Date:           Fri, 8 Nov 1996 21:53:44 -0500
From:           RKetcheso@aol.com
To:             eppstein@ics.uci.edu
Subject:        Egyptian fractions

Do you know if it is possible for a group of Egyptian fractions with odd,
distinct denominators to add up to 1?

Dave Ketcheson

To:             RKetcheso@aol.com
Subject:        Re: Egyptian fractions
Date:           Sat, 09 Nov 1996 10:37:08 -0800
From:           David Eppstein <eppstein@ICS.UCI.EDU>

1 = 1/3 + 1/5 + 1/7 + 1/9 + 1/11 + 1/13 + 1/23 + 1/721 + 1/979007 +
1/661211444787 + 1/622321538786143185105739 +
1/511768271877666618502328764212401495966764795565 +
1/209525411280522638000804396401925664136495425904830384693383280180439963265695525939102230139815

I found this  by applying EgyptOddGreedy[2/3,5] from my Egyptian fractions
notebook, http://www.ics.uci.edu/~eppstein/numth/egypt/.
For a less horrible example, found with more work by hand, try

1 = 1/3 + 1/5 + 1/7 + 1/9 + 1/15 + 1/21 + 1/35 + 1/45 + 1/55 +
1/77 + 1/165 + 1/275 + 1/385 + 1/495 + 1/825 + 1/1925 + 1/2475

--
David Eppstein      UC Irvine Dept. of Information & Computer Science
eppstein@ics.uci.edu  http://www.ics.uci.edu/~eppstein/

From:           mlerma@pythagoras.ma.utexas.edu (Miguel Lerma)
Date:           9 Nov 1996 06:09:43 GMT
Newsgroups:     sci.math
Subject:        Re: Egyptian fractions

rketcheso@aol.com wrote:
> Can someone tell me if it is known whether a group of Egyptian fractions
> with odd, distinct denominators can add up to 1?

1/3 + 1/5 + 1/7 + 1/9 + 1/11 + 1/13 + 1/23 + 1/721 + 1/979007
+ 1/661211444787 + 1/622321538786143185105739
+ 1/511768271877666618502328764212401495966764795565
+ 1/209525411280522638000804396401925664136495425904
830384693383280180439963265695525939102230139815
= 1

The last denominator has 96 digits and is written in two lines.

A slightly simpler example:

1/3 + 1/5 + 1/7 + 1/9 + 1/11 + 1/13 + 1/25 + 1/207 + 1/28307
+ 1/24439202289 + 1/398183072324002690725
= 1

Another one:

1/3 + 1/5 + 1/7 + 1/9 + 1/11 + 1/15 + 1/19 + 1/403 + 1/103237
+ 1/6779784609 + 1/52906135762881315915
= 1

Miguel A. Lerma

From:           eppstein@ics.uci.edu (David Eppstein)
Date:           9 Nov 1996 11:45:50 -0800
Newsgroups:     sci.math
Subject:        Re: Egyptian fractions

> Can someone tell me if it is known whether a group of Egyptian fractions
> with odd, distinct denominators can add up to 1?

to which I responded
> 1 = 1/3 + 1/5 + 1/7 + 1/9 + 1/15 + 1/21 + 1/35 + 1/45 + 1/55 +
>     1/77 + 1/165 + 1/275 + 1/385 + 1/495 + 1/825 + 1/1925 + 1/2475

I now have an even better solution:

1 = 1/3 + 1/5 + 1/7 + 1/9 + 1/15 + 1/21 + 1/35 + 1/45 + 1/55 + 1/77 +
1/165 + 1/231 + 1/385 + 1/495 + 1/693

...anyone have further improvements?
--
David Eppstein      UC Irvine Dept. of Information & Computer Science
eppstein@ics.uci.edu  http://www.ics.uci.edu/~eppstein/

From:           cet1@cus.cam.ac.uk (Chris Thompson)
Date:           10 Nov 1996 02:39:56 GMT
Newsgroups:     sci.math
Subject:        Re: Egyptian fractions

|> Can someone tell me if it is known whether a group of Egyptian fractions
|> with odd, distinct denominators can add up to 1?

Yes. Try

1/3 + 1/5 + 1/7 + 1/9 + 1/15 + 1/21 + 1/27 + 1/35 + 1/63 + 1/105 + 1/135

The lcm of the denominators in this example is 945, the smallest odd
pseudoperfect number [a pseudoperfect number is one which is the sum
of *some* of its proper divisors] - the connection should be obvious.

Problems in Number Theory" (R.K.Guy, Springer, 1994).

