Let $m,n$ be natural numbers, and let $S\_{m,n}$ be the set of all the natural number solutions $\mathbf x=(x\_1,x\_2,\cdots,x\_m)$ to the following equation :
$$\sum\_{j=1}^m\frac{1}{x\_j}=\frac1n.$$
Also, letting $$k\_{m,n}=\max\_{\mathbf x\in S\_{m,n}}\left(\max\_{1\le j\le m}x\_j\right)$$


then, here is my question.



> 
> **Question** : Is the following true for $m\ge 2$?
> 
> 
> $$k\_{m,n}=k\_{m-1,n}\left(k\_{m-1,n}+1\right).$$
> 
> 
> 


**Remark** : This question has been [asked previously on math.SE](https://math.stackexchange.com/questions/507903/about-the-set-of-all-the-solutions-mathbf-x-x-1-x-2-cdots-x-m-to-sum-j) without receiving any answers:


**Motivation** : I've been asking the following question on MO.


[What is the max of $n$ such that $\sum\_{i=1}^n\frac{1}{a\_i}=1$ where $2\le a\_1\lt a\_2\lt \cdots\lt a\_n\le m$?](https://mathoverflow.net/questions/142300/what-is-the-max-of-n-such-that-sum-i-1n-frac1a-i-1-where-2-le-a-1-l)


This got me interested in $S\_{m,n}.$


In the following, I'm going to prove that $S\_{m,n}$ is a finite set.


**Proof** : Let $S\_{m,n}^{\gt}$ be the set of all solutions $\mathbf x=(x\_1,x\_2,\cdots,x\_m)$ to the above equation such that $x\_1\ge x\_2\ge \cdots\ge x\_m$. Since any solution $\mathbf x$ is one that we can get from the elements of $S\_{m,n}^{\gt}$ by exchanging its coordinates, we get
$$|S\_{m,n}|\le m! |S\_{m,n}^{\gt}|$$
where $|S|$ represents the number of the elements of a set $S$. Hence, in the following, let's prove that $S\_{m,n}^{\gt}$ is a finite set for any natural number $n$ by induction on $m$.


The $m=1$ case is obvious. Then, let's suppose that $S\_{m-1,n}^{\gt}$ is a finite set. Since
$$\sum\_{j=1}^{m-1}\frac1{x^j}=\frac1n-\frac1{x\_m}=\frac{x\_m-n}{nx\_m},$$we get
$$\sum\_{j=1}^{m-1}=\frac{1}{(x\_m-n)x\_j}=\frac{1}{nx\_m}\ \ \ \ \cdots(\star).$$
Now, $x\_m$ satisfies the following inequality :
$$n+1\le x\_m\le mn\ \ \ \ \ \cdots(\star\star).$$
This is because if we deny $(\star\star)$, then $\mathbf x$ cannot be a solution of the equation. For a $x\_m$ which satisfies $(\star\star)$, the number of solutions $\mathbf x^{\prime}=\left((x\_m-n)x\_1,(x\_m-n)x\_2,\cdots,(x\_m-n)x\_{m-1}\right)$ which satisfies $(\star)$ is finite by the supposition. Since the number of $x\_m$ which satisfies $(\star\star)$ is finite as well, $|S\_{m,n}^{\gt}|$ is finite. Now the proof is completed.


In the following, I'm going to represent $k\_{3,n}$ by $n$.


