Very recently I was going through my previous MSE posts and I stumbled upon some of them regarding the Second Hardy-Littlewood Conjecture which states that,

For all $x,y\ge 2$ we have, $$\pi(x)+\pi(y)\ge \pi(x+y)$$where $\pi$ is the Prime Counting Function.

Observing that the Second Hardy-Littlewood Conjecture is equivalent to the following,

For all $k\ge 1$ and $y\in \mathbb{R}$ the following holds, $$\pi(ky)+\pi(y)\ge \pi((k+1)y)$$where $\pi$ is the Prime Counting Function.

I was wondering whether proving the following weak version of the conjecture is something significant or is it already known,

Proposition. For all $k\ge 1$ there exists $M_{k}>0$ such that for all $y\ge M_{k}$ we have, $$π(ky)+π(y)>π((k+1)y)$$

I searched the internet for something similar to this but even though I found some results closely related to the above, I could't find this exact result.

So my questions are,

  • Is the above proposition well-known? If so can anyone point me out to the paper/book that contains a proof of it or even better to some theorem from which this result follows?

  • If not then is the proof of this result considered a significant result?

I agree that the second question may seem to be more opinionated but currently this is the best version that I can come up with. I will be glad to receive constructive suggestions on making this question more specific and answerable

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    $\begingroup$ It seems that the first sentence in your Proposition is meant to be a condition, so you should start it with "Assume that". However, this condition follows from the Prime Number Theorem and the asymptotic expansion of $\mathrm{li}(y)$. So it seems that you are claiming the second sentence in your Proposition unconditionally, which as you remark is the second Hardy-Littlewood conjecture. Note also that this conjecture is believed to be false, as it contradicts the Hardy-Littlewood conjecture on prime tuples. $\endgroup$ – GH from MO Oct 10 '18 at 5:37
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    $\begingroup$ I suppose not equivalent, but regardless this 'weaker' form of the Hardy-Littlewood conjecture is still enough to contradict the Prime Tuple conjecture (see e.g. Theorem 7.16 in Montgomery and Vaughan) $\endgroup$ – Thomas Bloom Oct 10 '18 at 13:09
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    $\begingroup$ It is not clear what the last occurrence of $y_0$ in the proposition means. If we assume that $y_0$ may depend on $k$, i.e. $\forall k\exists y_0\forall y>y_0: \pi(ky)+\pi(y)\geq\pi((k+1)y)$, then the statement does not contradict the Prime tuple conjecture. In fact, it follows from any version of the prime number theorem with an error term better than $\frac{x}{\log^2 x}$. $\endgroup$ – Jan-Christoph Schlage-Puchta Oct 12 '18 at 14:09
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    $\begingroup$ But the last inequality does not contain $\epsilon$. $\endgroup$ – Jan-Christoph Schlage-Puchta Oct 12 '18 at 16:57
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    $\begingroup$ See my response below. $\endgroup$ – GH from MO Nov 1 '18 at 23:03

I have not seen your proposition before, but it follows routinely from the prime number theorem.

Indeed, there exists an absolute constant $c>0$ such that $$\pi(ky)+\pi(y)-\pi((k+1)y)=\mathrm{Li}(ky)+\mathrm{Li}(y)-\mathrm{Li}((k+1)y)+O_k\left(y e^{-c\sqrt{\log y}}\right),$$ where $$\mathrm{Li}(ky)+\mathrm{Li}(y)-\mathrm{Li}((k+1)y)=\int_0^y\left(\frac{k}{\log(kt)}+\frac{1}{\log t}-\frac{k+1}{\log(k+1)t}\right)dt.$$ For fixed $k$ and $t\to\infty$, the integrand is \begin{align*}&\frac{k}{\log t}\left(1-\frac{\log k+o(1)}{\log t}\right)+\frac{1}{\log t}-\frac{k+1}{\log t}\left(1-\frac{\log(k+1)+o(1)}{\log t}\right)\\[8pt]&=\frac{(k+1)\log(k+1)-k\log k+o(1)}{\log^2 t},\end{align*} whence there exists $C_k$ such that $$\mathrm{Li}(ky)+\mathrm{Li}(y)-\mathrm{Li}((k+1)y)\gg_k\frac{y}{\log^2 y},\qquad y\geq C_k.$$ As $\log^2 y$ grows much slower than $e^{c\sqrt{\log y}}$, we conclude that there exists $M_k$ such that $$\pi(ky)+\pi(y)-\pi((k+1)y)\gg_k\frac{y}{\log^2 y},\qquad y\geq M_k.$$

  • $\begingroup$ Do the subscripts $O_k$ and $\gg_k$ just indicate that the implicit constants depend on $k$? $\endgroup$ – Mike Miller Eismeier Nov 1 '18 at 23:18
  • $\begingroup$ @MikeMiller: Yes. This is standard notation in analytic number theory. $\endgroup$ – GH from MO Nov 1 '18 at 23:46
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    $\begingroup$ I assumed but wanted to check. Thanks for clarifying for an outsider! $\endgroup$ – Mike Miller Eismeier Nov 2 '18 at 2:11
  • $\begingroup$ What is the best known expression for $M_k$? $\endgroup$ – user57432 Nov 2 '18 at 12:18
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    $\begingroup$ @user170039: All the constants in my argument (including $c$) are effective and can be calculated by adding more detail. I leave this to you or other interested readers. $\endgroup$ – GH from MO Nov 2 '18 at 18:31

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