# On the Second Hardy-Littlewood conjecture

I have recently known about the Second Hardy-Littlewood conjecture, and I was struck when I noticed that this conjecture was very related with this question I asked in this site.

So, my question is, is the inequality at the end of this post sufficient to show that the Second Hardy-Littlewood conjecture is correct? Is there any way to show that it is false? I have checked and found no positive integer values $$x,y$$ other than $$x=2, y=2$$ and $$x=837, y=77$$ for which the inequality holds, but I am not able to show that it does not hold for some $$x,y\geq N_0$$.

I provide some context on how the inequality was obtained.

The conjecture states that $$\pi(x+y)\leq \pi(x)+\pi(y)$$ for integers $$x,y\geq 2$$, where $$\pi(n)$$ denotes the prime-counting function.

Without loss of generality, we have that $$x\geq y$$. Therefore, we have that $$\pi(x)\geq \pi(y)$$, and $$\pi(2x)\geq \pi(x+y)\geq \pi(2y)$$.

As proved by user @saulspatz in the MSE post mentioned, we have that, for $$x,y\geq 60184$$, $$2\pi(x)\geq \pi(2x)+\log_2{x}$$ $$2\pi(y)\geq \pi(2y)+\log_2{y}$$ Therefore, we have that $$2(\pi(x)+\pi(y))\geq \pi(2x)+\pi(2y)+\log_2{x}+\log_2{y}$$ Assume that the conjecture is false and that $$\pi(x+y)> \pi(x)+\pi(y)$$. Then, we would have that $$2(\pi(x+y))\geq \pi(2x)+\pi(2y)+\log_2{x}+\log_2{y}$$ For $$x,y\geq 60184$$, and as mentioned in the MSE post, applying Pierre Dusart's bounds, we have that $$\frac {2x}{\log (2x)-1}<\pi(2x)<\frac {2x}{\log (2x) -1.1}$$ $$\frac {2y}{\log (2y)-1}<\pi(2y)<\frac {2y}{\log (2y) -1.1}$$ $$\frac {x+y}{\log (x+y)-1}<\pi(x+y)<\frac {x+y}{\log (x+y) -1.1}$$ Therefore, if the conjecture were false, we could be able to show that $$2\frac {x+y}{\log (x+y) -1.1}\geq\frac {2x}{\log (2x)-1}+\frac {2y}{\log (2y)-1}+\log_2{x}+\log_2{y}$$ Operating, we have that $$\frac {2(x+y)}{\log (x+y) -1.1}\geq\frac {2x}{\log (2x)-1}+\frac {2y}{\log (2y)-1}+\log_2{x}+\log_2{y}$$ $$\frac {2x}{\log (x+y) -1.1}+\frac {2y}{\log (x+y) -1.1}\geq\frac {2x}{\log (2x)-1}+\frac {2y}{\log (2y)-1}+\log_2{x}+\log_2{y}$$ $$2x\left(\frac {1}{\log (x+y) -1.1}-\frac{1}{\log (2x) -1}\right)+2y\left(\frac {1}{\log (x+y) -1.1}-\frac {1}{\log (2y) -1}\right)\geq\log_2{x}+\log_2{y}$$ I am not able to simplify the last inequality further to obtain something helpful to disprove it for some $$x,y\geq N_0$$.

$$2\frac {x+y}{\log (x+y) -1.1}\geq\frac {2x}{\log (2x)-1}+\frac {2y}{\log (2y)-1}+\log_2{x}+\log_2{y}$$

Let $$f(x,y):=\frac {2x+2y}{\log (x+y) -1.1}-\frac {2x}{\log (2x)-1}-\frac {2y}{\log (2y)-1}-\frac{\log(x)}{\log(2)}-\frac{\log(y)}{\log(2)}$$

Let us consider the case $$y=x$$.