Chris Thompson
Email: cet1@cam.ac.uk

From:           ksbrown@seanet.com (Kevin Brown)
Date:           Sun, 10 Nov 1996 03:31:36 GMT
Newsgroups:     sci.math
Subject:        Re: Egyptian fractions

rketcheso@aol.com wrote:
> Can someone tell me if it is known whether a group of Egyptian fractions
> with odd, distinct denominators can add up to 1?

eppstein@ics.uci.edu (David Eppstein) wrote:
> to which I responded
>> 1 = 1/3 + 1/5 + 1/7 + 1/9 + 1/15 + 1/21 + 1/35 + 1/45 + 1/55 +
>>     1/77 + 1/165 + 1/275 + 1/385 + 1/495 + 1/825 + 1/1925 + 1/2475
> I now have an even better solution:
> 1 = 1/3 + 1/5 + 1/7 + 1/9 + 1/15 + 1/21 + 1/35 + 1/45 + 1/55 + 1/77 +
>    1/165 + 1/231 + 1/385 + 1/495 + 1/693
>...anyone have further improvements?

Here's another 15-term expansion with a smaller max denominator:

1 = 1/3 + 1/5 + 1/7 + 1/11 + 1/15 + 1/23 + 1/35 + 1/39 +

1/55 + 1/65 + 1/91 + 1/115 + 1/161 + 1/195 + 1/253

In Richard Guy's "Unsolved Problems In Number Theory" John Leech is
credited with the 11-term expansion

1 = 1/3 + 1/5 + 1/7 + 1/9 + 1/15 + 1/21 + 1/27 +

1/35 + 1/63 + 1/105 + 1/135

although it isn't stated whether this is optimal, either for fewest
terms or for smallest max denominator.  I wouldn't be surprised if
this was the optimum in both senses, because it comes from the
partition of 945 into its divisors, and recall that 945 is the
smallest odd abundant number

315 + 189 + 135 + 105 + 63 + 45 + 35 + 27 + 15 + 9 + 7
1 =    ------------------------------------------------------
945

In general I think these expansions can be generated from partitions
of any odd abundant number.  My 15-term example with max denoninator
of 253 is based on the odd abundant number

345345 = (3)(5)(7)(11)(13)(23)

Similarly, if we take the odd abundant number

15015 = (3)(5)(7)(11)(13)

we can partition 15015 into distinct divisors of itself as follows

15015 = 5005+3003+2145+1365+1155+1001+455+385+231+165+105

so we have another 11-term expansion

1 = 1/3 + 1/5 + 1/7 + 1/11 + 1/13 + 1/15 + 1/33 +

1/39 + 1/65 + 1/91 + 1/143

___________________________________________________________________
|                /*\                                                |
|   MathPages   /   \    http://www.seanet.com/~ksbrown/            |
|______________/_____\______________________________________________|

From:           fredh@ix.netcom.com (Fred W. Helenius)
Date:           Sun, 10 Nov 1996 21:33:42 GMT
Newsgroups:     sci.math
Subject:        Re: Egyptian fractions

ksbrown@seanet.com (Kevin Brown) wrote:

>rketcheso@aol.com wrote:
>> Can someone tell me if it is known whether a group of Egyptian fractions
>> with odd, distinct denominators can add up to 1?

>In Richard Guy's "Unsolved Problems In Number Theory" John Leech is
>credited with the 11-term expansion
>
>   1 = 1/3 + 1/5 + 1/7 + 1/9 + 1/15 + 1/21 + 1/27 +
>
>                                   1/35 + 1/63 + 1/105 + 1/135

>although it isn't stated whether this is optimal, either for fewest
>terms or for smallest max denominator.

Fewest terms [9]:

1 = 1/3 + 1/5 + 1/7 + 1/9 + 1/11 + 1/15 + 1/33 + 1/45 + 1/385.

Smallest maximum denominator [105]:

1 = 1/3 + 1/5 + 1/7 + 1/9 + 1/11 + 1/33 + 1/35 + 1/45 + 1/55 +
1/77 + 1/105.

Both are given as lower bounds in UPiNT, and it turns out they are
achievable.

--
Fred W. Helenius    <fredh@ix.netcom.com>

From:           cet1@cus.cam.ac.uk (Chris Thompson)
Date:           13 Nov 1996 00:44:35 GMT
Newsgroups:     sci.math
Subject:        Re: Egyptian fractions

In article <563g2p\$4e6@q.seanet.com>, ksbrown@seanet.com (Kevin Brown) writes:
[...]
|>
|> In general I think these expansions can be generated from partitions
|> of any odd abundant number.

In the nomenclature of UPiNT this conjecture is "every odd abundant number
is pseudoperfect", or even more charmingly "there are no odd weird numbers".
[A weird number is a primitive abundant number that is not pseudoperfect.]
Sadly, this seems too good to be true... Counterexample, anyone?

