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
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| Apr 19, 2022 at 12:57 | history | edited | Seva | CC BY-SA 4.0 |
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| Apr 4, 2022 at 6:15 | history | rollback | Seva |
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| Apr 3, 2022 at 20:35 | history | edited | Seva | CC BY-SA 4.0 |
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| Mar 31, 2022 at 19:43 | history | edited | Seva | CC BY-SA 4.0 |
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| Mar 31, 2022 at 18:54 | history | edited | Seva | CC BY-SA 4.0 |
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| Mar 31, 2022 at 17:39 | history | edited | Seva | CC BY-SA 4.0 |
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| Mar 31, 2022 at 17:20 | comment | added | David E Speyer | The conclusion has the same order of magnitude I suggest in my comment above, though: By Stirling's approximation, $\sqrt{(2n)!} \approx \sqrt{(2n/e)^{2n}} = 2^n (n/e)^n \approx 2^n n!$. | |
| Mar 31, 2022 at 16:21 | history | answered | Seva | CC BY-SA 4.0 |