
Re: Reciprocals of integers summing to 1
Posted:
Nov 22, 2012 7:07 AM


Thanks for doing this David. I too did a list of the first few, by laborious hand. So I'm glad to have my results confirmed. I have *some* unit fraction programs in Maple, I guess I should get to work myself and modify them for this task.
On Nov 22, 5:49 pm, david petry <david_lawrence_pe...@yahoo.com> wrote:
> > > For each n, what are the solutions in positive integers > > > to (1/X1)+(1/X2) + . . . + (1/Xn)=1 ?
> > 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.
> 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)
Seems to be so. For completeness I note here that for n = 1,
1 is the sole entry. For n = 2 we have also one entry,
2 2 and for n = 3 there are 3:
2 3 6 2 4 4 3 3 3 .
I confirm your 14 cases for n = 4 .
If we make a separate list, d(n), for those with *distinct* denominators, then trolling through these lists gives
n d(n) f(n) 1 1 1 2  1 3 1 3 4 6 14 5 68 ??
I have tenuous faith in the d(5) = 68 figure, which I also did by hand, but it should be fairly close. I have not got a figure for f(5), which is rather hard, but it looks like it should be about double of d(5), I guess.
The 68 cases have minimax denominator at 2 4 10 12 15, and largest denominator at 2 3 7 43 1806.
Doing it by hand, one finds that almost all cases for n+1 distinct summands are obtainable by taking some (nondistinct) case for n and splitting one of the entries into two. ALL of the above cases can be obtained that way except
3 3 3 and 2 5 5 10 (which are repeating anyway).
I nearly missed the 2 5 5 10 case because of this! (I use the term nondistinct to be the disjoint union of the distinct and the repeating cases.)
None of the k k k ... k cases can be obtained by such monoterm splitting, for k > 2; and in general it is often impossible to so obtain cases with lots of repeats, like 3 3 9 9 9. But to compile a list of distinct cases is not so hard, given that most (so far, all) of them are obtainable via monoterm binary splitting, often nonuniquely.
To this end, it may be convenient to have handy a list of binary splits of 1/m.
1 = 2 2 2 = 3 6 = 4 4 3 = 4 12 = 6 6 4 = 5 20 = 6 12 = 8 8 5 = 6 20 = 10 10 .... and so on.
If anyone wants to check my hand work, the number of ways of splitting 1/m is (I suggest)
m: 1 2 3 4 5 6 7 8 9 10 12 15 18 20 24 42 #(m) 1 2 2 3 2 5 2 4 3 5 4 5 8 8 11 14
(the gaps are where I didn't need them for the above work).
This table at least is easy to do by computer. It seems that #(p) = 2 for prime p, and in general the # function bears a close resemblance to d, the number of factors of m; though it tends to be a bit bigger, and is not multiplicative like d.
Have fun!
> 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.
Cool.
 Tediously tabulating Taylor

