Multiplicity of minimal Egyptians sums 
This Q. is an extension of Egyptian representations of $1$

Let $\ \emptyset\ne A\subseteq\mathbb N\ $ be a finite set. Then let


*

*$\ \ E_n\ :=\ \{ A :\ \sum_{x\in A} \frac 1x
               \ =\ 1\quad\&\quad \min\,A\ =\ n\} $

*$\ \ M_n\ :=\ \{A\in E_n :\ |A|\ =\ \min(|B|:\ B\in E_n)\} $


Questions:


*

*What is the minimal $\ n\ $ such that $\ |M_n|>1\ ?$

*Let $\ \mu_n:=\max( |M_k|:\ k=1\ldots n).\ $ What growth does function $\ (\mu_n)\ $ exhibit?

*How often $\ |M_n|>|M_{n+1}|\ ?$



PS. It goes without saying that everybody--be it a pedantic professor or a crazy computer hacker--is welcome to provide computer computations and/or info from the literature, encyclopedias, etc. (You never know, your reputation may gain dramatically :) ).

EDIT:
$$ 1\ \ =\ \ \frac 13\ +\ \frac 14\ +\ \frac 15\ + \frac 16\ +\ \frac 1{20} $$
is the unique shortest of its type, $\ F_3=5.\ $ Thus,
$$ |M_3|=1 $$
and the requested minimal $\ n\ $ with $\ |M_n|> 2\ $ is GREATER than 3 (against a comment below).
 A: 
This time I got some numerical data by writing a simple Perl program (I am willing to add its code below if asked). Within their scope, the provided answers are complete.


$F_2=\mathbf 3\ $ as illustrated by unique $\ 1\ =\ \frac 12+\frac 13+\frac 16$. There are also additional $6$ expressions of $4$ summands starting with $\frac 12$.

$F_3=\mathbf 5\ $ as llustrated by unique
  $\ 1\ =\ \frac 13+\frac 14+\frac 15+\frac 16+\frac 1{20}.\ $ There are also $27$ expressions of $6$ summands (starting with $\frac 13$).

$F_4=\mathbf 8,\ $ where there are $77$ expressions of $8$ summnads. One could say that there are so many of them because an expected shorter expression is missing--it feels like these minimal expressions are not minimal. Can one turn this intuition into a theorem?

$F_5=\mathbf{10},\ $ where there are $161$ minimal length expressions.
Again $\ \frac{F_n}n = 2\ $ (which seems to be high).

$F_6=\mathbf{11}.\ $ This time $\ \frac{F_n}n < 2,\ $ and the number of different minimal expressions is only $4$.

$F_7=\mathbf{13},\ $ and there are $\ 4\ $ minimal expressions. The situation is similar as for $\ F_6$.

$F_8=\mathbf{15}\ $ is served by $\ 19\ $ minimal expressions. It's an inbetween case--indeed, $\ \frac{F_8}8<2\ $ but by now, it's quite close to $\ 2.$


Etc. Have fun. I can copy in more specific results like some actual expressions, etc.

A: I'll prove, step by step,
$$ |M_n|>1\quad\Rightarrow\quad n>3 $$

-

THEOREM 1
$$ F_3 > 4 $$
PROOF   Let $\ 3<a<b<c$ be integers. Then
$$ \frac 13+\frac 1a+\frac 1b+\frac 1c\,\ \le
    \,\ \frac 13+\frac 14+\frac 15+\frac 16\,\ =
     \,\ \frac{19}{20}\,\ <\,\ 1 $$
END of Proof

THEOREM 2
$$ F_3=5 $$
PROOF
$$ \frac 13+\frac 14+\frac 15+\frac 16+\frac 1{20}\,\ =\,\ 1 $$
END of Proof

THEOREM 3   If integers $\,\ 3<a<b<c<d\,\ $ are such that
$$ \frac 13+\frac 1a+\frac 1b+\frac 1c+\frac 1d\ =\ 1 $$
then
$$\ (3\,\ a\,\ b\,\ c\,\ d)\ \ =\ \ (3\,\ 4\,\ 5\,\ 6\,\ 20)$$.
PROOF   Let $\ (a\ b\ c\ d)\ $ be as above except for
$$ (a\ b\ c\ d)\ \ne\ (4\ 5\ 6\,\ 20).\ $$
Since the respective Egyptian sum is still $\ 1,\ $ we get $\ d<20.\ $
Call integers $\ 3\ a\ b\ c\ d,\ $ to be seds (sed = a selected
denominator). No sed can be a prime $\ p\ $ such that
$\ p>\frac {19}2)\ $ (because there cannot be only one sed which
divides $p$).
Next, $\ 7\ $ is not among seds because the only other possible multiple of $\ 7\ $ among seds would be $\ 14.\ $ However the denominator of the sum
$$ \frac 17+\frac 1{14}\ =\ \frac 3{14} $$
is divisible by $\ 7.\ $ Thus, $\ 7\ $ is out. Then also $\ 14\ $ is out because it'd be the only sed divisible by $\ 7.$
Also, the only potential seds divisible by
$\ 5\ $ are $\ 5\,\ 10\,\ 15,\ $ while there can be no sed divisible
by $\ 5\ $ without no other seds divisible by $\ 5.\ $ More generally, every but one Egyptian sum of inverses of seds divisible by $\ 5\ $ has its (reduced) denominator divisible by $\ 5:$


*

*$\ \frac 15+\frac 1{10}\ =\ \frac 3{10} $

*$\ \frac 15+\frac 1{15}\ =\ \frac 4{15} $

*$\ \frac 1{10}+\frac 1{15}\ =\ \frac 16\qquad $ (the only exception!)

*$\ \frac 15+\frac 1{10}+\frac 1{15}\ =\ \frac {11}{30} $


In the above exceptional case we have:
$$ \frac 13+\frac 1a+\frac 1b+\frac 1c+\frac 1d\,\ \le\,\
   \frac 13+\frac 14+\frac 16+\frac 1{10}+\frac 1{15}\,\ =
   \,\ \frac{11}{12}\,\ <\,\ 1 $$
Thus, after all, $\ 5\ $ and its multiples are out as seds. This leaves potential seds to by divisible only by $2$ and $3$, and by no other primes. Hence
$$ \frac 13+\frac 1a+\frac 1b+\frac 1c+\frac 1d\,\ \le\,\
   \frac 13+\frac 14+\frac 16+\frac 18+\frac 1d\,\ =
   \,\ \frac 78+\frac 1d\,\ <\,\ 1 $$
(because $\ d>8$).   END of Proof

Brute force? Can you do it simpler? (... so that question
  $\ |M_4|\ =\ 1\ $ would be a snap?)

