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Is there a good characterization of the set $S$ of positive integers $n$ such that $\frac{1}{n}$ can be represented as a difference of Egyptian fractions with all denominators $< n$? For example, $44 \in S$ because $$ \dfrac{1}{44} = \left( \frac{1}{33} + \frac{1}{12}\right) - \frac{1}{11} $$

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 $n \in S$, then $mn \in S$ for any positive integer $m$.
  2. $mn \in S$ for integers $m,n$ with $n < m < 2 n$, because $$\dfrac{1}{mn} = \dfrac{1}{n(m-n)} - \dfrac{1}{m(m-n)}$$
  3. $S$ contains no prime or prime power.
  4. There are no members of the form $2p^k$ where $p$ is a prime $> 3$.
  5. There are no members of the form $3p^k$ where $p$ is a prime $> 11$.
Jack M
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Robert Israel
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  • If you want to describe the solutions of this equation, using this formula. http://math.stackexchange.com/questions/419766/number-of-solution-for-xy-yz-zx-n/713998#713998 Or use a different approach. http://math.stackexchange.com/questions/450280/erd%C5%91s-straus-conjecture/831870#831870 – individ Nov 28 '16 at 05:39
  • An interesting subset are made the numbers n where $1/n$ is given by only two terms. I believe for the OP’s sequence from 6 to 45 that is the case for most numbers except 21, 33 and the PO’s example 44. Examples for 21 and 33 are: $1/21= 1/7+1/14 – (1/10+1/15)$, and $1/33= 1/15+1/10-(1/22+1/11)$. – Mikael Jensen Jan 10 '18 at 22:27
  • One thing to note: Given property 1 it is useful to consider the set $S'$ such that for all $n\in S'$, there is no proper divisor of $n$ in $S'$ (then $S=\cup_{n\in\mathbb{N}}nS'$). Then the proper generalization of properties 4-5 is there are no members of $S'$ of the form $np^k$ where $p$ is a prime $>H_n \text{lcm}(1,\dots, n)$ where $H_n$ is the $n$-th harmonic number. – Will Fisher Jun 18 '18 at 16:24
  • Can someone explain the meaning of 'characterisation' in this context. As, it seems to me that the question already lists a plethora of properties of $\mathbb{S}$. – user0 Aug 23 '18 at 07:56
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    @DevashishKaushik characterization = necessary and sufficient condition (ideally one that can be easily tested). – Robert Israel Aug 23 '18 at 14:31
  • @Robert Thanks. So, the question is basically asking for a necessary and sufficient condition for a natural number, $n$ to be in the set, $\mathbb{S}$. Is this correct? – user0 Aug 23 '18 at 15:47
  • Yes, that is correct. – Robert Israel Aug 23 '18 at 18:15
  • @RobertIsrael: Are the (even) perfect numbers a subsequence of OEIS A278638? – Jose Arnaldo Bebita Dris Dec 31 '20 at 05:41
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    @ArnieBebita-Dris Yes because (regardless of whether $p$ or $2^p-1$ is prime) $2^{p-1}(2^p-1) = m n$ where $m = 2^{p-1}$ and $n = 2^p-1$, and $m < n < 2m$. – Robert Israel Dec 31 '20 at 07:21
  • That is interesting! How about the odd perfect numbers (OPNs)? I know that OPNs must have the Eulerian form $q^k t^2$, where $q$ is the special prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,t)=1$. In particular, I can take $q^k t^2 = mn$, where $m = q^k$ and $n = t^2$, and $m < n$. Alas, I am not so sure about $t^2 = n < 2m = 2q^k$. – Jose Arnaldo Bebita Dris Dec 31 '20 at 08:00

1 Answers1

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Partial answer. It shows for example that if $N$ is large enough and satisfies $P^*(N)^2\leq N$ then $N\in S$. (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, $\pi^*(y)$ 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 $\frac{a}{b}$ 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 $[1, 2y^4]$;
  • (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 $.

Elaqqad
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