Moderators: gmalivuk, Prelates, Moderators General
tomtom2357 wrote:The odd greedy expansion does not necessarily work, the expansion could be infinite.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
Nitrodon wrote:I took some ideas from that other thread, and 2 can be expressed as the sum of 1/n for all n>1 dividing 1310112879075 except ...(some of them)
skullturf wrote:
1 = 1/5 + 1/7 + 1/9 + 1/11 + 1/13 + 1/15 + 1/21 + 1/27 + 1/33 + 1/35 + 1/39 + 1/45 + 1/55 + 1/63 + 1/65 + 1/77 + 1/91 + 1/99 + 1/105 + 1/117 + 1/135 + 1/143 + 1/273 + 1/2079 + 1/135135
Afif_D wrote:But i still believe nobody can produce a 2.
Afif_D wrote:Nice finding. But i still believe nobody can produce a 2.
I think we should al just try to work out a proof
No. For example, suppose r = 1/a_{1} + 1/a_{2} + ... + 1/a_{n}. Write 1/a_{n} = 1/(a_{n}+2) + 2/(a_{n}(a_{n} + 2)), then expand 2/(a_{n}(a_{n}+2)) as a sum of odd Egyptian fractions using any technique. Of course all these fractions will be smaller than 1/(a_{n}+2), so you will end up with a new set of distinct odd Egyptian fractions summing to r.Afif_D wrote:Ya . Good one. Thanks. And ya i am an idiot.
Now tell me. Do there exist only finite number of solutions for this kind of problem?
Users browsing this forum: letterX and 1 guest