Timeline for Splitting the integers from $1$ to $2n$ into two sets with products as close as possible
Current License: CC BY-SA 4.0
9 events
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| Jun 18, 2022 at 16:32 | comment | added | Max Alekseyev | @PeterTaylor: I've extended A352813 to $n\leq 70$ terms, and confirmed that the equality ${\tt A352813}(n) = {\tt A038667}(2n)$ continues to hold. There is a counterexamples along these lines with the split of 39!, where the smaller factor can be represented as the product of $\lfloor 39/2\rfloor = 19$ numbers but not of $\lceil 39/2\rceil = 20$ numbers. Similarly, the split of $51!$ does not work in the opposite way around. | |
| Apr 5, 2022 at 9:42 | history | edited | Peter Taylor | CC BY-SA 4.0 |
n=29 is an exception to the previous observation that the difference doesn't exceed n!
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| Apr 4, 2022 at 21:02 | comment | added | Peter Taylor | @MaxAlekseyev and others, if you want to add anything before I propose the current draft for publishing then it's A352813. | |
| Apr 2, 2022 at 23:20 | comment | added | Peter Taylor | @GregMartin, as explicitly noted, that was always my intention. | |
| Apr 2, 2022 at 23:19 | history | edited | Peter Taylor | CC BY-SA 4.0 |
Extend table; discuss asymptotics
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| Apr 2, 2022 at 22:19 | vote | accept | Bernardo Recamán Santos | ||
| Apr 1, 2022 at 17:21 | comment | added | Greg Martin | It would be better if you edited the additional computations into the answer, rather than leaving them as a list of comments. | |
| Mar 31, 2022 at 21:55 | comment | added | Peter Taylor |
Still no counterexample at $n=20$, where optimal solutions at 470500040794291200 include [2, 3, 6, 9, 11, 12, 13, 15, 18, 19, 21, 23, 24, 26, 27, 29, 30, 31, 37, 39]. (I'll merge these comments into the answer at some point, but I don't want to bump every time my search program spits out another result).
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| Mar 31, 2022 at 13:32 | history | answered | Peter Taylor | CC BY-SA 4.0 |