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 30, 2012 at 15:18 | comment | added | user9072 | @Gerhard Paseman: 10 is not prime, alas. So I do not see how this 'explains the specific result seen by the poster', which are quite focused on the fact that for primes with an even index (in the sequence of primes) the result is (allegedly) always 1, and never -1. | |
| Jul 30, 2012 at 14:58 | comment | added | Gerhard Paseman | Quid, (-1) can and does occur for some start values, such as 10. You can work backwards to see which values occur. It should be clear using the above result that the partial sum before -1 will be 1, and then one has a choice between 4 and -2 which can come from -1 and 3 respectively. Using the invariant above, one can limit the possibilities by hand pretty quickly. Gerhard "Ask Me About System Design" Paseman, 2012.07.30 | |
| Jul 30, 2012 at 13:54 | comment | added | Furlox | $-1$ cannot occur, is more personal opinion and evidence based guessing than proven fact. Maybe I didn't make that clear. Making the assumption seems to make the problem 'almost' susceptible to inductive methods, heuristics, or some witch's brew of everything we need :) | |
| Jul 30, 2012 at 13:44 | comment | added | Furlox | @quid: Yes, conjecture is equivalent to $Z_f(P({2n}))=1$ Also, so far, $-1$ hasn't shown up for me (and I assume, for others who checked via programs, otherwise they would have posted). It is very possible that there is some prime number with $Z_f$ equal to $-1$. However, I find it unlikely as every odd number so far has terminated in $\{0,1,2\}$ only. Check the page on M.SE, I think it has a better explanation. | |
| Jul 30, 2012 at 12:14 | comment | added | user9072 | I am a bit confused. Sure this will depend on pi(x) but the question says so (implicitly) when talking about any other prime. For 3 and evry other prime there after...so precisely if one starts at the k-th prime with k even! Yet, then, I still do not see an argument why -1 cannot occur. So I am even more confused. | |
| Jul 30, 2012 at 11:51 | comment | added | Furlox | About the whole $-2$ thing, I applied the algorithm wrong for $9$. And didn't double check. I'm sorry guys! | |
| Jul 30, 2012 at 11:48 | comment | added | Furlox | I would like a formal proof. Knowing the initial parity cannot be used to predict a result. The final parity is still dependent on $\pi(x)$. | |
| Jul 30, 2012 at 5:22 | history | answered | Gerhard Paseman | CC BY-SA 3.0 |