71
$\begingroup$

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 $> 11$.
$\endgroup$
  • $\begingroup$ If you want to describe the solutions of this equation, using this formula. math.stackexchange.com/questions/419766/… Or use a different approach. math.stackexchange.com/questions/450280/… $\endgroup$ – individ Nov 28 '16 at 5:39
  • $\begingroup$ 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 $1/33= 1/15+1/10-(1/22+1/11)$. $\endgroup$ – Mikael Jensen Jan 10 '18 at 22:27
  • $\begingroup$ 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. $\endgroup$ – Will Fisher Jun 18 '18 at 16:24
  • $\begingroup$ 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 $\mathbb{S}$. $\endgroup$ – Devashish Kaushik Aug 23 '18 at 7:56
  • 1
    $\begingroup$ @DevashishKaushik characterization = necessary and sufficient condition (ideally one that can be easily tested). $\endgroup$ – Robert Israel Aug 23 '18 at 14:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.