Timeline for Is $1 = \sum_{n=1}^{\infty} \frac{\pi(p_n^2)-n+2}{p_n^3-p_n}$ , where $\pi$ denotes the prime counting function and $p_n$ denotes the $n$-th prime?
Current License: CC BY-SA 4.0
20 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 1, 2023 at 19:26 | comment | added | mathoverflowUser | @mick: Do not think it is trivial, just because you understand it. | |
| Nov 28, 2023 at 19:43 | comment | added | mick | @mathoverflowUser but that one is so trivial. Not surprising or complex. My point remains. | |
| Nov 28, 2023 at 13:16 | comment | added | mathoverflowUser | @mick: The equation in the "edit" section, can be proven. See my remark on the edit section, and the comments from fedja and GHfromMO. | |
| Nov 28, 2023 at 12:25 | comment | added | mick | I doubt any such type of equation could be true. Let alone provable. If it looks to good to be true it usually is. | |
| Nov 27, 2023 at 17:36 | history | edited | mathoverflowUser | CC BY-SA 4.0 | Corrected formula and added explanations from the comments. |
| Nov 27, 2023 at 13:13 | history | edited | LSpice | CC BY-SA 4.0 | Comma inside displayed equation |
| Nov 27, 2023 at 1:38 | comment | added | mathoverflowUser | @GHfromMO:Thanks for the clarification. | |
| Nov 27, 2023 at 1:27 | comment | added | GH from MO | @mathoverflowUser fedja's formula follows from the Chinese Remainder Theorem. Indeed, the integers counted by $a_p(n)$ have zero residue modulo the prime $p$, and nonzero residue modulo any prime $q<p$. Hence these integers have $\prod_{q<p}(q-1)$ possible residues modulo $p\prod_{q<p}q$. The result follows by decomposing $\{1,\dotsc,n\}$ into these admissible residue classes modulo $p\prod_{q<p}q$. | |
| Nov 27, 2023 at 1:04 | comment | added | mathoverflowUser | @GHfromMO: Thanks again for the answer. | |
| Nov 27, 2023 at 1:03 | comment | added | mathoverflowUser | @fedja: How do you see your formula? $\frac{1}{p}\prod_{q<p,q \text{prime}} ( 1-\frac{1}{q})$? | |
| Nov 27, 2023 at 1:01 | vote | accept | mathoverflowUser | ||
| Nov 27, 2023 at 1:00 | comment | added | GH from MO | The sum is less than $1$. See my calculation below. | |
| Nov 27, 2023 at 0:54 | answer | added | GH from MO | timeline score: 18 | |
| Nov 26, 2023 at 21:56 | comment | added | fedja | Perhaps I misunderstand your definition of $a_p(n)$, but, as written, the limit of the ratio $a_{p}(n)/n$ is, clearly, not what you wrote but $\frac 1p\prod_{q<p, q\text{ prime}}(1-\frac 1q)$, isn't it? | |
| Nov 26, 2023 at 21:14 | comment | added | YCor | It's sensitive to the choice of enumeration of primes which makes it very unlikely (and makes the sum of little relevance). Apparently there's no serious indication at all that this should hold. | |
| Nov 26, 2023 at 20:05 | comment | added | mathoverflowUser | @JoshuaZ: It is a division of the unit, so if the two first half-guessed / half-proven formulas are correct, it should converge to 1. But I am not sure, so this is why I am asking this question. | |
| Nov 26, 2023 at 19:57 | comment | added | JoshuaZ | I would be very surprised if this ended up converging to exactly 1 though. | |
| Nov 26, 2023 at 19:56 | comment | added | JoshuaZ | When summed up to n=16000, the sum is around 0.95228. It seems hard to tell from the slow growth rate if it is likely to go beyond 1 or stay below 1. | |
| Nov 26, 2023 at 19:35 | history | edited | mathoverflowUser | CC BY-SA 4.0 | typo |
| Nov 26, 2023 at 19:14 | history | asked | mathoverflowUser | CC BY-SA 4.0 |