Timeline for A prime number pattern
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 29, 2012 at 16:38 | comment | added | Gerhard Paseman | Using domotorps argument above, one can show that -2 is not an ending value. More specifically, each partial sum lies in the closed interval [-p-1,p], and Doug Zare's observation gives you the parity of the final term. Gerhard "Ask Me About System Design" Paseman, 2012.07.29 | |
| Jul 29, 2012 at 16:24 | comment | added | Furlox | Also, $Z_f(5)=0$ and $Z_f(31)=2$ | |
| Jul 29, 2012 at 16:19 | comment | added | Furlox | At math.stackexchange.com/questions/176394/a-prime-number-pattern @alex has already shown (using above parity argument) that starting positions between -2 to 2 are valid. To resolve the conjecture as proposed, only the -1 case needs scrutiny. | |
| Jul 29, 2012 at 15:55 | comment | added | Gerhard Paseman | I think you will find it of interest to work the sequence backward from a small number of ending positions to see how quickly you jump out of the range of a suggested starting position. For example, one should be able to prove that the sums will not end in ...,10,5,2,0. Gerhard "Ask Me About System Design" Paseman, 2012.07.29 | |
| Jul 29, 2012 at 15:40 | comment | added | Gerhard Paseman | But it is the first step in following the idea of Douglas Zare. After you bound the size, you note that the parity of the answer differs from (the start + 1) by the parity of pi(x). Note that one arrives at (-1) by starting from 10, so I think the range of values will be from -1 to 2. Gerhard "Ask Me About System Design" Paseman, 2012.07.29 | |
| Jul 29, 2012 at 15:20 | comment | added | Furlox | 'If n is prime, it is assumed accounted for by the first step' $19-17-13+11+7-5-3+2=1$ | |
| Jul 29, 2012 at 15:12 | comment | added | Will Sawin | We can extend that example to a counterexample to the main claim! $19 - 19 = 0 + 17 = 17 - 11 = 6 - 7 = -1 + 5 = 4 - 3 = 1 - 2 = -1.$ | |
| Jul 29, 2012 at 14:40 | comment | added | Niemi | Plus, the claim is simply not true for $Z=4 < 2 \cdot 3$. We obtain $4-3-2 = -1$. | |
| Jul 29, 2012 at 14:20 | comment | added | Furlox | Also, while we inductively prove for the next prime $q$, we also need to prove that $z \in \{0,1,2\}$ for all $2r>k>2q$ where $r$ is the next prime (before we attempt induction on $r$). | |
| Jul 29, 2012 at 13:50 | comment | added | Furlox | The above induction wouldn't suffice to prove the regularity in pattern, i.e. only every other prime, beginning with 3, reaches 1. | |
| Jul 29, 2012 at 9:12 | history | answered | domotorp | CC BY-SA 3.0 |