1n\frac{1}{n} as a difference of Egyptian fractions with all denominators

Is there a good characterization of the set S of positive integers n such that 1n can be represented as a difference of Egyptian fractions with all denominators <n?
For example, 44S because
144=(133+112)111

If I'm not mistaken, the first few members of S are
6,12,15,18,20,21,24,28,30,33,35,36,40,42,44,45
This does not appear to be in the OEIS yet; I intend to submit it soon.
[ EDIT: It is now in OEIS as A278638.]

Here are some things I know so far:

  1. If nS, then mnS for any positive integer m.
  2. mnS for integers m,n with n<m<2n, because 1mn=1n(mn)1m(mn)
  3. S contains no prime or prime power.
  4. There are no members of the form 2pk where p is a prime >3.
  5. There are no members of the form 3pk where p is a prime >11.

Answer

Partial answer. It shows for example that if N is large enough and satisfies P(N)2N then NS. (I think that the methods used in the paper are powerful enough to answer the question completely, In Progress)


The condition that the exceptions are tiny multiple of prime powers reminded me of the studies of representing the unity as the sum of Egyptian fractions by P. Erdös, E. S. Croot, and G. Martins.

After some google search, I found the following paper by Greg Martin, Denser Egyptian fractions, Acta Arith.

Let for any y>1 a real number, π(y) denote the number of prime powers less than y, and for any positive integer n, P(n) denote the largest prime power dividing n.

In the above paper, it is proven in proposition 7 that,

Proposition 7. Let y be a sufficiently large real number, and let ab be
a rational number satisfying alog(y)<ab<1 and P(b)y.
Then there is a set S of integers satisfying:

  • (i) S is contained in [1,2y4];
  • (ii) |S| = 2π^∗(y);
  • (iii) a/b = \sum_{s\in S} \frac 1 s

After reviewing the proof of this proposition, if we don't care about the size, then we can actually prove the following (already contained in the proof of proposition 7).

Proposition : Let y be a sufficiently large real number, and let \frac{a}{b} be
a rational number satisfying \frac{a}{\log(y)} < \frac{a}{b} < 1 and P^∗(b) \leq y.
Then there is a set S of integers satisfying:

  • (i) S is contained in [1, y^2];
  • (iii) a/b = \sum_{s\in S} \frac 1 s

As a corollary of this proposition, one can prove that for any large number N such that
P^*(N)^2\leq N there exists a set S\subset [1,N-1] such that :
\frac 1 n =(\frac 12+\frac 13 +\frac 1 6)- \sum_{s\in S} \frac 1 s

By taking \frac a b = \frac {n-1} n , y=\sqrt n .

Attribution
Source : Link , Question Author : Robert Israel , Answer Author : Elaqqad

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