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‎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 ‎$‎D(1,1)=1, D(2,2)=2, D(3,3)=5,‎\cdots‎‎$‎\‎

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 paper (arxiv.org/pdf/1612.08697.pdf), 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 idea or comment for prove it?

Thank you so much for any helps or comment.

‎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.

‎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 ‎$‎D(1,1)=1, D(2,2)=2, D(3,3)=5,‎\cdots‎‎$‎\‎

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 paper arxiv.org/pdf/1612.08697.pdf, 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 idea or comment for prove it?

Thank you so much for any helps or comment.

‎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 ‎$‎D(1,1)=1, D(2,2)=2, D(3,3)=5,‎\cdots‎‎$‎\‎

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 paper (arxiv.org/pdf/1612.08697.pdf), 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 idea or comment for prove it?

Thank you so much for any helps or comment.

‎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 ‎$‎D(1,1)=1, D(2,2)=2, D(3,3)=5,‎\cdots‎‎$‎\‎ 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 idea or comment for prove it?

Thank you so much for any helps or comment.

‎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 ‎$‎D(1,1)=1, D(2,2)=2, D(3,3)=5,‎\cdots‎‎$‎\‎

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 paper arxiv.org/pdf/1612.08697.pdf, 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 idea or comment for prove it?

Thank you so much for any helps or comment.