Reciprocals of integers summing to 1

[Charlie-Boo]

For each n, what are the solutions in positive integers (or in
integers) to (1/X1)+(1/X2) + . . . + (1/Xn)=1 ?

C-B

[William Elliot]

On Thu, 15 Nov 2012, Charlie-Boo wrote:

> For each n, what are the solutions in positive integers (or in
> integers) to (1/X1)+(1/X2) + . . . + (1/Xn)=1 ?

x1 = x2 =..= x_n = n

[Charlie-Boo]

But there is also 1/2 + 1/3 + 1/6 = 1. I am asking for all solutions.

C-B

[William Elliot]

On Thu, 15 Nov 2012, Charlie-Boo wrote:
> On Nov 16, 12:21 am, William Elliot <ma...@panix.com> wrote:
> > On Thu, 15 Nov 2012, Charlie-Boo wrote:

> > > For each n, what are the solutions in positive integers (or in
> > > integers) to (1/X1)+(1/X2) + . . . + (1/Xn)=1 ?
> >
> > x1 = x2 =..= x_n = n
>
> But there is also 1/2 + 1/3 + 1/6 = 1. I am asking for all solutions.
>

1/3 + 1/2 + 1/6
1/6 + 1/3 + 1/2,
etc.,
for a total of 6 + 1.

Do I hear more?

[red...@siu.edu]

Here's another

1/2 + 1/3 + 1/7 + 1/42 = 1.

don

[gus gassmann]

What about 1/1 + 1/1 + 1/(-1)?

Seems to me that the original question was not very well stated.

Maybe you want x_i =/= x_j for i =/= j? That is still quite a bit more
work than I am willing to invest.

On the other hand,

M = {n | exist 0 < x_1 < x_2 < ... < x_n with sum 1/x_i = 1}

would be interesting. I offer k=2 as one positive integer that is not
contained in M. Are there others? Is there a nice characterization?

[billh04]

On Nov 16, 8:57 am, gus gassmann <g...@nospam.com> wrote:

Since 1 = 1/2 + 1/3 + 1/6, then 1/2 = 1/4 + 1/6 + 1/12. Hence 1 = 1/2
+ 1/2 + 1/4 + 1/6 + 1/12. We probably want to rule out cases like this
also. More generally, 1/n = 1/2n + 1/3n + 1/6n and then 1 = 1/n + 1/n
+ 1/n + ... + 1/2n + 1/3n + 1/6n.

[lui...@yahoo.com]

El viernes, 16 de noviembre de 2012 00:39:24 UTC-4:30, Charlie-Boo 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

[Charlie-Boo]

There are an infinite number of sets of positive integers whose
reciprocals sum to 1:

N = # of terms. Term # M where M = 1 to N = ( If M < N then (M+1)! /
M Else M ! )

C-B

[Charlie-Boo]

On Nov 16, 2:52 am, William Elliot <ma...@panix.com> wrote:
> On Thu, 15 Nov 2012, Charlie-Boo wrote:
> > On Nov 16, 12:21 am, William Elliot <ma...@panix.com> wrote:
> > > On Thu, 15 Nov 2012, Charlie-Boo wrote:
> > > > For each n, what are the solutions in positive integers (or in
> > > > integers) to (1/X1)+(1/X2) + . . . + (1/Xn)=1 ?
>
> > > x1 = x2 =..= x_n = n
>
> > But there is also 1/2 + 1/3 + 1/6 = 1.  I am asking for all solutions.
>
> 1/3 + 1/2 + 1/6
> 1/6 + 1/3 + 1/2,
> etc.,
> for a total of 6 + 1.
>
> Do I hear more?
>
>

Start with {1}, a set containing N elements including N! . Replace N!
with (N+1)!/N and (N+1)! to create another such set.

C-B

[Bill Taylor]

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 no-one can get anywhere much theoretically,
can some computer whizz at least produce a list of
the first several values of f please?

-- Whacking Willy.

** No-one has any right to not be offended.
** Neither on their own behalf, their religion, country,
** football team nor anything else!

[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 ?

[doumin]

>"Ludovicus" a écrit dans le message de groupe de discussion :
>d8fcd1cb-615d-49ab...@googlegroups.com...

