I am interested in having an upper bound for the cardinality of

$\#\left\{n\leq x\,:\quad\omega(n)=k, \omega(n+2)=\ell\right\}$ for $k,\ell\geq 1$,

where $\omega(n)=\sum_{p\vert n}1$ counts the number of (distinct) prime factors of the integer $n$.
I would actually like a sharp upper bound (up to a constant) for $k=\ell=\log\log x$. I thought this was known, but I can't find anything in the literature. For $k\ll\sqrt{\log\log x}$ and $\ell\ll\log\log x$ I can have a good upper bound, but I can't reach $k=\ell=\log\log x$.

Does anyone has a reference or an idea please ?

Thank you very much !

  • 1
    $\begingroup$ Do you really mean $=\log\log x$ there? The double logarithm is almost always a non-integer, so $\omega(n)$ won't take such value. $\endgroup$ – Wojowu Feb 26 '16 at 12:47
  • 1
    $\begingroup$ Hello, thank you for your interest. I actually mean $\lfloor \log\log x\rfloor$. I wrote this because around this integer, $\pi_k(x)\sim\pi_{k+1}(x)$ so we don't really care where $k$ is plus or minus 1. $\endgroup$ – elie520 Feb 26 '16 at 12:57
  • 2
    $\begingroup$ For each prime $p$ we consider a two-dimensional vector $\xi_p(n)=({\mathbf 1}_{p|n},{\mathbf 1}_{p|n+2})$. Then $(w(n),w(n+2))=\sum_p \xi_p(n)$. Next, if we choose $n\leqslant x$ at random, $\xi_p$ is a random vector with known distribution, and they almost do not correlate. It suggests that their sum satisfies some sort of CLT, as in Erdos-Kac theorem. $\endgroup$ – Fedor Petrov Feb 26 '16 at 18:41
  • $\begingroup$ well, the fact they do not correlate mutually is too weak, but we may compute all moments with high acccuracy, it is already enough for CLT $\endgroup$ – Fedor Petrov Feb 26 '16 at 19:41
  • 1
    $\begingroup$ Thank you. We indeed have "Erdos-Kac"-like theorems for (f(n),g(n+2)) for f, g additive (or maybe strongly additive), but they do not allow, to my knowledge, to recover the local laws. $\endgroup$ – elie520 Feb 26 '16 at 20:11

The problem as been solved and generalized. Thread can be closed.

  • 1
    $\begingroup$ Questions that have been answered on this site are not "closed"; rather, the asker can "accept" an answer by clicking the checkmark on the left. $\endgroup$ – j.c. Nov 27 '17 at 12:56

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.