Timeline for Is there any formula to find number of Pythagorean triplets between two integers 2 and j, j>2?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Jul 21, 2016 at 18:12 | comment | added | Benjamin Dickman | @YaakovBaruch Ah, yes... something obvious! Thanks. | |
| Jul 21, 2016 at 18:06 | comment | added | Yaakov Baruch | yes, but as I mentioned in the previous comment, another factor of 2 must come from switching $r$ and $s$. My counting of positive triples assumed $0< r < s < n$. | |
| Jul 21, 2016 at 15:47 | comment | added | Benjamin Dickman | @YaakovBaruch (Perhaps I am missing something obvious but...) shouldn't allowing $r$ and $s$ to be negative create a factor $4$ discrepancy? There is an $n \geq 1$ requirement... | |
| Jul 20, 2016 at 12:38 | comment | added | Yaakov Baruch | to be precise: $n$ is positive in the paper above, while $r$ and $s$ can be negative and can be switched with each other. | |
| Jul 20, 2016 at 8:47 | comment | added | Yaakov Baruch | I counted 1216894 triples below 637460, of which 101461 are prime. The latter number seems consistent with $\frac{1}{2\pi}N$ rather than with $\frac{4}{\pi}N$. The number 1216894 for all triples in turn seems consistent with $\frac{1}{2\pi}N \log{N}$ rather than $\frac{4}{\pi} N\log{N}$. The factor of $8$ discrepancy clearly comes from one paper counting integer triples and the other POSITIVE integer triples. | |
| Jul 19, 2016 at 14:54 | history | answered | Benjamin Dickman | CC BY-SA 3.0 |