Egyptian fraction decomposition of $1$ and the sequence A007018.

Question

While answering a question, I have observed an interesting property of the matrix introduced there.

Consider positive integer solutions of the equation:

$$ \begin{vmatrix} -x_1&1&1&\dots&1\\ 1&-x_2& 1&\dots&1\\ \vdots&\ddots&\ddots&\dots&\vdots\\ 1&1&1&\ddots&1\\ 1&1&1&\dots&-x_n \end{vmatrix}=0. $$

As was noted in comments with the help of Sylvester determinant theorem the problem boils down to finding the positive integer solutions to the Diophantine equation $$ \sum_{i=1}^n\frac{1}{1+x_i}=1, $$ which can be recognized as egyptian fraction decomposition of $1$.

As the equation is obviously symmetric with respect to variables, any permutation of a solution is again a valid solution. Therefore, let us consider only solutions with

$$x_1\le x_2\le \dots \le x_n.$$

The following statement seem by numerical evidence to be true:

There is always a solution in form $(A_1,A_2,\dots,A_{n-1},A_n-1)$, where $A_i$ are numbers of A007018 sequence. This solution is the largest one (in the sense of having the largest element).

How to prove this statement? How many distinct (up to permutations) solutions does the equation have?

Answer 1

Below is a quotation to a paper giving the fastest-growing solution. Oddly enough (to me), you refered to it as A014117. The sequence you are interested in is in fact the Sylvester's sequence (A000058): 2,3,7,43,1807,...

The terms of the Sylvester's sequence satisfy $\frac{1}{a_1}+\cdot\cdot\cdot+\frac{1}{a_{n-1}}+\frac{1}{a_n-1}=1$. Thus for all $n$, $(x_1,...,x_n)=(a_1-1,...,a_{n-1}-1,a_n-2)$ - with $(a_n)$ the Sylvester's sequence - is a solution to your problem (the link between your question and egyption fractions is developed in the comments and in Will Jagy's answer).

Now you can find here a proof by Curtis that it is the only solution making $x_n$ maximal.

Thus the behaviour of the maximal solution is similar to that of the Sylvester's sequence, which is quite well-known.

Answer 2

It appears user1551 has answered fully in comments, but this is material I did not know.

A simple version of the Sylvester Determinant Theorem is this. Suppose $A$ is a square invertible matrix. Suppose $u,v$ are column vectors. Then $$ \det \left( A + u v^T \right) = \left( 1 + v^T A^{-1} u \right) \det A$$

For us, $u=v$ are both the vector with all entries equal to $1.$ Therefore $v^T A^{-1} u$ is just the sum of all the elements of $A^{-1}.$ For that matter, $A$ and $A^{-1}$ are going to be diagonal, so $v^T A^{-1} u$ will become the trace of $A^{-1}.$

For emphasis, let $J$ be the square matrix of all $1$'s, and $D$ an invertible diagonal matrix. Then $$ \det \left( D + J \right) = \left( 1 + \; \operatorname{trace} D^{-1} \right) \det D$$

We have positive integers $x_j.$ We define, illustration for $n=4,$ $$ A = \left( \begin{array}{cccc} -1 - x_1 & 0 & 0 & 0 \\ 0 & -1 - x_2 & 0 & 0 \\ 0 & 0 & -1 - x_3 & 0 \\ 0 & 0 & 0 & -1 - x_4 \end{array} \right) $$ while $$ u v^T = \left( \begin{array}{cccc} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{array} \right) $$ while $$ A = \left( \begin{array}{cccc} -\frac{1}{ \,1 \; + \; x_1 \;} & 0 & 0 & 0 \\ 0 & -\frac{1}{ \,1 \; + \; x_2 \;} & 0 & 0 \\ 0 & 0 & -\frac{1}{ \,1 \; + \; x_3 \;} & 0 \\ 0 & 0 & 0 & -\frac{1}{ \,1 \; + \; x_4 \;} \end{array} \right) $$

$$ \det \left( A + u v^T \right) = \left( 1 + v^T A^{-1} u \right) \det A$$ gives us $$ \det \left( A + u v^T \right) = \left( 1 - \sum_{j=1}^n \frac{1}{ \,1 \; + \; x_j \;} \right) \det A$$ We know that $\det A \neq 0,$ so the condition that $ \det \left( A + u v^T \right) = 0$ is precisely the condition that $$ \sum_{j=1}^n \frac{1}{ \,1 \; + \; x_j \;} = 1 $$