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
7 events
| when toggle format | what | by | license | comment | |
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| Oct 16, 2017 at 20:06 | comment | added | Max Alekseyev | I assume it's a common knowledge. | |
| Oct 16, 2017 at 20:04 | comment | added | Fedor Petrov | Ok, but this could be mentioned in the answer, I think. | |
| Oct 16, 2017 at 19:59 | history | edited | Max Alekseyev | CC BY-SA 3.0 |
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| Oct 16, 2017 at 19:36 | history | edited | Max Alekseyev | CC BY-SA 3.0 |
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| Oct 16, 2017 at 19:24 | comment | added | Max Alekseyev | @FedorPetrov: Notice that $\binom{s}{i}^2 = [x^iz^s]\ ((1+xz)(1+z))^s$ and use Lagrange inversion w.r.t. variable $z$. | |
| Oct 16, 2017 at 19:21 | comment | added | Fedor Petrov | How do you get the formula for $F$? | |
| Oct 16, 2017 at 16:24 | history | answered | Max Alekseyev | CC BY-SA 3.0 |