First of all, note that $k\_{2,n}=n(n+1)$. Next, we know that
$$\left(n(n+1)(n(n+1)+1),n(n+1)+1,n+1\right)\in S\_{3,n}^{\gt}.$$
Hence, by the definition of $k\_{3,n}$, we get
$$k\_{3,n}\ge n(n+1)(n(n+1)+1).$$
Now, for any $\mathbf x=(x\_1,x\_2,x\_3)$, we get $(rx\_1,rx\_2)\in S\_{2,n(n+r)}^{\gt}$ by $(\star)$ where $r=x\_3-n$ and $1\le r\le 2n$ by $(\star\star).$
Hence, by the definition of $k\_{m,n}$, we get
$$rx\_1\le k\_{2,n(n+r)}=n(n+r)(n(n+r)+1).$$
Hence, we get
$$x\_1\le \frac 1r n(n+r)(n(n+r)+1).$$
However, since $$n(n+1)(n(n+1)+1)-\frac 1r n(n+r)(n(n+r)+1)=\frac{1}{r}n^2(n^2-n+1)(r-1)\ge \frac 1r n^2(n^2-2n+1)(r-1)\ge 0,$$
we get
$$x\_1\le n(n+1)(n(n+1)+1).$$


After observing this question, I reached the above expectation. However I can neither prove that this is true nor find any counterexample. Can anyone help?



About the set of all the solutions $\mathbf x=(x_1,x_2,\cdots,x_m)$ to $\sum_{j=1}^m\frac{1}{x_j}=\frac1n$
I've just been able to prove that my expectation is true.


In order to prove this, let us define the following sequence for $n\in\mathbb N$ : 
$$e\_{1,n}=n+1,\ \ e\_{m,n}=ne\_{1,n}e\_{2,n}\cdots e\_{m-1,n}+1\ \ (m=2,3,4,\cdots).$$


**Lemma 1** : 
$$\begin{align}\frac1{e\_{1,n}}+\frac1{e\_{2,n}}+\cdots+\frac1{e\_{m,n}}=\frac1n-\frac1{e\_{m+1,n}-1}.\end{align}$$


**Proof for lemma 1** : By the definition, we get
$$(e\_{m-1,n}-1)e\_{m-1,n}=ne\_{1,n}\cdots e\_{m-2,n}\cdot e\_{m-1,n}=e\_{m,n}-1.$$
Since 
$$\frac1{e\_{m-1,n}}=\frac1{e\_{m-1,n}-1}-\frac1{e\_{m,n}-1},$$
we get
$$\sum\_{i=2}^{m+1}\frac1{e\_{i-1,n}}=\sum\_{i=2}^{m+1}\left(\frac1{e\_{i-1,n}-1}-\frac1{e\_{i,n}-1}\right)=\frac1{e\_{1,n}-1}-\frac1{e\_{m+1,n}-1}=\frac1n-\frac1{e\_{m+1,n}-1}.$$
Now the proof for lemma 1 is completed.


**Lemma 2** : 
$$ x\_1y\_1+x\_2y\_2+\cdots+x\_my\_m $$$$=(x\_1-x\_2)y\_1+(x\_2-x\_3)(y\_1+y\_2)+\cdots+(x\_{m-1}-x\_m)(y\_1+y\_2+\cdots+y\_{m-1})+x\_m(y\_1+y\_2+\cdots+y\_m)$$


**Proof for lemma 2** : Let $S\_i=y\_1+y\_2+\cdots+y\_i\ (i\in\mathbb N)$.


Then, we get
$$\begin{align}\sum\_{i=1}^mx\_iy\_i & = x\_1y\_1+\sum\_{i=2}^mx\_i(S\_i-S\_{i-1}) \\
 & = x\_1y\_1+\sum\_{i=2}^mx\_iS\_i-\sum\_{i=1}^{m-1}x\_{i+1}S\_i \\ 
 & = x\_1y\_1+\sum\_{i=2}^{m-1}(x\_i-x\_{i+1})S\_i+x\_mS\_m-x\_2S\_1 \\
 & = (x\_1-x\_2)y\_1+\sum\_{i=2}^{m-1}(x\_i-x\_{i+1})(y\_1+\cdots+y\_i)+x\_m(y\_1+\cdots+y\_m).
\end{align}$$
Now the proof for lemma 2 is completed.


