Timeline for A geometric proof that there are infinitely many primes?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 1, 2023 at 16:53 | history | edited | mathoverflowUser | CC BY-SA 4.0 |
added current status of proof.
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| Jul 1, 2023 at 16:51 | vote | accept | mathoverflowUser | ||
| Jul 1, 2023 at 16:23 | answer | added | Mark Schultz-Wu | timeline score: 5 | |
| Jul 1, 2023 at 6:17 | comment | added | mathoverflowUser | @Mark: Thanks for your comment. However I get that: $d(n)=\prod_{k=1}^n \prod_{p|k} (1-1/p^2) = \prod_{p \le n} (1-1/p^2)^{\operatorname{floor}(n/p)}$ which I can verify empirically: sagecell.sagemath.org/?q=yuaowe (d1=dd) and it should follow theoretically. If it is not asked too much: It would be nice to promote your comment into an answer, so that I can follow your reasoning. | |
| Jul 1, 2023 at 6:04 | comment | added | Mark Schultz-Wu | Your display states "we want to look at ... $d(n):=\det(G_n) = \prod_{i = 1}^n\frac{h(i)}{i^2} = \prod_{k = 1}^n\prod_{p\mid k}1-p^{-2}$. This seemed to be where $d(n)$ was defined, and by this definition the result seemed obvious. | |
| Jul 1, 2023 at 3:16 | comment | added | mathoverflowUser | @Mark: I am not sure how you arrive at this expression, therefore I programmed it to see if $d(n+1)/d(n)$ is equal to your expression, and it seems that it is not: sagecell.sagemath.org/?q=rosgpt | |
| Jun 30, 2023 at 22:26 | history | edited | Michael Hardy | CC BY-SA 4.0 |
edited body
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| Jun 30, 2023 at 20:42 | comment | added | Mark Schultz-Wu | Each $\prod_{p\mid n+1}(1-p^{-2})$ is the product of many terms in $(0,1)$, and is therefore in $(0,1)$ itself (and notably $<1$). It appears that $d(n+1)/d(n)$ is precisely this quantity, so therefore $d(n+1)<d(n)$ is strictly decreasing. | |
| Jun 30, 2023 at 19:44 | comment | added | mathoverflowUser | @Mark: I am not sure how you mean this comment? | |
| Jun 30, 2023 at 19:31 | comment | added | Mark Schultz-Wu | Isn't $d(n+1)/d(n) = \prod_{p\mid n+1} (1-p^{-2}) < 1$? | |
| Jun 30, 2023 at 18:53 | comment | added | mathoverflowUser | related: mathoverflow.net/questions/373475/… | |
| Jun 30, 2023 at 18:45 | history | asked | mathoverflowUser | CC BY-SA 4.0 |