Let $n$ be a positive integers and $T=T_{n,n}$ be the $n\times n$ table in the first quadrant composed of $n^2$ unit squares, whose $(x,y)$-blank is locate in the $x^{th}$-column from the left and the $y^{th}$-row from the bottom hand side of $T_{n,n}$ . Put $D(n,n)$ be the number of all lattice path from the first column to entry $(n,n)$ of the table $T$ which steps comes from the set $S=\{(1,0), (1,1),(1,-1)\}$.(we allowed to move only to the right (up, down or straight) ). It is easy to see for $n\geq 2$ $$D(n,n)=D(n-1,n)+D(n-1,n-1)$$ where $D(1,1)=1, D(2,2)=2, D(3,3)=5, D(4,4)=13, \cdots$.

Notice, the entry $(x,y)$ means cordinate $x$ and $y$ in the table $T$ not the row $x$ and column $y$.

For example $$D(3,3)=D(2,3)+D(2,2)=2+3=5$$ and $$D(2,3)=D(1,3)+D(1,2)=1+1=2.$$ For calculating $D(2,3)$ you must consider the table with 3 rows and columns and by using this table calculate all lattice paths reach to entry $(2,3)$ in this table!! in my arXiv paper, there are some references for this sequence!! I think these lattice paths very interesting and obtained some results about them. Put $D(n,n)=d_n$, I check and known that the Hankel determinant evaluation of $D(n,n)$ is

$$ \det(H_n^1)=\det \begin{bmatrix} d_{1} & d_{2} & d_{3} & \dots & d_{n} \\ d_{2} & d_{3} & d_{4} & \dots & d_{n+1} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ d_{n} & d_{n+1} & d_{n+2} & \dots & d_{2n-1} \end{bmatrix} =1. $$ Do you have ideas or comments for proving it?

Thank you so much for any help or comment.

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