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!



How many solutions does $\frac{1}{x_1}+\frac{1}{x_2}+\cdots +\frac{1}{x_n}=1$ have?
N. Burshtein has published several papers on this problem, [On distinct unit fractions whose sum equals 1](http://ac.els-cdn.com/0012365X73901362/1-s2.0-0012365X73901362-main.pdf?_tid=2e819cfe-c5bc-11e5-b495-00000aacb35f&acdnat=1453984953_26680765f7196bbca82cbd943b861d18), and [a more recent paper,](http://www.sciencedirect.com/science/article/pii/S0022314X07000510) 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.](http://arxiv.org/abs/1108.6118)


