
Re: Reciprocals of integers summing to 1
Posted:
Nov 19, 2012 11:16 AM


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 7, 2,4,616,51,944,6018 8, 2,4,6,17,44,658,3828,648788
Can someone continue this table ?

