Timeline for Is there a combinatorial interpretation of the identity $\sum_{k=0}^m 2^{-2k} \binom{2k}{k} \binom{2m-k}{m} =4^{-m} \binom{4m+1}{2m}$?
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 14, 2010 at 19:16 | comment | added | Jason Dyer | @miwalin: I'll think about it more then, and edit if a proof comes to me. | |
| Apr 14, 2010 at 19:13 | comment | added | Jason Dyer | Michael, regarding the Haar measure on G(n,1): take the reciprocal of the rational part of every 4th term (there's an offset but I don't know offhand, I think 2) and you get the sequence. | |
| Apr 14, 2010 at 19:00 | comment | added | Sunni | @ Jakobsen: That is true. Dyer's explanation is insufficient. Also, I don'y know whether we shall introduce Haar measure to this problem. | |
| Apr 14, 2010 at 18:54 | comment | added | Sune Jakobsen | I think that OP wanted a "counting in two ways"-proof of the equality, and not just a interpretation of LHS. (Did you?) | |
| Apr 14, 2010 at 18:34 | comment | added | Michael Lugo | As a non-layman, I'm curious: what do you have in mind in terms of the Haar measure on G(n,1)? | |
| Apr 14, 2010 at 18:02 | history | answered | Jason Dyer | CC BY-SA 2.5 |