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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 !

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    $\begingroup$ 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, 2016 at 12:47
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    $\begingroup$ 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, 2016 at 12:57
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    $\begingroup$ 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. $\endgroup$ Feb 26, 2016 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 $\endgroup$ Feb 26, 2016 at 19:41
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    $\begingroup$ 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, 2016 at 20:11

1 Answer 1

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

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    $\begingroup$ 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, 2017 at 12:56

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