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Topic: Reciprocals of integers summing to 1
Replies: 21   Last Post: Nov 23, 2012 8:44 PM

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Bill Taylor

Posts: 186
Registered: 11/17/10
Re: Reciprocals of integers summing to 1
Posted: Nov 22, 2012 7:07 AM
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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 non-uniquely.

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



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