‎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](http://arxiv.org/abs/1612.08697), 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.



Hankel determinant evaluation of special lattice paths
The $d(n):=D(n,n)$ is OEIS sequence [A005773](http://oeis.org/A005773). The Hankel determinant property is given in the sequence entry. Also the recursion $\;nd\_{n}=2nd\_{n-1}+3(n-2)d\_{n-2}$. A proof could come from a similar proof of the Hankel determinant property of the Catalan numbers. A related property is to use $d\_0=1$ and take $\det(H\_n^0)$ with $d\_0$ in the first row and column giving OEIS sequence [A163806](http://oeis.org/A163806) $(1,1,1,0,-1,-1,0,1,1,0,-1,-1,0,1,1,\dots)$ with a period of $6$ after the initial $d\_0$.


The proof of the Hankel properties could probably use the [Lindstrom-Gessel-Viennot Lemma](https://en.wikipedia.org/wiki/Lindstr%C3%B6m%E2%80%93Gessel%E2%80%93Viennot_lemma). This can be used to show that the unique solution to $\det(H\_n^0)=\det(H\_n^1)=1$ for all $n$ is the Catalan numbers. For a reference look at Aigner, Catalan-like numbers and determinants, J. Comb. Th. A 87 (1999), 33-51.


By the way, notice how the diagram for $D(4,4)=13$ is
$$
\begin{matrix}
 1 & 2 & 5 & 13 \\
 1 & 3 & 8 & 21 \\
 1 & 3 & 8 & 21 \\
 1 & 2 & 5 & 13 
\end{matrix}‎‎$$
where each entry is the sum of two or three entries in the preceding column.


Catalan and related sequence proofs are given by J, Cigler in [Some nice Hankel determinants](http://homepage.univie.ac.at/johann.cigler/preprints/hankel-conjectures.pdf).


