# Is the following weak version of second Hardy-Littlewood conjecture already known?

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

• 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. – GH from MO Oct 10 '18 at 5:37
• 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) – Thomas Bloom Oct 10 '18 at 13:09
• 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 $\frac{x}{\log^2 x}$. – Jan-Christoph Schlage-Puchta Oct 12 '18 at 14:09
• But the last inequality does not contain $\epsilon$. – Jan-Christoph Schlage-Puchta Oct 12 '18 at 16:57
• See my response below. – GH from MO Nov 1 '18 at 23:03

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 $$\pi(ky)+\pi(y)-\pi((k+1)y)\gg_k\frac{y}{\log^2 y},\qquad y\geq M_k.$$
• Do the subscripts $O_k$ and $\gg_k$ just indicate that the implicit constants depend on $k$? – Mike Miller Eismeier Nov 1 '18 at 23:18
• What is the best known expression for $M_k$? – user57432 Nov 2 '18 at 12:18
• @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. – GH from MO Nov 2 '18 at 18:31