Timeline for A curious property of Ramanujan's function $\tau(n)$
Current License: CC BY-SA 3.0
16 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Sep 26, 2016 at 5:00 | history | edited | Tito Piezas III | CC BY-SA 3.0 |
Deleted phrase.
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| Apr 30, 2015 at 14:33 | history | edited | Tito Piezas III | CC BY-SA 3.0 |
Removed sentence made obsolete by update.
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| Apr 30, 2015 at 14:10 | comment | added | Tito Piezas III | I have found my error. Surprisingly, I still ended up with an integer. Kindly see update above. | |
| Apr 30, 2015 at 14:07 | history | edited | Tito Piezas III | CC BY-SA 3.0 |
Added Update.
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| Apr 30, 2015 at 13:28 | comment | added | Zurab Silagadze | I have problems to login in the chat. We can continue by email: [email protected] Here, I'm afraid, such type of comments may be not appropriate. | |
| Apr 30, 2015 at 13:18 | comment | added | Zurab Silagadze | Let us continue this discussion in chat. | |
| Apr 30, 2015 at 13:16 | comment | added | Tito Piezas III | @ZurabSilagadze: This is strange. We get different integers, but they differ only by $15$ $$19344611534578279675385464545221922990834243736585-\\19344611534578279675385464545221922990834243736570 = 15$$ | |
| Apr 30, 2015 at 13:08 | comment | added | Zurab Silagadze | I have checked again. If I subtract 42, maxima gives 19344611534578279675385464545221922990834243736570 and if I add 96723057672891398376927322726109614954171218682856/5. | |
| Apr 30, 2015 at 13:04 | comment | added | Tito Piezas III | If I subtract 42, I get a rational number. We cannot both be correct. You started with $c(1) = 196884$? | |
| Apr 30, 2015 at 13:02 | comment | added | Zurab Silagadze | No. I have subtracted 42. If I add, the result is not a multiple of 70. I used maxima. | |
| Apr 30, 2015 at 12:56 | comment | added | Tito Piezas III | @ZurabSilagadze: Regarding the mod, for the j-function, I have $$\frac{\sum_{n=1}^{24}c(n)^2-(\color{red}{-}42)}{70} =19344611534578279675385464545221922990834243736585$$ whereas, $$\frac{\sum_{n=1}^{24}c(n)^2-(42)}{70} =\frac{96723057672891398376927322726109614954171218682781}{5}$$hence $$ \sum_{n=1}^{24}c(n)^2 \equiv -42 \;\; (\mathrm{mod} \;70)$$ One must add (not subtract) $42$ to the sums, correct? (I've double-checked my results with Mathematica.) | |
| Apr 30, 2015 at 12:18 | comment | added | Zurab Silagadze | Interesting! However, I have checked that in all four cases the correct number is 42, not -42. Note the definition of $a\equiv b (\mathrm{mod}\;\;n)$: $a-b$ is divisible by $n$. Please check and correct. Also I noticed that all four sums are congruent to zero (mod 14)! I suggest you to publish your findings on arxiv. | |
| Apr 30, 2015 at 5:49 | history | edited | Tito Piezas III | CC BY-SA 3.0 |
Missing summation symbol.
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| Apr 29, 2015 at 16:47 | history | edited | Tito Piezas III | CC BY-SA 3.0 |
Details.
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| Apr 29, 2015 at 16:23 | history | answered | Tito Piezas III | CC BY-SA 3.0 |