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Re: Reciprocals of integers summing to 1
Posted:
Nov 19, 2012 11:16 AM
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El sábado, 17 de noviembre de 2012 12:22:34 UTC-4:30, Ludovicus escribió:
> El viernes, 16 de noviembre de 2012 00:39:24 UTC-4:30, Charlie-Boo escribió:
> > For each n, what are the solutions in positive integers (or in
> > integers) to (1/X1)+(1/X2) + . . . + (1/Xn)=1 ?> There are infinitely many different solutions. But I am not sure that it is valid for all n.
>
> Take k primes and do the following sum: 1/p1 + 1/p2 + 1/p3 + ...+ 1/pk = S < 1
> Now make Q = 1 - S.
> By the theorem of the Egyptian fractions, Q always can be decomposed as:
> Q = 1/x1 + 1/x2 + 1/x3 +....+ 1/xj.
> I am not sure that it is possible, ever, that k+j = n.
> Ludovicus
The six first decompositions with different least denominators Xi are:
n Xi
3 2,3,6
4 2,4,5,20
5 2,4,5,30,60
6, 2,3,18,115,414
7, 2,4,616,51,944,6018
8, 2,4,6,17,44,658,3828,648788
Can someone continue this table ?
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