>El sábado, 17 de noviembre de 2012 12:22:34 UTC-4:30, Ludovicus escribió:
>> El viernes, 16 de noviembre de 2012 00:39:24 UTC-4:30, Charlie-Boo
>> 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

1/2 + 1/3 + 1/18 + 1/115 +1 /414= 0,9

> 7, 2,4,616,51,944,6018

2,4,6,16,51,944,6018 looks better

[david petry]

On Sunday, November 18, 2012 4:45:05 AM UTC-8, 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 no-one 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.

[Bill Taylor]

On Nov 20, 5:16 am, Ludovicus <luir...@yahoo.com> wrote:

> 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 ?

This list seems to be badly wrong.

> 4 2,4,5,20

There is also, as David Petry noted:

2 3 9 18
2 3 10 15

2 4 6 12

> 5 2,4,5,30,60

2 3 6 12 20
2 4 5 36 45
3 4 5 6 20 and many more.

One thus has little faith in the rest of the list.

-- Baffled Bill

** There is no moral right to not be offended.
** There is no moral right to not be offended.
** There is no moral right to not be offended.

[Bill Taylor]

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

[Ludovicus]

This is my last computation (22/11/2012), of first seven decompositions of unit, as sum of Egyptian fractions with n different minimum denominators Xi :

n Xi
3 2,3,6
4 2,4,6,12
5 3,4,5,6,20
6, 2,4,6,14,120,280
7, 2,4,6,16,51,944,6018
8, 2,4,6,16,59,262,17346,102704
9, 2,4,5,26,97,892,9313,1391520,86722657

Can someone to extend or correct this table ?

[david petry]

On Thursday, November 22, 2012 4:07:09 AM UTC-8, Bill Taylor wrote:
>
> 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.

The formula 1/(a*b) = 1/(a*(a+b)) + 1/(b*(a+b)) shows why there is a close relation between your function and the number of ways the number can be factored into two factors.

[david petry]

I'm not sure what you're trying to compute.

Bill Taylor gave us 2 4 10 12 15. Is that a correction to your entry number 5 ? If it is, then 3 4 6 10 12 15 would be a correctionn to your entry number 6.

Could you explain better what you're computing?

[Bill Taylor]

On Nov 23, 3:08 pm, david petry <david_lawrence_pe...@yahoo.com>
wrote:

> On Thursday, November 22, 2012 4:07:09 AM UTC-8, Bill Taylor wrote:
>
> > 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
>

> > It seems that  #(p) = 2  for prime p, and in general

In fact this is simply routinely proved, unsurprisingly.

> > 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.
>
> The formula  1/(a*b) = 1/(a*(a+b))  +  1/(b*(a+b))   shows why
> there is a close relation between your function and
> the number of ways the number can be factored into two factors.

Of course! Neat triviality. I wonder if the surplus ways
are also simply characterizable. Must investigate.

-- Bumbling Bill

** There is no moral right to "not be offended".

** There should be no legal right to "not be offended" !

[Musatov]

0

[Bill Taylor]

On Nov 24, Bill Taylor <wfc.tay...@gmail.com> wrote:

> david petry <david_lawrence_pe...@yahoo.com> wrote:

> > > 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   8   5   8   8  11  14

With David Petry's encouragement, I have found the formula.

Recall:- #(n) = # of distinct ways of expressing 1/n as
the sum of two (possibly equal) positive unit fractions.

It is (d(n^2) + 1)/2 , as can be verified above.

I do not give a full proof here, so as to leave folk some fun
trying to prove it for themselves, [hint below].

David was correct that it had a close relation to d(n),
the number of factors of n, a standard num-theoretic function.
The result #(p) = 2 for prime p is thus verified also.
One may also note that numbers with the same "factor structure",
like 12, 18 and 20, must necessarily have the same #.
(This was handy in detecting an error in the original list!)

The result is a somewhat unusual one, and quite intriguing IMHO.

If one rejects the cases of two equal denominators, as is
common practice in Egyptian fraction work, then the formula
is the same with + replaced by - .

This whole thing might make a neat advanced-problem for
a class in basic number theory. :-)

The hint: to express 1/n as 1/(n+k) + 1/(etc),
for k from 1 to 2n (WLOG), find conditions on k that
allow (etc) to be an integer.

-- Battling Bill

Q: Which is worse, apathy or ignorance?
A: I don't know and I don't care...