Timeline for The diophantine equation $ \sum_{n=1}^{N} \frac{1}{x_{n}} = \prod_{k=1}^{N} \left(1-\frac{1}{x_{k}} \right) $
Current License: CC BY-SA 4.0
28 events
| when toggle format | what | by | license | comment | |
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| Jan 23 at 21:11 | answer | added | Max Alekseyev | timeline score: 2 | |
| Jan 22 at 23:43 | answer | added | David desJardins | timeline score: 4 | |
| Jan 22 at 15:01 | answer | added | Oscar Lanzi | timeline score: 1 | |
| Jan 22 at 11:18 | answer | added | Fred Hucht | timeline score: 2 | |
| Jan 22 at 8:40 | history | edited | Max Lonysa Muller | CC BY-SA 4.0 |
Added information on the findings by David desJardins
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| Jan 22 at 5:10 | comment | added | David desJardins | When N=6, there are exactly 9219 solutions. In ascending order, the lexicographically first is (3, 6, 29, 803, 643727, 414383582242) and the last is (6, 9, 10, 12, 13, 60). In descending order, the lexicographically first is (33, 21, 15, 10, 7, 6) and the last is (414383582242, 643727, 803, 29, 6, 3). | |
| Jan 22 at 1:06 | comment | added | David desJardins | If you sort the solutions in descending order, then the largest observed value by position is (802, 46, 10, 5) when N=4, and (643726, 910, 74, 12, 5) when N=5. So it seems likely possible to find most solutions for the "next few" values of N by constraining the last several variables to be "small". I don't know how to prove you have them all, though. It's easy to show that any solution with N=5 must have x5 < 8, because 1/8+1/9+1/10+1/11+1/12 < (1-1/8)(1-1/9)(1-1/10)(1-1/11)(1-1/12). | |
| Jan 21 at 22:07 | answer | added | Gareth McCaughan | timeline score: 11 | |
| Jan 21 at 22:01 | comment | added | Max Lonysa Muller | @DaviddesJardins Thank you for running the Mathematica algorithm. Is there a chance you could personally send me a file containing the 293 solutions for the $N=5$ case? It would be interesting to compare them with the 24 solutions for the $N=4$ case, also found by Max Alekseyev | |
| Jan 21 at 21:50 | comment | added | David desJardins | Mathematica quickly finds the 24 solutions for N=4, and confirms there are no more. It also finds exactly 293 solutions for N=5. Lexicographically first is (48,31,13,6,4) and last is (643726,803,29,6,3). All of the N=5 solutions contain 3, 4, or 5. Larger values of N would probably require a more sophisticated search algorithm. | |
| Jan 21 at 19:43 | comment | added | Fred Hucht | Maybe it helps to look at the (characteristic) polynomial $P_{x}(\xi)=\prod_{n=1}^N(x_n-\xi)$. The OPs condition (1) is fulfilled iff $P_{x}'(0) + P_{x}(1) = 0$. | |
| Jan 21 at 18:19 | comment | added | Max Lonysa Muller | @DanielAsimov That is the third question indeed. I'm curious about your thoughts on this equation | |
| Jan 21 at 17:56 | comment | added | Daniel Asimov | Does question 3. ask for a countably infinite set of distinct positive integers x_j such that the sum of all 1/x_j equals the product of all (1 - 1/x_j) ? (In any case, that would be an interesting equation to solve.) | |
| Jan 21 at 15:59 | history | edited | Max Lonysa Muller | CC BY-SA 4.0 |
Added information about the solutions for $N=4$ and $N=8, ... , 13$, as obtained by Max Alekseyev
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| Jan 21 at 14:51 | comment | added | Max Alekseyev | More solutions: $$N=10: [8, 9, 12, 20, 21, 28, 29, 30, 31, 174]$$ $$N=11: [9, 10, 11, 21, 23, 24, 25, 28, 30, 99, 184]$$ $$N=12: [10, 13, 15, 16, 19, 21, 24, 26, 27, 30, 420, 771]$$ $$N=13: [11, 15, 16, 20, 21, 22, 23, 24, 25, 27, 28, 5635, 38124]$$ | |
| Jan 21 at 14:43 | comment | added | Max Alekseyev | @MaxMuller: No problem. I'm pretty confident that the above list for $N=4$ is the complete set of solutions. | |
| Jan 21 at 14:41 | comment | added | Max Lonysa Muller | @MaxAlekseyev I've corrected it, thank you for pointing it out, and sorry for misspelling your name. | |
| Jan 21 at 14:40 | history | edited | Max Lonysa Muller | CC BY-SA 4.0 |
added 7 characters in body
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| Jan 21 at 14:37 | history | edited | Max Alekseyev | CC BY-SA 4.0 |
spelling
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| Jan 21 at 14:31 | history | edited | Max Lonysa Muller | CC BY-SA 4.0 |
Wrote more on Max Alexeyev's findings
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| Jan 21 at 14:24 | comment | added | Max Alekseyev | There are many more solutions for $N=4$: [3, 6, 29, 802], [3, 6, 30, 415], [3, 6, 31, 286], [3, 6, 34, 157], [3, 6, 37, 114], [3, 6, 46, 71], [3, 7, 17, 549], [3, 7, 18, 194], [3, 7, 19, 123], [3, 7, 24, 52], [3, 8, 13, 480], [3, 9, 11, 457], [3, 9, 12, 100], [3, 9, 15, 37], [3, 9, 16, 32], [3, 10, 15, 26], [4, 5, 11, 340], [4, 5, 12, 93], [4, 5, 15, 36], [4, 5, 17, 28], [4, 6, 8, 297], [4, 6, 9, 56], [4, 7, 8, 35], [5, 6, 10, 12] | |
| Jan 21 at 14:23 | comment | added | Max Lonysa Muller | @MaxAlekseyev Thanks! I've added both your and Brendan McKay's findings to the question body | |
| Jan 21 at 14:22 | history | edited | Max Lonysa Muller | CC BY-SA 4.0 |
Added solutions for N=5 and N=6, as found by Brendan McKay and Max Alekseyev, respectively; deleted 6 characters in body
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| Jan 21 at 14:16 | comment | added | Max Alekseyev | There are a lot of solutions for $N=6,7,8,9$ as well. Here are a few of examples: [3, 7, 27, 50, 336, 1060], [6, 7, 13, 15, 16, 35, 96], [8, 9, 10, 16, 17, 18, 40, 51], [9, 10, 12, 16, 18, 19, 20, 25, 266] | |
| Jan 21 at 14:10 | comment | added | Brendan McKay | $(3,10,11,73,37050)$, $(3,9,11,458,209146)$, seems like a lot of solutions. | |
| Jan 21 at 14:10 | comment | added | Max Lonysa Muller | @BrendanMcKay Thank you! It's an interesting solution, because the rational number it corresponds to is bigger than $7/13$, for instance (which corresponds to a solution in the $N=4$ case). This means the solutions don't necessarily monotonically decrease towards $1/2$ as $N$ increases | |
| Jan 21 at 13:55 | comment | added | Brendan McKay | $(4,5,11,341,115820)$ | |
| Jan 21 at 13:14 | history | asked | Max Lonysa Muller | CC BY-SA 4.0 |