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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 23, 2012 8:44 PM
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On Nov 24, Bill Taylor <> wrote:
> david petry <> 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...

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