expressed as an infinite product over the primes, that only depends on curly H. (This comment is too short to copy the precise definition. I'm sorry, sometimes clicking the link is the only way.) $\endgroup$
    – Vincent
    Commented Oct 30, 2015 at 11:17
  • 2
    $\begingroup$ Next time please also tell us that you ask the same question elsewhere, e.g. on Tao's blog (to avoid duplicate answers, time wasted etc.). Actually, Tao responded there. $\endgroup$
    – GH from MO
    Commented Oct 30, 2015 at 15:43
  • $\begingroup$ $C_2\le3.91045$, so $2!2^2=2\cdot4=8$ isn't the best upper bound for the particular case $k_0=2$. $\endgroup$
    – user178594
    Commented Apr 2, 2023 at 14:13

2 Answers 2


This is explained for example in Iwaniec & Kowalski's "Analytic Number Theory", as an standard application of Selberg's $\Lambda^2$ sieve. See chapter 6, Elementary sieve methods. In particular you get $C_{k_0}=2^{k_0}k_0!$, for an upper bound, using your notation:

$${(2^{k_0}k_0! {\mathfrak G} + o(1)) \frac{x}{\log^{k_0} x}}$$

Of course, the conjecture predicts that you can take $C_{k_0}=1$, but as far as I know the $2^{k_0}k_0!$ bound still is the best one known unconditionally.

Probably the most complete classic reference is

  • Atle Selberg, Lectures on Sieves (1991)

You can also prove the same bound using Montgomery and Vaughan's large sieve. For this and all other things sieve-related, Opera de Cribro is he best source.

  • $\begingroup$ The case $k_0=2$ is presented as Theorem 3.2 of these short notes by Heath-Brown: arxiv.org/pdf/math/0209360v1.pdf. (Note that the twin-prime constant $C_2$ appearing in the formula for $\mathfrak{G}$ in our case is defined in these notes slightly differently than usual - it is multiplied by 2 compared to the usual definition - see page 14). $\endgroup$ Commented Oct 30, 2015 at 16:32

The best proved bound of $C_{k_0}$ for $k_0=2$ is in https://arxiv.org/abs/0705.1652, which gives


giving the best current bound


This is an improvement of the well-known upper bound $C_2\le8$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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