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Topic: Reciprocals of integers summing to 1
Replies: 21   Last Post: Nov 23, 2012 8:44 PM

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David Petry

Posts: 1,103
Registered: 12/8/04
Re: Reciprocals of integers summing to 1
Posted: Nov 21, 2012 11:49 PM
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On Sunday, November 18, 2012 4:45:05 AM UTC-8, Bill Taylor wrote:
> Quite clearly a lot of respondents didn't seem to read the question!
>
>
>

> > For each n, what are the solutions in positive integers
>
> > to (1/X1)+(1/X2) + . . . + (1/Xn)=1 ?
>
>
>
> Admittedly no comment was made on permutations of solutions,
>
> but in such contexts they are almost always regarded as the same.
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>
>
> Clearly repeats among the x_i are allowed.
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> (Though one could also answer with them disallowed.)
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>
>
> But most important, A FUNCTION OF n IS REQUIRED.
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>
>
> i.e. what is f(n) = card({x1, x2, ... , xn} | etc)
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>
>
> where the curly brackets denote unordered multisets.
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>
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> So far we have f(1) = 1, f(2) = 1, f(3) = 3 (seemingly)
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> and no great effort on any higher values.
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>
>
> If no-one can get anywhere much theoretically,
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> can some computer whizz at least produce a list of
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> the first several values of f please?


Clearly it grows quite fast.

For n = 4, we have the following sets.

2 3 7 42
2 3 8 24
2 3 9 18
2 3 10 15
2 3 12 12
2 4 5 20
2 4 6 12
2 4 8 8
2 5 5 10
2 6 6 6
3 3 4 12
3 3 6 6
3 4 4 6
4 4 4 4

So f(4) = 14 (I hope)

That's by hand, so I'm not going any further.

If someone does tabulate a few more values, it would probably be a new entry for that list of integer sequences someone has on the internet.



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