18
$\begingroup$

Let $\mathcal{H}=\left\{ h_{i}\right\} _{i=1}^{k}$ be an admissible set, and define $$\pi_{\mathcal{H}}(x)=\left|\left\{ n\leq x\ :\ \exists\ i,j\leq k,\ i\neq j\ \text{such that both }n+h_{i},\ n+h_{j}\ \text{are prime}\right\} \right|.$$ The Hardy-Littlewood $k$-tuple conjecture implies that $$\pi_{\mathcal{H}}(x)\sim C_{\mathcal{H}}\frac{x}{\log^{2}x},$$ where $C_{\mathcal{H}}$ is a constant depending on $\mathcal{H},$ and the Selberg sieve can be used to prove that $$\pi_{\mathcal{H}}(x)\ll_{\mathcal{H}}\frac{x}{\log^{2}x}.$$ Yitang Zhang recently proved that for any sufficiently large admissible set $\mathcal{H}$, we have $$\lim_{x\rightarrow\infty}\pi_{\mathcal{H}}(x)\rightarrow\infty.$$ (the current lower bound on $\mathcal{H}$ can be found here)

Question: Does Zhang's work give a lower bound on $\pi_{\mathcal{H}}$ with the correct order of magnitude? That is, can we prove that $$\pi_{\mathcal{H}}(x)\gg_{\mathcal{H}}\frac{x}{\log^{2}x}?$$

$\endgroup$
  • 3
    $\begingroup$ In particular, Zhang's theorem implies the existence of a positive integer $k$ such that there exist infinitely many primes $p$ such that $p + 2k$ is also prime. Does Zhang's theorem imply a similar lower bound for $\pi_{2k}(x)$ as indicated above? $\endgroup$ – Stanley Yao Xiao Oct 17 '13 at 5:56
17
$\begingroup$

In the main theorem of this recent paper of Pintz, Zhang's method is used to show that (for $k$ large enough), there are $\gg_{\mathcal H} \frac{x}{\log^k x}$ values of $n \le x$ such that two of the $n+h_i$ are prime and the remaining $n+h_j$ are almost prime (have $O_k(1)$ prime factors, all of which are $\gg x^{c_k}$ for some $c_k>0$ independent of $x$). This is the correct asymptotic (up to multiplicative constants) if one enforces the almost primality conditions, but has the wrong power of log in the denominator if those conditions are removed.

In a slightly different direction, in this paper of Goldston, Pintz, and Yildirim (a followup to their most famous paper), it is shown that for any fixed $\varepsilon>0$, there are $\gg_\varepsilon \frac{x}{\log x}$ primes $p_n$ less than $x$ such that $p_{n+1}-p_n \leq \varepsilon \log p_n$. Again, this is the correct asymptotic up to multiplicative constants.

I have a graduate student who has recently started looking at improvements to these results, with some encouraging preliminary findings, but the work is not yet completed. It is indeed unlikely that one will be able to obtain the optimal asymptotic here unconditionally, but some improvement upon the partial results given above look plausible.

$\endgroup$
  • $\begingroup$ Does Zhang's result have any useful implications for previous polymath project, "Deterministic way to find primes"? $\endgroup$ – Nurdin Takenov Oct 23 '13 at 2:19
  • $\begingroup$ Not as far as I know; that project could use uniform upper bounds on prime gaps $p_{n+1}-p_n$, but not upper bounds on a sparse set of prime gaps (i.e. it uses bounds of lim sup type rather than lim inf type). $\endgroup$ – Terry Tao Oct 28 '13 at 21:43

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.