**Lemma 3** : If 
$$\frac1{x\_1}+\frac1{x\_2}+\cdots+\frac1{x\_m}\lt\frac1n$$
where $x\_i\ (i=1,2,\cdots,m)$ are natural numbers, then
$$\begin{align}\frac1{x\_1}+\frac1{x\_2}+\cdots+\frac1{x\_m}\le\frac1{e\_{1,n}}+\frac1{e\_{2,n}}+\cdots+\frac1{e\_{m,n}}\qquad(\star)\end{align}$$


**Proof for lemma 3** : Let us prove this by induction on $m$. We'll use the proof by contradiction. Also, by the symmetry about the letters, we may suppose that $x\_1\le x\_2\le \cdots\le x\_m.$


If $m=1$, then since $x\_1\gt n$ and $x\_1\ge n+1=e\_{1,n}$, we get $(\star).$ 


In the following, let's suppose $(\star)$ for any less than or equal to $m-1$.


Now, for $m$, let's suppose that there exists $(x\_1,x\_2,\cdots,x\_m)$ which does not satisfy $(\star)$. (we'll use the proof by contradiction)


This means $(x\_1,x\_2,\cdots,x\_m)$ satisfies
$$\begin{align}\frac1{e\_{1,n}}+\frac1{e\_{2,n}}+\cdots+\frac1{e\_{m,n}}\lt \frac1{x\_1}+\frac1{x\_2}+\cdots+\frac1{x\_m}\lt \frac1n\qquad(\star\star)\end{align}$$


Hence, from the lemma 1, we know
$$0\lt\frac1n-\left(\frac1{x\_1}+\frac1{x\_2}+\cdots+\frac1{x\_m}\right)\lt\frac1n-\left(\frac1{e\_{1,n}}+\frac1{e\_{2,n}}+\cdots+\frac1{e\_{m,n}}\right)=\frac1{e\_{m+1,n}-1}.$$


Here, we know, by reducing the fractions to a common denominator,
$$\frac1n-\left(\frac1{x\_1}+\frac1{x\_2}+\cdots+\frac1{x\_m}\right)\ (\gt 0)$$
can be represented as
$$\frac{M}{nx\_1x\_2\cdots x\_m}$$
where $M\in\mathbb N.$
By the definition of the sequence $\{e\_{m,n}\}$, we get
$$\frac{1}{nx\_1x\_2\cdots x\_m}\le\frac{M}{nx\_1x\_2\cdots x\_m}\lt\frac{1}{ne\_{1,n}e\_{2,n}\cdots e\_{m,n}}.$$
Hence, we get
$$\begin{align}x\_1x\_2\cdots x\_m\gt e\_{1,n}e\_{2,n}\cdots e\_{m,n}\qquad (1)\end{align}$$
Here, letting 
$$P=\frac{x\_1}{e\_{1,n}}+\frac{x\_2}{e\_{2,n}}+\cdots+\frac{x\_m}{e\_{m,n}},$$
we get the following by AM–GM inequality and $(1)$ : 
$$\begin{align} P\ge m\sqrt[m]{\frac{x\_1x\_2\cdots x\_m}{e\_{1,n}e\_{2,n}\cdots e\_{m,n}}}\gt m\qquad (2)\end{align}$$
On the other hand, we get the following by lemma 2 : 
$$\begin{align} P= & (x\_1-x\_2)\frac1{e\_{1,n}}+(x\_2-x\_3)\left(\frac1{e\_{1,n}}+\frac1{e\_{2,n}}\right)+\cdots \\ & +(x\_{m-1}-x\_m)\left(\frac1{e\_{1,n}}+\frac1{e\_{2,n}}+\cdots+\frac{1}{e\_{m-1,n}}\right) \\ & +x\_{m}\left(\frac1{e\_{1,n}}+\frac1{e\_{2,n}}+\cdots+\frac1{e\_{m,n}}\right) \end{align}$$


