Timeline for are there infinitely many triples of consecutive square-free integers?
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21 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Sep 28, 2013 at 20:29 | history | edited | user9072 |
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| Apr 1, 2011 at 15:58 | comment | added | GH from MO | @Junkie: You are right. | |
| Mar 28, 2011 at 10:32 | answer | added | Marius Overholt | timeline score: 8 | |
| Mar 27, 2011 at 23:07 | comment | added | Junkie | @GH: I guess I don't understand why it is difficult to get something good enough. Only 1/9 of the numbers are divisible by $3^2$, and so on, with an error (remembering mod 4) of no more than 4. Don't bother with fancy inclusion and exclusion, and directly sum $\sum_{n<\sqrt{X}} X/n^2+O(4)$ for $n>1$ odd as an upper bound for counting of those divisible by odd squares. The sum is easily bounded by $0.24X$ without much work, so is less than 1/3. The upper bound here is not close to $1−8/\pi^2$ but that is not needed | |
| Mar 27, 2011 at 20:34 | comment | added | George Lowther | @Junkie: Yes, sorry, my mistake. I deleted that comment now. | |
| Mar 27, 2011 at 20:25 | comment | added | GH from MO | @Junkie: Your argument is OK, but proving your $8/\pi^2$ is no easier then proving the density in the original problem. In fact it is harder, because $8/\pi^2$ comes from evaluating explicitly an Euler product. Well, you only need that the density is less than $1/3$, but even that requires some numerics. | |
| Mar 27, 2011 at 20:07 | comment | added | Junkie | @George Lowther: I removed the integers that are 0 mod 4, since we can ignore them in any case. The proportion is then $8/\pi^2$, as $p=2$ does not contribute. | |
| Mar 27, 2011 at 20:03 | comment | added | GH from MO | @Frank Thorne: I don't think you can truncate at $Y$ as large as $X^{1/10}$ without including some fancy sieving weights. You need that the product of squares of primes up to $Y$ is much smaller than $X$, e.g. $Y=(\log X)/10$ is fine. See my response below. | |
| Mar 27, 2011 at 19:47 | vote | accept | Ewan Delanoy | ||
| Mar 27, 2011 at 19:41 | comment | added | Junkie | A lot of the comments seem to invoke unnecessary sieve theory. Does this simpler argument work? Look at the integers that are not 0 mod 4. Among these, easy argument gives $8/\pi^2$ squarefree, or 19% with a nontrivial square factor. As 19% < 1/3, the proposition follows by pigeonholes, with some density. For more precision, extra work is needed. | |
| Mar 27, 2011 at 19:21 | answer | added | George Lowther | timeline score: 32 | |
| Mar 27, 2011 at 19:02 | answer | added | GH from MO | timeline score: 12 | |
| Mar 27, 2011 at 18:42 | comment | added | Frank Thorne | It might be added to George's comments that you can truncate his product to $p < Y$ with error roughly on the order of $1/Y$ (imagine you are looking for triples in $[X, 2X]$ and take, say, $Y = X^{1/10}$). This is why his proof works here but you can't prove, say, the Twin Prime Conjecture using a similar argument. | |
| Mar 27, 2011 at 18:24 | history | edited | Ewan Delanoy | CC BY-SA 2.5 |
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| Mar 27, 2011 at 18:15 | comment | added | George Lowther | @Qiaochu: The probability of none of $4a+1,4a+2,4a+3$ being a multiple of $p^2$ is $1−3/p^2$, for odd primes $p$ (and it is 1 for $p=2$). Treating these as independent events, you multiply the probabilities together. This is just a heuristic, of course, but I don't think it's hard to turn into a rigorous proof. A triple of consecutive integers has zero probability of all being even, just looking at it mod 2, so the same kind of heuristic gives the proportion of triples of consecutive even integers to be zero, as you'd hope. | |
| Mar 27, 2011 at 18:12 | history | edited | Ewan Delanoy | CC BY-SA 2.5 |
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| Mar 27, 2011 at 17:55 | answer | added | Zander | timeline score: 18 | |
| Mar 27, 2011 at 17:54 | comment | added | Qiaochu Yuan | @George: I don't follow. The proportion of even integers is nonzero but there are no triples of consecutive even integers... | |
| Mar 27, 2011 at 17:35 | comment | added | George Lowther | Surely the proportion of integers $a$ satisfying this property is $\prod_p(1-3/p^2)$, with the product over odd primes $p$? So, yes, there are infinitely many. | |
| Mar 27, 2011 at 17:11 | history | asked | Ewan Delanoy | CC BY-SA 2.5 |