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
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 16, 2017 at 1:46 | answer | added | Robin Houston | timeline score: 19 | |
| Oct 25, 2017 at 20:04 | comment | added | Jeff Egger | I think that your question has been adequately answered below, but I can't help wondering what motivated it. Are you able to tell us what prompted you to consider the LHS? | |
| Oct 17, 2017 at 14:51 | comment | added | Sergey Dovgal | As I know from some talks, WZ can be sometimes inefficient, but recently guys invented a new more efficient algorithm to do the job. arxiv.org/abs/1510.07487 arxiv.org/abs/1404.5069 Offtopic: Is there a simple explanation why such kinds of questions "combinatorial identity of type sum of product of binomials having linear combinations as arguments" are so popular on MO? | |
| Oct 17, 2017 at 5:04 | history | edited | j.c. | CC BY-SA 3.0 |
put identity in title
|
| Oct 16, 2017 at 16:24 | answer | added | Max Alekseyev | timeline score: 12 | |
| Oct 16, 2017 at 8:50 | comment | added | darij grinberg | Crossposted at artofproblemsolving.com/community/… | |
| Oct 16, 2017 at 8:17 | history | edited | Shahrooz | CC BY-SA 3.0 |
improved formatting
|
| Oct 16, 2017 at 7:18 | comment | added | Peter Heinig | It would be very interesting if someone knowing much about the Wilf-Zeilberger method would write a relevant comment on whether this identity nowadays is regarded as 'automatically provable'. | |
| Oct 16, 2017 at 7:16 | history | edited | Peter Heinig | CC BY-SA 3.0 |
The grammar in the title was clearly wrong. I only changed the adjective, from an (almost) inexistant one into a usual one. Usefully relevant tags added. Light grammatical corrections in body text.
|
| Oct 16, 2017 at 7:00 | answer | added | Fedor Petrov | timeline score: 25 | |
| Oct 16, 2017 at 0:25 | comment | added | MTyson | Note that without the squares, the LHS counts the number of paths down Pascal's triangle from $(n,k)=(0,0)$ to $(a+b,a)$ passing through a marked point $(i+j,i)$. Since each path contains $a+b+1$ points, that sum equals $(a+b+1){a+b\choose a}$. | |
| Oct 15, 2017 at 14:44 | comment | added | Somos | The numbers you have come from the OEIS sequence A091044 but I don't see anything there now that would lead to a proof. | |
| Oct 15, 2017 at 13:25 | history | edited | Martin Sleziak | CC BY-SA 3.0 |
typo in the title
|
| Oct 15, 2017 at 13:24 | review | First posts | |||
| Oct 15, 2017 at 14:01 | |||||
| Oct 15, 2017 at 13:23 | history | asked | ken | CC BY-SA 3.0 |