Multiplicity of minimal Egyptians sums

Question

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).

Answer 1

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.

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$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 $\frac 12$.

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$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 $\frac 13$).

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$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?

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

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

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

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$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.

Answer 2

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

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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

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THEOREM 2 $$ F_3=5 $$

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

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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?)