# sum of unit fractions

David Wilson davidwwilson at comcast.net
Sat Dec 18 13:42:10 CET 2004

```I have been thinking about this problem of expressing an integer as a
sum of reciprocals of distinct positive integers with least possible
maximum denominator.  I'm sure the ideas here have all been explored
before, but here goes:

Let S be a set of positive integers.  Let p be a prime dividing some
element of S, and let S_p be the set of elements of S divisible by p.
For example, if

S = { 1, 2, 3, 4, 5, 6, 8, 9, 10, 15, 18, 20, 24 }

then p must be one of { 2, 3, 5 }, and

S_2 = { 2, 4, 6, 8, 10, 18, 20, 24 }
S_3 = { 3, 6, 9, 15, 18, 24 }
S_5 = { 5, 10, 15, 20 }

Letting R(X) be the sum of reciprocals of elements of X, we have

R(S_2) = 58/45
R(S_3) = 31/40
R(S_5) = 5/12

In each case, p does not divide the denominator of R(S_p).

It turns out that

Above Theorem: R(S) is an integer iff, for each p dividing some
element of S, R(S_p) does not have p in its denominator.  Proof is
left as an exercise.

The theorem is actually merely saying that R(S) is an integer iff
we can eliminate each possible prime from the denominator of R(S),
which is "obvious on reflection(c)".

Im not sure, but it is possible that the Above Theorem may be
leveraged to reduce computation when deciding if a sum of reciprocals
is an integer.
------------------------------------------------------------------------

At any rate, I had an interesting idea (I think).

Let m be good if it is the largest element of a set whose reciprocals
add to an integer.  More precisely, letting M(S) be the maximum element
of S, m is good if there exists S with R(S) integer and M(S) = m.

The sets

S = { 1 }
S = { 1, 2, 3, 6 }
S = { 1, 2, 3, 4, 5, 6, 8, 9, 10, 15, 18, 20, 24 }

demonstrate that 1, 6 and 24 are good.

Primes, however, are bad.  For suppose prime p is good.  Then there exists
S with R(S) integer and M(S) = p. We must then have S_p = { p }, giving
R(S_p) = 1/p, and we cannot eliminate p from the denominator.  The
Above Theorem then implies R(S) is not an integer.

The primes are not the only bad numbers, for instance, 10 is also bad.
For suppose 10 is good.  Then S exists with R(S) integer with M(S) = 10.
Then S_5 = { 5, 10 } or { 10 }, giving R(S_5) = 3/10 or 1/10.  Either
way, 5 appears in the denominator, and the Above Theorem implies that
R(S) is not an integer.

These observations beg the questions:

(1) Which integers are good?  This has the potential for an OEIS
sequence.

(2) Is the Above Theorem sufficient to show an integer is bad?

```