Timeline for How to represent integers using Egyptian fractions?
Current License: CC BY-SA 3.0
21 events
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| Mar 6, 2018 at 12:33 | history | edited | J. M.'s missing motivation♦ |
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| Dec 1, 2017 at 23:03 | history | tweeted | twitter.com/StackMma/status/936732639775592450 | ||
| Nov 30, 2017 at 19:39 | answer | added | kirma | timeline score: 4 | |
| Nov 29, 2017 at 7:19 | comment | added | kirma |
This problem is essentially the same as computing the OEIS sequence A101877. With a little bit of cleverness one can search the values up to at least 300 (which would correspond with $N(300)=\{0,1,2,3,4,5\}$), just by using straight-forward Solve and constraints on the problem size based on solutions already found. Going up to the sum corresponding to 6 (at 469) might be infeasible using this method (probably takes at least several days of computing time, possibly a lot more).
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| Nov 27, 2017 at 11:33 | answer | added | dibbidabbi | timeline score: 0 | |
| Nov 27, 2017 at 6:39 | comment | added | kirma |
Straight-forward method using Solve (and not suffering from exponential memory usage) would be something like this: n0 /. (With[{n = 50}, Solve[{Sum[a[i]/i, {i, n}] == n0, Array[a, n] \[Element] Cuboid[ConstantArray[0, n]]}, {n0}, Integers]] /. ConditionalExpression[v_, ___] :> v) (* {0, 1, 2, 3} *) ... this doesn't solve the exponential time complexity, though.
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| Nov 27, 2017 at 5:39 | comment | added | asad | @evanb, thanks, I corrected. Also I like to have representation for each of them in output as you see in P.S. | |
| Nov 27, 2017 at 5:39 | history | edited | asad | CC BY-SA 3.0 |
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| Nov 26, 2017 at 21:44 | comment | added | ybeltukov | I changed the title to make the question to be more attractive. I hope, you don't mind. | |
| Nov 26, 2017 at 21:44 | history | edited | ybeltukov | CC BY-SA 3.0 |
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| Nov 26, 2017 at 16:50 | answer | added | evanb | timeline score: 3 | |
| Nov 26, 2017 at 16:31 | comment | added | Carl Woll | You can brute force it for $n \lesssim 25$. Will that work, or are you really interested in $n=1000$? | |
| Nov 26, 2017 at 16:26 | comment | added | evanb |
A solution that scales (very, very) poorly is Union[Select[Total /@ Subsets[1/Range[6]], IntegerQ]] (for n=6, for example).
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| Nov 26, 2017 at 16:23 | comment | added | evanb | Doesn't N(n) always contain 0? For example, for N(6), 0 = 0/1 + 0/2 + 0/3 + 0/4 + 0/5 + 0/6. | |
| Nov 26, 2017 at 15:41 | history | edited | yarchik | CC BY-SA 3.0 |
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| Nov 26, 2017 at 10:49 | comment | added | aardvark2012 |
Your A function takes e == 0 and runs through all the i-values, and then takes e == 1 and runs through all the i-values, giving 0/1 + 0/2 + ... + 0/n + 1/1 + 1/2 + ... + 1/n. Try Table[{e, i}, {e, 0, 1}, {i, 1, 6}] to see what the indices are doing.
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| Nov 26, 2017 at 9:07 | history | edited | asad | CC BY-SA 3.0 |
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| Nov 26, 2017 at 9:02 | history | edited | asad | CC BY-SA 3.0 |
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| Nov 26, 2017 at 8:50 | comment | added | Sumit |
A[e,n] does not depend on e. So you can use A[n_]:=... Otherwise it looks good.
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| Nov 26, 2017 at 8:45 | history | edited | Sumit | CC BY-SA 3.0 |
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| Nov 26, 2017 at 7:56 | history | asked | asad | CC BY-SA 3.0 |