Timeline for A prime number pattern
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Jul 31, 2012 at 17:16 | history | edited | Will Sawin | CC BY-SA 3.0 |
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| Jul 31, 2012 at 17:16 | comment | added | Will Sawin | Yes, that's basically why. There are lots of solutions, and a solution implies that the interval is quite large. Then it's possible for there to be too subtractions or two additions in a row, meaning the interval can grow faster and can continue to be quite large. | |
| Jul 31, 2012 at 16:54 | comment | added | user9072 | I did not check the entire argument in detail, but for the contrast of the two situations: for 1, we can not 'manually check there are no sulotions up till then'. | |
| Jul 31, 2012 at 15:43 | comment | added | Gerhard Paseman | I think you mean [1-p,p]. Also, I worry that the same argument can apply for starting with (ending with) 1. If you can contrast the two situations, that will make a compelling argument. Gerhard "Ask Me About System Design" Paseman, 2012.07.31 | |
| Jul 31, 2012 at 12:14 | comment | added | Will Sawin | I believe I clarified that. | |
| Jul 31, 2012 at 12:14 | history | edited | Will Sawin | CC BY-SA 3.0 |
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| Jul 31, 2012 at 5:14 | comment | added | Furlox | @Will: I am totally lost as to what you mean by 'smallest subinterval $[p-1,p]$ ... which after $p$ is that number ends at $-1$.' Also, you refer to primes where $Pi(x=p)$ is odd. However, the unresolved 'pattern' relates to $Pi(x=p)$ being even. If thats what $p_1$ and $p_2$ deal with, I'm really sorry. May I ask for a complete example, so I can understand better? Thanks! | |
| Jul 30, 2012 at 16:47 | history | answered | Will Sawin | CC BY-SA 3.0 |