Timeline for Hankel determinant evaluation of special lattice paths
Current License: CC BY-SA 3.0
Post Revisions
16 events
| when toggle format | what | by | license | comment | |
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| S Aug 15, 2017 at 2:38 | history | suggested | Somos | CC BY-SA 3.0 |
Improved ref to arXiv paper. Punctuation improved.
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| Aug 15, 2017 at 2:28 | review | Suggested edits | |||
| S Aug 15, 2017 at 2:38 | |||||
| Aug 14, 2017 at 23:47 | vote | accept | d.y | ||
| Aug 14, 2017 at 23:36 | history | edited | d.y | CC BY-SA 3.0 |
added 8 characters in body
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| Aug 14, 2017 at 23:35 | answer | added | Somos | timeline score: 6 | |
| Aug 14, 2017 at 23:35 | history | edited | d.y | CC BY-SA 3.0 |
added 447 characters in body
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| Aug 14, 2017 at 22:33 | comment | added | d.y | Let us continue this discussion in chat. | |
| Aug 14, 2017 at 22:32 | comment | added | d.y | I want to attach one image of this lattice path but I don't know how!! | |
| Aug 14, 2017 at 22:25 | comment | added | d.y | No, $D(3,3)=5$, consider the table with $3$ rows and columns, and the number of lattice paths from the first columns to the entry $(3,3)$ is equal 5. | |
| Aug 14, 2017 at 22:07 | comment | added | Simon Willerton | You've now changed you relation in the question so it doesn't involve $D(n-1, n+1)$. Can you explain what $D(2,3)$ is? You have not defined it. With the definition I would guess it seems to be 2. That would mean that $D(3,3) = 2 + 2 =4 \ne 5$. | |
| Aug 14, 2017 at 22:05 | comment | added | d.y | I check the relation in question and fixed it. actually, if we have a table with $n$ rows and $m$ columns and let $D(x,y)$ denoted the number of lattice paths from first column to the entra$(x,y)$ we have $$D(x+1,y)=D(x,y-1)+D(x,y)+D(x,y+1)$$ where $D(x,0)=D(x,m+1)=0$ for all $x\geq 1$ and $y=1,2,\cdots , m$. In the above qustion I consider for specail case $m=n$. Also, $D(n-1,n+1)$ is the number of lattice paths with three steps $(1,1),(1,-1),(1,0)$ from the first columns to the entry $(n-1,n+1)$. | |
| Aug 14, 2017 at 21:56 | comment | added | Simon Willerton | You have not read my comment. You have not said what $D(n-1, n+1)$ is. The relation you have written in your comment is not the same as the relation you have written in the question. | |
| Aug 14, 2017 at 21:56 | history | edited | d.y | CC BY-SA 3.0 |
deleted 12 characters in body
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| Aug 14, 2017 at 21:51 | comment | added | d.y | $D(n,n)$ is the number of all lattice paths from the first column to the entry $(n,n)$, then, $$D(n,n)=D(n-1,n)+D(n-1,n-1)$$. | |
| Aug 14, 2017 at 21:45 | comment | added | Simon Willerton | What is $D(n-1, n+1)$? You seem only to have defined $D(n,n)$. If it means what it seems to mean then how can your relation for $D(n,n)$ involve $D(n-1,n+1)$ as that would seem to involve your path leaving the square $T_{n,n}$. | |
| Aug 14, 2017 at 20:52 | history | asked | d.y | CC BY-SA 3.0 |