We have $$f(x,x)=\frac {0.4x\log(2)-2\log(x)(\log (2x) -1.1)(\log (2x)-1)}{(\log (2x) -1.1)(\log (2x)-1)\log(2)}$$ which can be written as $$f(x,x)=\frac {x\bigg(0.4\log(2)-2\cdot\dfrac{\log(x)}{x^{1/3}}\cdot\dfrac{\log (x)+\log(2) -1.1}{x^{1/3}}\cdot\dfrac{\log (x)+\log(2)-1}{x^{1/3}}\bigg)}{(\log (2x) -1.1)(\log (2x)-1)\log(2)}$$

Since $$\displaystyle\lim_{x\to\infty}\dfrac{\log(x)}{x^{1/3}}=0$$, we can say that there is a positive integer $$N$$ such that for every $$x$$ satisfying $$x\ge N$$, $$f(x,x)\gt 0$$.

This means that your idea does not work to prove that the conjecture is true.

Let us consider the case $$x=y+10$$.

$$f(y+10,y)=\frac {4y+20}{\log (2y+10) -1.1}-\frac {2y+20}{\log (2y+20)-1}-\frac {2y}{\log (2y)-1}-\frac{\log(y^2+10y)}{\log(2)}$$

$$=\frac{\color{red}{g(y)}}{(\log (2y+10) -1.1)(\log (2y+20)-1)(\log (2y)-1)\log(2)}$$

where

$$\color{red}{g(y)}=(4y+20)(\log (2y+20)-1)(\log (2y)-1)\log(2)-(2y+20)(\log (2y+10) -1.1)(\log (2y)-1)\log(2)-2y(\log (2y+10) -1.1)(\log (2y+20)-1)\log(2)-\log(y^2+10y)(\log (2y+10) -1.1)(\log (2y+20)-1)(\log (2y)-1)$$

$$=(4y+20)(\log (2y+20)-1)(\log (2y)-1)\color{blue}{h(y)}$$

and $$\color{blue}{h(y)}=\log(2)-\frac{(y+10)(\log (2y+10) -1.1)}{(2y+10)(\log (2y+20)-1)}\log(2)-\frac{y(\log (2y+10) -1.1)}{(2y+10)(\log (2y)-1)}\log(2)-\frac{\log(y^2+10y)(\log (2y+10) -1.1)}{4y+20}$$

$$=\underbrace{\bigg(1-\frac{(y+10)(\log (2y+10) -1.1)}{(2y+10)(\log (2y+20)-1)}-\frac{y(\log (2y+10) -1.1)}{(2y+10)(\log (2y)-1)}\bigg)}_{i(y)}\log(2)-\underbrace{\frac{\log(y^2+10y)(\log (2y+10) -1.1)}{4y+20}}_{j(y)}$$

$$=i(y)\bigg(\log(2)-\frac{j(y)}{i(y)}\bigg)$$

WolframAlpha says

• For $$y\gt 2.96567$$, $$i(y)\gt 0$$ (see here)

• $$\displaystyle\lim_{y\to\infty}\dfrac{j(y)}{i(y)}=0$$ (see here)

So, this means that there is a positive integer $$N_1$$ such that for every $$(x,y)$$ satisfying $$x=y+10\ge N_1$$, $$f(x,y)\gt 0$$ holds.

• thanks for your answer! Indeed, by substitution and simplification, $f(x,x)=\frac{2x}{\log (2x)-1.1} -\frac{2x}{\log (2x)-1}-\log_2 x$, and it can be checked that $f(x,x)>0$ for all $x\geq 3638$. However, if $y=x$, the conjecture is true for that case just by the inequality $2\pi(x)\geq \pi(2x)+\log_2{x}$. Could the formula be used restricting the values of $x$ and $y$ that can be plugged in just for those for which the conjecture is not known to be true? Commented Jun 1, 2023 at 19:58
• @Juan Moreno : I added another example. Commented Jun 3, 2023 at 10:53