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!

1more comment