I need to find the determinant of matrix defined by
\begin{align\*}
& a\_{i,1}=a\_{1,j}=1,\quad \forall 1\leq i,j\leq n,\\ & a\_{i,j}=a\_{i-1,j}+a\_{i,j-1}+i-j, \quad \forall 1< i,j\leq n.
\end{align\*}


Numerically, for $n=1$ to $12$; I found that $\det(A)=F\_n$ where $F\_n$ is the Fibonacci sequence.


How to prove it if true?


Addition: edit I correct the formula


I found


$$a\_{i,j}=\displaystyle{\binom{i+j-1}{j}}-{\binom{i+j-2}{i}}+j-i\quad \forall 1\leq i,j\leq n.$$


Let $\Delta\_n=\det(A)$, it's clear that $\Delta\_1=\Delta\_2=1$.


I must prove that $\Delta\_n=\Delta\_{n-1}+\Delta\_{n-2}, \forall n\ge 3$.


**Addition 2**


**My friend jandri gave me this proof**: It proves that $\det(A\_n)=F\_n$ with $A\_n$ is the matrix of order $n$ defined by:
$a\_{i,1}=a\_{1,j}=1$ for $1\leq i,j\leq n$ and $a\_{i,j}=a\_{i-1,j}+a\_{i,j-1}+i-j$ for $1< i,j\leq n$.


We introduce the matrix $B\_n $ whose general term is defined by $b\_{i,j}=a\_{i,j}+i-j$ for $1\leq i,j\leq n$.


We have $b\_{i,1}=i$, $b\_{1,j}=2-j$ and for $i,j\geq2$ : $b\_{i,j}=b\_{i-1,j}+b\_{i,j-1}$.


We perform on the matrix $A\_n$ the operations $C\_j\leftarrow C\_j-C\_{j-1}$ for $j$ from $n$ to $2$ and after the operations $L\_i\leftarrow L\_i-L\_{i-1}$ for $i$ from $n$ to $2$. This gives a matrix that has $0$ in row $1$ and column $1$ (except for the first term which is équal to $1$) and the rest is the matrix $B\_{n-1}$, so $\det( A\_n)=\det(B\_{n-1})$.


Then we perform the same operations on the matrix $B\_n+xJ\_n$ where $J\_n$ is the matrix of order $n$ such that all the terms of which are $1$.
We obtain the matrix written in 4 blocks: $\left (\begin{array}{c|c} 1+x & L \\ \hline C& B\_{n-1} \end{array}\right)$ where $ L$ is the row in which all the terms are $-1$ and $C$ is the column in which all the terms are $1$. We deduce $\det(B\_n+xJ\_n)=(1+x)\det(B\_{n-1}+\frac1{x+1}J\_{n-1})$.


We then obtain by induction on $n$ that $\det(B\_n+xJ\_n)=F\_{n+1}+xF\_n$ whence $\det(B\_n)=F\_{n+1}$ then $\det( A\_n)=F\_n$.



Determinant equal to Fibonacci sequence
The idea of Jandri can also be applied to a slightly more general case:


Let $$ a(i,j,x,y,z,t)= x\binom{i+j-1}{j}-y \binom{i+j-2}{i}+z(j-i)+t,$$
$A\_n(x,y,z,t)$ be the matrix with entries $a(i,j,x,y,z,t)$ and $D\_n(x,y,z,t)=\det(A\_n(x,y,z,t).$


We apply the following operations on the matrix $A\_n(x,y,z,t):$ We change column $C\_j$ to $C\_j-C\_ {j-1}$ from $j=n$ to $j=2.$ Then we change row $R\_i$ to $R\_i-R\_{i-1}$ from $i=n$ to $i=2.$ We then get the block matrix
$$\begin{pmatrix}x+t&L \\ C & A\_{n-1}(x,y,0,0)\end{pmatrix}.$$
Here $L$ is a row vector all of whose entries are $z-y$ and $C$ is a column vector with entries $x-z$.
Observing that
$$\begin{pmatrix}x+t&L \\ C & A\_{n-1}(x,y,0,0)\end{pmatrix}=\begin{pmatrix}1&0 \\{\frac{C}{x+t}} & I\end{pmatrix}\begin{pmatrix}x+t&0 \\ 0& A\_{n-1}(x,y,0,0)-\frac{C}{x+t}L \end{pmatrix}\begin{pmatrix}1& \frac{L}{x+t} \\ 0 & I\end{pmatrix} $$
and that $CL=(x-z)(z-y)J\_{n-1}$ we get taking determinants
$$D\_n(x,y,z,t)=(x+t)D\_{n-1}(x,y,0,\frac{(x-z)(y-z)}{x+t}).$$


Let now $G\_n(x,y)=\sum\_{i=0}^{\lfloor{n/2}\rfloor}{n-j\choose j}y^jx^{n-j},$
which satisfies $G\_n(x,y)=xG\_{n-1}(x,y)+xy G\_{n-2}(x,y)$ and $G\_n(1,1)=F\_{n+1}$
and define
$$d\_n(x,y,z,t)=(x+t)G\_{n-1}(x,y)+(x-z)(y-z) G\_{n-2}(x,y).$$
These polynomials satisfy $d\_n(x,y,z,t)=(x+t)d\_{n-1}(x,y,0,\frac{(x-z)(y-z)}{x+t})$ and the same initial values as $D\_n(x,y,z,t).$
Therefore we get $$D\_n(x,y,z,t)=d\_n(x,y,z,t).$$
For example $D\_n(1,1,z,0)=F\_{n}+(1-z)^2F\_{n-1},$


