Timeline for Does listing the prime factors always stop?
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
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| Jul 12, 2010 at 16:59 | comment | added | O.R. | whoops & whoops | |
| Jul 11, 2010 at 23:31 | comment | added | Johannes Hahn |
Also: "Probability" not always refers to the stochastic interpretation as measure space with $\mu(\Omega)=1$. When talking about subsets of $\mathbb{N}$ it seems to be very common to call the density $\lim_{n\to\infty}\frac{|A\cap\{1,\ldots,n\}|}{n}$ (if it exists) the probability of a natural number being in $A$.
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| Jul 11, 2010 at 15:49 | comment | added | JBL | No, it doesn't: "happens with probability zero" is not the same as "never happens." | |
| Jul 11, 2010 at 14:45 | comment | added | O.R. | Ah, OK. It is Conway being funny, because proving that the provability is "exactly zero" does prove it. | |
| Jul 11, 2010 at 14:43 | comment | added | Nurdin Takenov | It means that if we will use non-rigorous arguments, then we can deduce that for every initial number this sequence ends in a prime number. But, it's not a proof. | |
| Jul 11, 2010 at 14:36 | vote | accept | O.R. | ||
| Jul 11, 2010 at 14:30 | comment | added | O.R. | What does the Conways comment mean? "Probabilistic arguments give exactly zero for the chance that the sequence of integers starting at n contains no prime, the expected number of primes being given by a divergent sequence - John Conway (conway(AT)math.princeton.edu)" He said exactly zero? Is it them proven? | |
| Jul 11, 2010 at 14:16 | history | undeleted | Nurdin Takenov | ||
| Jul 11, 2010 at 14:16 | history | deleted | Nurdin Takenov | ||
| Jul 11, 2010 at 14:15 | history | answered | Nurdin Takenov | CC BY-SA 2.5 |