Timeline for What did Yu Jianchun discover about Carmichael numbers?
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14 events
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| Aug 2, 2016 at 3:37 | comment | added | Zhiyun Cheng | @Charles: $=\frac{1}{8}\times3^{3n}(3^n+1)^3(3^n+2)^3(3^{2n-1}+2\times3^{n-1}+1)^3$ $=[\frac{1}{2}\times 3^n(3^n+1)(3^n+2)(3^{2n-1}+2\times3^{n-1}+1)]^3$ The number is $y-x+1=(3^n+1)^3$ Therefore $\sum\limits_{i=0}^{(3^n+1)^3-1}(\frac{1}{2}\times3^{n-1}(3^{3n}+3^{2n}-5\times3^n-9)+i)^3=\frac{1}{2}\times3^n(3^n+1)(3^n+2)(3^{2n-1}+2\times3^{n-1}+1)$ | |
| Jul 27, 2016 at 17:48 | comment | added | Charles | @ZhiyunCheng: Thank you! Would you translate the other image as well, the one that starts with $\frac183^{3n}$ fragment that I included in the question? It looks like just a few words interspersed with formulas. | |
| Jul 26, 2016 at 9:45 | history | edited | Myshkin | CC BY-SA 3.0 |
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| Jul 23, 2016 at 6:37 | comment | added | Zhiyun Cheng | Here is a translation of the picture above: PS: Absolute (Fermat) pseudoprimes that can be written as the product of three (prime?) factors can be completely classified. For example, the two simplest formulas which contain the quadratic term are 1. $(6n+1)(18n+1)(54n^2+12n+1)$; 2. $(1764n-139)(2268n-179)(1000188n^2-157752n+6221))$. 7 is a magic number. A deformation of the Fermat's little theorem (with the same base) $\frac{(\frac{N^{p_1}-N}{N}-\frac{N^{p_1-p_2+1}-N}{p_1-p_2})(p_1-p_2)}{p_2}(N, p_1, p_2$ | |
| Jul 22, 2016 at 2:14 | vote | accept | Charles | ||
| Jul 22, 2016 at 0:56 | comment | added | Gerry Myerson | I can confirm that the number you ask about is not Carmichael. | |
| Jul 21, 2016 at 20:38 | comment | added | Ben Green | It's certainly a very difficult unsolved problem to show that there are infinitely many n for which each of the three factors in (1), or in (2), are prime - if that is what is required here. | |
| Jul 21, 2016 at 19:48 | history | edited | Myshkin | CC BY-SA 3.0 |
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| Jul 21, 2016 at 19:31 | history | edited | Myshkin | CC BY-SA 3.0 |
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| Jul 21, 2016 at 19:01 | history | edited | Myshkin | CC BY-SA 3.0 |
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| Jul 21, 2016 at 17:59 | comment | added | Sylvain JULIEN | A little more information: chinatopix.com/articles/95664/20160718/… | |
| Jul 21, 2016 at 17:36 | history | edited | Myshkin | CC BY-SA 3.0 |
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| Jul 21, 2016 at 15:56 | comment | added | Charles | Is it? Banks mentions the AGP result, but earlier in the article it's said to be a method to verify Carmichael numbers, and Yu refers to an "algorithm", which sounds more like "is n a Carmichael number or not" than "there are infinitely many Carmichael numbers". | |
| Jul 21, 2016 at 15:51 | history | answered | Myshkin | CC BY-SA 3.0 |