Timeline for Combinatorial identity: $\sum_{i,j \ge 0} \binom{i+j}{i}^2 \binom{(a-i)+(b-j)}{a-i}^2=\frac{1}{2} \binom{(2a+1)+(2b+1)}{2a+1}$
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 23, 2017 at 15:38 | comment | added | Robin Houston | @AlexanderBurstein If it can, I don’t yet see how. I’ll let you know if I think of anything. | |
| Nov 21, 2017 at 10:24 | history | edited | Robin Houston | CC BY-SA 3.0 |
remove tangential incorrect proof
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| Nov 17, 2017 at 11:52 | history | edited | Robin Houston | CC BY-SA 3.0 |
small clarification
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| Nov 17, 2017 at 8:43 | history | edited | Robin Houston | CC BY-SA 3.0 |
corrected proof of Andrews-Paule identity; added references
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| Nov 17, 2017 at 1:44 | comment | added | Alexander Burstein | This is really nice. Can this argument be modified somehow to show that, for any permutation $\pi$ of $\{0,1,\dots,n\}$, we have $\sum_{i,j=0}^{n}\binom{i+j}{i}\binom{2n-i-j}{n-i}\binom{\pi(i)+\pi(j)}{\pi(i)}\binom{2n-\pi(i)-\pi(j)}{n-\pi(i)}>\binom{2n+1}{n+1}^2$? I had to prove this once a while ago, but could not find any argument using a natural lattice path interpretation, like yours above. | |
| Nov 16, 2017 at 12:49 | history | edited | Robin Houston | CC BY-SA 3.0 |
added combinatorial interpretation of the Andrews-Paule identity
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| Nov 16, 2017 at 11:17 | history | edited | Robin Houston | CC BY-SA 3.0 |
describe the inverse of the GKS bijection
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| Nov 16, 2017 at 10:59 | history | edited | Robin Houston | CC BY-SA 3.0 |
notationally distinguish column vectors from binomial coefficients
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| Nov 16, 2017 at 2:26 | history | edited | Robin Houston | CC BY-SA 3.0 |
clarification
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| Nov 16, 2017 at 2:03 | history | edited | Robin Houston | CC BY-SA 3.0 |
more explanation of last part
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| Nov 16, 2017 at 1:56 | history | edited | Robin Houston | CC BY-SA 3.0 |
correct final expression
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| Nov 16, 2017 at 1:46 | history | answered | Robin Houston | CC BY-SA 3.0 |