It is well known that $\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1$ and this is the only solution to $\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}=1$
with $2\leq x_1<x_2<x_3$.
My question is:

Let $n\in \mathbb{N}$ and $x_i\in \mathbb{N} ,1\leq i\leq n$.

How many $n$-tuples $(x_1,x_2,...,x_n)$ exist such that $\frac{1}{x_1}+\frac{1}{x_2}+\cdots+\frac{1}{x_n}=1$ with $2\leq x_1<x_2<...<x_n$ ?

Any reference would be appreciated!

  • 3
    $\begingroup$ See this paper with upper and lower bounds: link.springer.com/article/10.1134%2FS0001434614010295 $\endgroup$
    – Ian Agol
    Jan 28, 2016 at 15:33
  • 1
    $\begingroup$ See also: oeis.org/A006585 $\endgroup$
    – Ian Agol
    Jan 28, 2016 at 16:24
  • $\begingroup$ math.stackexchange.com/questions/290435/… started out as a question about the case $n=5$, but several of the answers turned to the more general case. $\endgroup$ Jan 28, 2016 at 21:44
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    $\begingroup$ The paper by Konyagin (mentioned above by Ian) gives the best known lower bound. The best known upper bound is in a paper by Browning and Elsholtz, The number of representations of rationals as a sum of unit fractions, Illinois J. Math. 55, Number 2 (2011), 685-696. maths.bris.ac.uk/~matdb/preprints/IJM343.pdf $\endgroup$ Jan 29, 2016 at 11:17
  • $\begingroup$ @ChristianElsholtz, do you think the asymptotics with distinct denominators is comparable to allowing repeats like in your and Sandor's papers? $\endgroup$
    – Ian Agol
    Jan 29, 2016 at 14:15

3 Answers 3


N. Burshtein has published several papers on this problem, On distinct unit fractions whose sum equals 1, and a more recent paper, making the problem more challenging by adding the restriction that the $n$ integers $x_i$ must be odd. There are no solutions for even $n$. The smallest odd $n=2k+1$ for which a solution exists is $k=4$, and the number of solutions is five.

For larger $n=2k+1$ the number of solutions (with the odd denominator restriction) is not known, but a lower bound of $(\sqrt 2)^{(k+1)(k-4)}$ was derived in Egyptian fractions with restrictions.


It might be worth remarking that if $G$ is a finite group with $n$ conjugacy classes and representatives $g_{1},g_{2}, \ldots ,g_{n}$, then the class equation for $G$ (divided through by $|G|$) gives $\sum_{i=1}^{n} \frac{1}{|C_{G}(g_{i})|} = 1$, though the $|C_{G}(g_{i})|$ are not usually distinct. Long ago (around 1895, I think) , E. Landau gave a bound on the number of solutions of $\sum_{i=1}^{n} \frac{1}{x_{i}} = 1$ with the $x_{i}$ (not necessarily distinct) positive integers. This allows a crude bound on the size of a finite group with $n$ conjugacy classes though this is quite far from the presently known bounds (eg of L. Pyber) which use much more group-theoretic information.

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    $\begingroup$ an upvote because of new (group theoretic) view. $\endgroup$
    – Shahrooz
    Jan 28, 2016 at 19:25
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    $\begingroup$ I wonder if any other kinds of finite algebraic structure might also generate solutions? I found this paper which seems relevant: math.ucr.edu/~jbergner/GpdEFrac.pdf $\endgroup$
    – Flounderer
    Jan 28, 2016 at 22:01

Any perfect number is an answer for your question. Also, Prof. Graham showed that $a(m)>0$ for $m>77$, where $a(m)$ denotes the total number of solution for this equation such that the sums of the denominators is $m$. You can see this page for the number of such solutions:


Also the below papers studied somehow the question:

[1] "Representation of One as the Sum of Unit Fractions" by Yuya Dan

[2] "UNIT FRACTIONS THAT SUM TO 1" by Yutaka Nishiyama

  • $\begingroup$ For clarity, since the OP used the variable $n$ in $\frac{1}{x_n}$, it might be better to use the variable $m$ as the sum of the denominators. $\endgroup$ Jan 28, 2016 at 14:36

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