Timeline for Dividing a finite arithmetic progression into two sets of same sum: always the same asymptotics?
Current License: CC BY-SA 3.0
8 events
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| Oct 3, 2017 at 8:23 | history | edited | Wolfgang | CC BY-SA 3.0 |
included Anthony Quas' comment, added prob tag
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| Oct 3, 2017 at 0:17 | comment | added | Anthony Quas | So I did the calculations with the CLT. This is still heuristic at this point, but I think the correct answer should be $\sqrt{d}\times \sqrt{6/\pi}2^m/n^{3/2}$. Are there any examples that suggest that this isn't right? By the way, it's probably more natural to express everything in terms of $m$ and $d$: $\sqrt{6/\pi}2^m/(m^{3/2}d)$. | |
| Oct 2, 2017 at 9:21 | comment | added | Wolfgang | @AnthonyQuas Maybe. I was already reminded of the central limit theorem because of the $1/\sqrt{\pi}$ lurking out, but I wonder if it is applicable given that the two subsets must have exactly the same sum. For "near-same" sum it would seem more plausible. Anyway, probability theory is not my domain. Would you think it's worth adding the probability tag? | |
| Oct 2, 2017 at 9:00 | comment | added | Wolfgang | @MaxAlekseyev Intuitively, I would expect the number of pluses and minuses for fixed d,r,m to behave itself like a gaussian curve (with fairly small variance that should tend to 0 for $m\to\infty$ with d,r fixed). That might be a different question. | |
| Oct 2, 2017 at 5:27 | comment | added | Anthony Quas | I haven’t tried calculating yet, but can you just get the constants out of the central limit theorem? I have in mind adding or subtracting each term with equal probability, and trying to estimate the probability that you obtain 0. The distribution of this sum should be very close to a normal random variable, probably with very nice “smooth” densities on Z. | |
| Oct 2, 2017 at 4:30 | comment | added | Max Alekseyev | Is there a chance that the number of solutions with equal number of pluses and minuses prevails in the asymptotic? That would eliminate $r$ from the main term, and essentially make it depend only on $n$ and $m\approx n/d$. | |
| Oct 1, 2017 at 14:58 | history | edited | Wolfgang | CC BY-SA 3.0 |
added 2 characters in body
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| Oct 1, 2017 at 14:51 | history | asked | Wolfgang | CC BY-SA 3.0 |