Chris Thompson
Email: cet1@cam.ac.uk

From:           cet1@cus.cam.ac.uk (Chris Thompson)
Date:           13 Nov 1996 11:35:27 GMT
Newsgroups:     sci.math
Subject:        Re: Egyptian fractions

In article <56b5lj\$kvt@lyra.csx.cam.ac.uk>, I wrote
>In article <563g2p\$4e6@q.seanet.com>, ksbrown@seanet.com (Kevin Brown) writes:
>[...]
>|>
>|> In general I think these expansions can be generated from partitions
>|> of any odd abundant number.
>
>In the nomenclature of UPiNT this conjecture is "every odd abundant number
>is pseudoperfect", or even more charmingly "there are no odd weird numbers".
>[A weird number is a primitive abundant number that is not pseudoperfect.]
>Sadly, this seems too good to be true... Counterexample, anyone?

Well, now I have UPiNT in front of me {moral: never post without it...] I
should correct this: primitiveness is not required for weirdness. Still,
"no odd primitive weird numbers" <=> "no odd weird numbers", obviously.

The existence of odd weird numbers is said to be open by UPiNT, but the
Erdos prize for settling this is a mere \$10. Nothing much seems to have
changed between the 1982 and 1994 editions, or indeed since:

S.J. Benkoski & P. Erdos
On weird and pseudoperfect numbers
Math. Comp. 28 (1974) 617-623   + corrigendum 29 (1975) 673

Chris Thompson
Email: cet1@cam.ac.uk

From:           ksbrown@seanet.com (Kevin Brown)
Newsgroups:     sci.math
Subject:        Re: Egyptian fractions
Date:           Sun, 17 Nov 1996 20:18:09 GMT
Organization:   Seanet Online Services, Seattle WA

ksbrown@seanet.com (Kevin Brown) writes:
> In general I think these expansions can be generated from partitions
> of any odd abundant number.

cet1@cus.cam.ac.uk (Chris Thompson) wrote:
> In the nomenclature of UPiNT this conjecture is "every odd abundant
> number is pseudoperfect", or even more charmingly "there are no odd
> weird numbers"...  The existence of odd weird numbers is said to be
> open by UPiNT, but the Erdos prize for settling this is a mere \$10.

I imagine he valued the answer so low because he believed there
DOES exist an odd weird number, and it would just take a little
number crunching to find it.  A proof that no such number exists
would (I think) be worth at least \$11.  [By the way, has a fund
been established to pay off the open "Erdos prizes" if/when they
are settled?  Seems like a good idea, and wouldn't cost much.]

Of course, the typical odd abundant number has not just one but
MANY distinct representations as a sum of its divisors.  In this
respect the question resembles Goldbach's conjecture (where we can
see empirically that an even number is generally expressible as a
sum of two primes in MANY different way, and yet we can't prove
that there must be at least one way).

It's interesting to note that if you apply the greedy algorithm
to express the odd abundant integer N as a sum of its divisors, then
it usually works.  Moreover, in all the cases where it doesn't work,
the left-over remainder is just 1, at least for all N < 40000.  In
that range the only odd obundant numbers for which the greedy
algorithm doesn't succeed are

4095 5775 9555 12915 21945 22275 24255 27405 29925 35805 38745

and in every one of these cases the remainder is 1.  What is the
smallest odd abundant number such that the remainder is greater
than 1?

_________________________________________________________________
|              /*\                                                |
|  MathPages  /   \     http://www.seanet.com/~ksbrown/           |
|____________/_____\______________________________________________|

From:           fredh@ix.netcom.com (Fred W. Helenius)
Date:           Mon, 18 Nov 1996 02:15:52 GMT
Newsgroups:     sci.math
Subject:        Re: Egyptian fractions

ksbrown@seanet.com (Kevin Brown) wrote:

>It's interesting to note that if you apply the greedy algorithm
>to express the odd abundant integer N as a sum of its divisors, then
>it usually works.  Moreover, in all the cases where it doesn't work,
>the left-over remainder is just 1, at least for all N < 40000.  In
>that range the only odd obundant numbers for which the greedy
>algorithm doesn't succeed are

> 4095 5775 9555 12915 21945 22275 24255 27405 29925 35805 38745

>and in every one of these cases the remainder is 1.  What is the
>smallest odd abundant number such that the remainder is greater
>than 1?

The greedy algorithm leaves a remainder greater than one for 94
values less than 2^25; the first few are:

207207, 223839, 279279, 1828827, 1851759,
2721411, 2909907, 2999997, 4603599, 5432427,
6467769, 6724809, 8027019, 8045037, 8646183,
8909901, 9124731, 9648639, 9993753, 10153143

In each case, with one exception, the remainder is two.  The
exceptional value is 15630225, for which the remainder is
three.

--
Fred W. Helenius    <fredh@ix.netcom.com>