
Re: Reciprocals of integers summing to 1
Posted:
Nov 21, 2012 11:49 PM


On Sunday, November 18, 2012 4:45:05 AM UTC8, Bill Taylor wrote: > Quite clearly a lot of respondents didn't seem to read the question! > > > > > For each n, what are the solutions in positive integers > > > to (1/X1)+(1/X2) + . . . + (1/Xn)=1 ? > > > > Admittedly no comment was made on permutations of solutions, > > but in such contexts they are almost always regarded as the same. > > > > Clearly repeats among the x_i are allowed. > > (Though one could also answer with them disallowed.) > > > > But most important, A FUNCTION OF n IS REQUIRED. > > > > i.e. what is f(n) = card({x1, x2, ... , xn}  etc) > > > > where the curly brackets denote unordered multisets. > > > > So far we have f(1) = 1, f(2) = 1, f(3) = 3 (seemingly) > > and no great effort on any higher values. > > > > If noone can get anywhere much theoretically, > > can some computer whizz at least produce a list of > > the first several values of f please?
Clearly it grows quite fast.
For n = 4, we have the following sets.
2 3 7 42 2 3 8 24 2 3 9 18 2 3 10 15 2 3 12 12 2 4 5 20 2 4 6 12 2 4 8 8 2 5 5 10 2 6 6 6 3 3 4 12 3 3 6 6 3 4 4 6 4 4 4 4
So f(4) = 14 (I hope)
That's by hand, so I'm not going any further.
If someone does tabulate a few more values, it would probably be a new entry for that list of integer sequences someone has on the internet.

