doumin
Posts:
1
Registered:
11/20/12


Re: Reciprocals of integers summing to 1
Posted:
Nov 20, 2012 3:02 PM


>"Ludovicus" a écrit dans le message de groupe de discussion : >d8fcd1cb615d49ab8316a6ed2bf9a7d3@googlegroups.com...
>El sábado, 17 de noviembre de 2012 12:22:34 UTC4:30, Ludovicus escribió: >> El viernes, 16 de noviembre de 2012 00:39:24 UTC4:30, CharlieBoo >> 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 ?

