Determinant equal to Fibonacci sequence 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 $\det( A_n)=F_n$.
 A: 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 $D_n(1,1,z,0)=F_{n}+(1-z)^2F_{n-1},$
A: An answer also by my friend Jandri  to the second problem posed by
Johann Cigler
The same reasoning also applies in this context.
Let $A_n(x)$ be the matrix,  verifying
$$\text{$a_{i,1}=a_{1,j}=1$ and $a_{i,j}=a_{i-1,j}+a_{i,j-1 }+x(i-j)$.}$$
By introducing $B_n$ whose general term is defined by $b_{i,j}=a_{i,j}+x(i-j)$ we obtain $\det(A_n(x))=\det(B_{n -1}(x))$ then $\det(B_n(x)+yJ_n)=(1+y)\det(B_{n-1}(x)+\frac{x^2}{y+1 }J_{n-1})$.
We deduce $\det(B_n(x)+yJ_n)=a_n(x)+y\;a_{n-1}(x)$ where the sequence $a_n(x)$ verifies $a_0=a_1=1$ and $a_{n+1}(x)=a_n(x)+x^2\;a_{n-1}(x)$.
Hence $\det(A_n(x))=a_{n-1}(x)$.
We can show by induction that $a_n(x)=\displaystyle \sum_{k=0}^{\lfloor n/2\rfloor}{\binom{n-k}k}x^{2k}$.
A: An answer also by my friend Jandri to the first  problem posed by Johann Cigler :
let $a_{i,j}=x(\binom{i+j-1}{j}+j-i)-y \binom{i+j-2}{i}$. Johann Cigler  conjectures that $\Delta_n= x \Delta_{n-1}+ x y\Delta_{n-2}, \forall n\ge 3.$
This last matrix to the second problem posed by Johann Cigler is very similar to the one I considered just above, it also satisfies $a_{i,j}=a_{i-1,j}+a_{i,j-1}+x(i-j) $. The only difference is that $a_{i,1}=x$ and $a_{1,j}=jx-(j-1)y$.
It is treated exactly the same way by introducing the matrix $B_n$ with general term $b_{i,j}=a_{i,j}+x(i-j)$.
We get $\det(A_n)=x\;\det(B_{n-1})$ then $\det(B_n+zJ_n)=(x+z)\det(B_{n-1}+\frac {xy}{x+z}J_{n-1})$.
We deduce $\det(B_n+zJ_n)=a_n+z\;a_{n-1}$ with $a_0=1$, $a_1=x$ and $a_{n+1}=x\;a_n+ xy\;a_{n-1}$ and hence $\det(A_n)=x\;a_{n-1}$.
The sequence $\det(A_n)$ therefore follows the same recurrence relation as the sequence $a_n$.
