Is there any result known for the number of different solutions of $1 = \sum\limits\_{k=0}^n \frac{1}{x\_k}$ in dependency of the length $n$ of this partition?
All I know, up to now, is that there are for every $n$ only finitely many different solutions and the maximal $x\_k$ is given by the $k$th element of the Sylvester-sequence.
Further, there is a result by Hofmeister/Stoll about the number of solutions up to length $n$ for $\frac{a}{b} = \sum\limits\_{k=0}^m \frac{1}{x\_k}$ with $m \leq n$ and $a \neq b$. So unfortunately this is not usable in my case too.



Function or bounds for the number of solutions of $\sum_{i=0}^k \frac{1}{x_i} = 1$
An upper bound of about $c\_0^{2^k}$ follows by elementary induction, (from your comment $x\_k$ is bounded by the Sylvester sequence).
Here $c\_0 \approx 1.264$ is $\lim u\_n^{\frac{1}{2^n}}$, ($u\_n$ the $n$-th term of the
Sylvester sequence).


An upper bound of $c\_0^{(1+\epsilon) 2^{k-1}}$ is in a paper by C.
Sándor. Sándor also gives a lower bound: for $k \geq 3$:
$\exp(c \frac{k^3}{\log k })$, for some positive constant $c$. His paper is:  

Periodica Mathematica Hungarica  

Volume 47, Numbers 1–2, 215–219.  

[On the number of solutions of the Diophantine equation $\sum\_{i=1}^n\frac{1}{x\_i}=1$](https://doi.org/10.1023/B:MAHU.0000010822.30987.e7 "zbMATH review at https://zbmath.org/?q=an:1047.11031")


An upper bound of $c\_0^{(\frac{5}{24} +\epsilon) 2^{k-1}}$
was proved by T. Browning and myself in our paper
"[The number of representations of rationals as a sum of unit fractions](https://doi.org/10.1215/ijm/1359762408 "zbMATH review at https://zbmath.org/?q=an:1306.11029")", which is to appear in the Illinois J of
Mathematics (vol. 55, no. 2, 685-696, 2011). It is online here:
[https://www.math.tugraz.at/~elsholtz/WWW/papers/papers33FINAL2013.pdf](https://www.math.tugraz.at/%7Eelsholtz/WWW/papers/papers33FINAL2013.pdf).
This paper gives an essentially best possible answer for a general fraction $\frac{m}{n}$ and $k=2$, then a nontrivial upper bound for $k=3$, and as a corollary lifts these results to general $k$. The case $\frac{m}{n}=1$ then follows.


All in all, there is a large gap between upper and lower bound for your original question.


