doumin
Posts:
1
Registered:
11/20/12
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Re: Reciprocals of integers summing to 1
Posted:
Nov 20, 2012 3:02 PM
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>"Ludovicus" a écrit dans le message de groupe de discussion :
>d8fcd1cb-615d-49ab-8316-a6ed2bf9a7d3@googlegroups.com...
>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
1/2 + 1/3 + 1/18 + 1/115 +1 /414= 0,9
> 7, 2,4,616,51,944,6018
2,4,6,16,51,944,6018 looks better
> 8, 2,4,6,17,44,658,3828,648788
>Can someone continue this table ?
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