According to the OEIS ([A002966](http://oeis.org/A002966)) there are 294314 solutions in positive integers to the equation
$$\sum\_{i=1}^7\frac{1}{x\_i}=1$$ assuming $x\_1\leq x\_2\leq\cdots\leq x\_7$.  

Similarly for 8 summands there are 159330691 solutions.


My question: What are they? Is there a way of counting them without knowing them?


The bound for $x\_n$ for $n$ summands is double exponential and I could only compute the solutions up to $n=6$ with Maple.



Diophantine equation: Egyptian fraction representations of 1
As far as I know, the only significant result to speed up these calculations is that $E\_2(\frac{p}{q}) = \frac{1}{2}|\lbrace d: d | q^2, d \equiv -q (mod p) \rbrace|$, where $E\_2(p/q)$ represents the number of decompositions into 2 unit fractions, and each matching $d$ represents the decomposition $\frac{p}{q} = \frac{qp}{q(q+d)} + \frac{dp}{q(q+d)}$. (Take floor() or ceil() depending on whether you want to allow repeats.)


When I've coded this in the past, I called one of 4 different functions depending on a) whether $p=1$ or not, and b) whether $q/p \ge min$ or not, where $min$ is the greatest denominator I'm already using. When $p=1$ and $q \ge min$, in particular, we can just calculate $\tau(q^2)/2$ from the factorisation of $q$; in the other cases I actually walked the factors from $q/p$ to $\sqrt{q}$.


So: yes, you can count the number of matching sets without generating the 7 elements of each set, but computationally the elements are just a whisker away.


