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Let there be a finite set of positive integers such that:

(a) no two members of the set are equal

(b) the sum of the inverse of each member of the set is equal to one

The smallest set (as defined by its number of members) is simply $[1]$. The next is $[2, 3, 6]$.

There are (by my count) six different sets which have four members. However, all of them include the number '$2$'. This is self-evident, as, if they included only numbers greater than $2$, the greatest that four members could sum to is $1/3 + 1/4 + 1/5 + 1/6 = 19/20$.

Consequently, there is a way to have a five member set which doesn't use $2$. $[3, 4, 5, 6, 20]$.

What is the smallest set which uses neither $2$ nor $3$? It would seem possible to do it with $7$ members, as $1/4 + 1/5 + ... 1/10 = 1.095...$ However, I could not find a way to sum to exactly one with $7$ members. The smallest set appears to have $8$ members, such as $[4,5,6,8,10,12,20,40]$.

The general question, then, is this. What is the relationship between $n$ (minimal number permitted to be a member of the set) and $m$ (members of the minimal set). So far, we have $n=1,m=1$; $n=2,m=3$; $n=3,m=5$; $n=4,m=8$.

  • Can anyone extend this series by considering $n=5$; or solve the general problem? Thanks
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    $\begingroup$ Observe that $\sum_{i=n}^{n+m} \frac{1}{i}$ is approximately $\log (n+m) - \log n$; thus asymptotically you need at least $m \geq (e-1)n \sim 1.7n$. $\endgroup$ – Aravind Jul 18 '16 at 13:58
  • $\begingroup$ Thanks Aravind. Excellent point. $\endgroup$ – Thomas Delaney Nov 14 '16 at 0:11
  • $\begingroup$ If you are still interested, the next several values I've got so far are $$[n,m]:[5,10],[6,12],[7,15]$$ $\endgroup$ – Yuriy S Feb 5 '17 at 13:28
  • $\begingroup$ See also oeis.org/A192881 $\endgroup$ – Yuriy S Feb 7 '17 at 14:13
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Aravind has correctly observed that asymptotically, one needs to go at least up to $\sim en$, so at least $\sim (e-1)n$ terms. It was an open question of Erdős and Graham whether you ever need substantially more than this, in particular whether $1$ can be represented using only denominators between $n$ and $en + o(n)$.

Ernie Croot settled this conjecture affirmatively in "On unit fractions with denominators in short intervals" (Acta Arithmetica, 2001). In particular, we can find Egyptian fractions summing to $1$ where all the denominators lie in the range $[n, en + O\big(\frac{n \log \log n}{\log n}\big)]$, which also sharply limits the number of terms used to at most $(e-1 )n + O(n \log \log n/\log n)$.

You may be interested in Greg Martin's "Denser Egyptian Fractions" (Acta Arithmetica, 1998) which solves a sort of dual problem of representing $1$ (or any positive fraction) using a prescribed number of Egyptian fractions with all denominators $\le n$. By taking the prescribed count up to the maximum allowable number this forces the vast majority (perhaps not all) of the denominators to lie in the range $[n/(e-1), n]$ by size considerations.

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    $\begingroup$ This is a great answer. Thank you for the links! $\endgroup$ – Yuriy S Feb 2 '17 at 17:56
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A very interesting question. I don't know if there is any way to answer it for general $n$ exactly, except for experimenting with particular values.

For the case $n=5$ the best I was able to achieve is $m=10$:

$$1=\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{14}+\frac{1}{18}+\frac{1}{20}+\frac{1}{24}+\frac{1}{28}$$

For the case $n=6$ the best result so far is $m=12$:

$$\{6,7,8,9,10,14,15,18,20,24,28,30\}$$

$$n=7, \qquad m=15$$

$$\{7,8,9,10,12,14,15,18,20,21,24,30,35,40,56\}$$

$$n=8, \qquad m=23$$

$$\{8,9,10,12,14,15,18,20,21,22,28,30,33,35,40,55,60,77,80,84,126,176,198\}$$

The last two are likely not the best values, because they were done by a relatively simple computer algorithm. The result depends on the 'seed' expansion we use. For $n=9$ the best I've got so far is $m=29$.

I have entertained a related question Expanding integers into distinct egyptian fractions - what is the optimal way?, however my case requires not using $any$ of the previous denominators in the next expansion, i.e. after $[2,3,6]$ we have to introduce a new expansion without using $2$ or $3$ or $6$.

I think your case is more simple, except for the fact that it's hard to be sure we obtained the least $m$ for larger $n$.

The following method is the most useful in transforming fractions:

$$\frac{1}{n}=\frac{1}{ab}=\frac{1}{a(a+b)}+\frac{1}{b(a+b)}$$

Where $a,b$ are two distinct divisors of $n$ closest in value.

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