Here, noting that $x\_i-x\_{i+1}\le0\ (i=1,2,\cdots,m-1)$ and that the inductive assumption leads
$$\frac1{x\_1}\le\frac{1}{e\_{1,n}},$$
$$\frac1{x\_1}+\frac{1}{x\_2}\le\frac{1}{e\_{1,n}}+\frac1{e\_{2,n}},$$
$$\vdots$$
$$\frac1{x\_1}+\cdots+\frac{1}{x\_{m-1}}\le\frac1{e\_{1,n}}+\cdots+\frac1{e\_{m-1,n}},$$
we get
$$\begin{align} P\le & (x\_1-x\_2)\frac1{x\_1}+(x\_2-x\_3)\left(\frac1{x\_1}+\frac1{x\_2}\right)+\cdots \\ & +(x\_{m-1}-x\_{m})\left(\frac1{x\_1}+\cdots+\frac1{x\_{m-1}}\right) \\ & +x\_m\left(\frac1{e\_{1,n}}+\frac1{e\_{2,n}}+\cdots+\frac1{e\_{m,n}}\right).\end{align}$$
Here, using $(\star\star)$, since we get
$$\begin{align}P\lt & (x\_1-x\_2)\frac1{x\_1}+(x\_2-x\_3)\left(\frac1{x\_1}+\frac1{x\_2}\right)+\cdots \\ & +(x\_{m-1}-x\_{m})\left(\frac1{x\_1}+\cdots+\frac1{x\_{m-1}}\right)\\ & +x\_m\left(\frac1{x\_1}+\frac1{x\_2}+\cdots+\frac1{x\_m}\right),\end{align}$$
we get $P\lt 1+1+\cdots+1=m$, which contradicts $(2)$. Hence, by the proof of contradiction, we now know that there does not exist $(x\_1,x\_2,\cdots,x\_m)$ which satisfies $(\star\star)$. Hence, the above leads $(\star)$ for $m$. The equality is attained when $x\_1=e\_{1,n}, x\_2=e\_{2,n},\cdots, x\_m=e\_{m,n}.$ Now the proof for lemma 3 is completed.


**Proof for my expectation** : Let us prove that 
$$\begin{align}\max\_{\mathbf x\in S\_{m,n}}\left(\max\_{1\le i\le m}x\_i\right)=e\_{m,n}-1\qquad (3)\end{align}$$
If we can prove $(3)$, then $k\_{m,n}=e\_{m,n}-1$ leads
$$k\_{m,n}=(k\_{m-1,n}+1)k\_{m-1,n}$$
by the relational expression about $\{e\_{m,n}\}.$


Let us prove $(3)$ by induction on $m$.


If $m=1$, then we leads $x\_1=n=e\_{1,n}-1.$


In the $m\ge 2$ case, we may suppose that $x\_1\le x\_2\le \cdots\le x\_m$.
Since
$$\sum\_{i=1}^{m-1}\frac1{x\_i}\lt\frac1n,$$
we get by lemma 3
$$\frac1{x\_1}+\frac1{x\_2}+\cdots+\frac1{x\_{m-1}}\le\frac1{e\_{1,n}}+\frac1{e\_{2,n}}+\cdots+\frac1{e\_{m-1,n}}.$$
Hence, we get by lemma 1
$$\frac1{x\_m}=\frac1n-\sum\_{i=1}^{m-1}\frac1{x\_i}\ge\frac1n-\sum\_{i=1}^{m-1}\frac1{e\_{i,n}}=\frac1{e\_{m,n}-1}.$$
Since $x\_m\le e\_{m,n}-1$, we get
$$\max\_{\mathbf x\in S\_{m,n}}\left(\max\_{1\le i\le m}x\_i\right)\le e\_{m,n}-1.$$
By the way, since we know by lemma 1
$$(e\_{1,n},e\_{2,n},\cdots,e\_{m-1,n},e\_{m,n}-1)\in S\_{m,0},$$
the equality of the above inequality is obtained. Now the proof for $(3)$ is completed. Hence, we now know that the proof for my expectation is completed.


