Timeline for On the connection between sums of prime numbers and distribution of prime numbers
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 10, 2021 at 23:00 | history | bounty ended | Juan Moreno | ||
| Jun 10, 2021 at 23:00 | vote | accept | Juan Moreno | ||
| Jun 9, 2021 at 17:41 | comment | added | GH from MO | Using that, for large $x$, there is always prime in $[x,x+x^{3/5}]$, it follows from my earlier bounds that $\pi(x)-S(x)>x^{0.7499}$ holds for infinitely many primes $x$. Also, I don't think that investigating $\pi(x)-S(x)$ is an interesting research direction. | |
| Jun 9, 2021 at 17:38 | comment | added | Juan Moreno | And, despite of the interest that the conjecture might have, I am more interested in knowing your thoughts about the possible interest of this connection in the study of the distribution of prime numbers, and its worthiness to continue investigating it or not... | |
| Jun 9, 2021 at 17:31 | comment | added | Juan Moreno | clear now, thanks for you explanation! Then it can be discarded to prove $\pi(x)-S(x)\leq{x^\frac{1}{2}}$ for all $x\in\mathbb{N}$. And it seems difficult to prove it for all $x\in\mathbb{P}$, or at least for those prime numbers $p_n$ inmediately lesser than $p_{k}^2$. What do you think about it? Or maybe there is another underlying reason for the conjecture to be true (or false!). | |
| Jun 9, 2021 at 16:33 | comment | added | GH from MO | For any $c<3/4$ we have $\limsup\frac{\pi(x)-S(x)}{x^c}>0>\liminf\frac{\pi(x)-S(x)}{x^c}$. In particular, we have this for $c=2/3$ , which contradicts your expectations. This is what I proved in my earlier post, I am sorry if that was not clear to you. Informally speaking, the difference of $\pi(x)$ and $S(x)$ is infinitely often as large as $x^{0.7499}$ (in both directions), and here you can replace $0.7499$ by anything less than $3/4$. | |
| Jun 9, 2021 at 15:11 | comment | added | GH from MO | @JuanMoreno: We have $\limsup\frac{\pi(x)-S(x)}{x^{2/3}}>0$, which is incompatible with $\pi(x)-S(x)\leq x^{1/2}$. Similarly, we have $\liminf\frac{\pi(x)-S(x)}{x^{2/3}}<0$, which is incompatible with $\pi(x)-S(x)\geq -x^{1/2}$ | |
| Jun 9, 2021 at 14:44 | comment | added | Juan Moreno | thanks for your contribution! However, if we have $f(x)=\pi(x)-S(x)$ and $g(x)=x^\frac{1}{2}$, and we state that $f(x)=\Omega_\pm(g(x)$, I understand it means that $\limsup_{x \to \infty} \frac{f(x)}{g(x)} > 0$ while $\liminf_{x \to \infty} \frac{f(x)}{g(x)} < 0$. Therefore, it could be the case that $\limsup_{x \to \infty} \frac{\pi(x)-S(x)}{x^\frac{1}{2}} < 1$ while $\liminf_{x \to \infty} \frac{\pi(x)-S(x)}{x^\frac{1}{2}} > -1$, which is compatible with both $\pi(x)-S(x)\leq{x^\frac{1}{2}}$ and $\pi(x)-S(x)\geq{-x^\frac{1}{2}}$. Or am I missing something? | |
| Jun 9, 2021 at 12:35 | history | answered | GH from MO | CC BY-SA 4.0 |