Timeline for Looking for a combinatorial proof for an identity involving $q$-Catalan triangles
Current License: CC BY-SA 4.0
13 events
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| Jul 28 at 18:44 | history | edited | Wolfgang |
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| Feb 16, 2021 at 14:56 | comment | added | T. Amdeberhan | @AlexanderBurstein: that looks fine and perhaps would help. | |
| Feb 16, 2021 at 7:08 | comment | added | Alexander Burstein | Not sure if this is of any help towards a combinatorial proof, but $1+q^n=[2n]/[n]$, so the identity may be rewritten as $$\sum_{k=1}^n q^{(k-1)^2}[2k] \left[\frac{[k]}{[n]} \binom{2n}{n-k}_q\right]^2 = \binom{2n}{n}_q \binom{2n-2}{n-1}_q.$$ | |
| Feb 11, 2021 at 17:27 | history | edited | T. Amdeberhan | CC BY-SA 4.0 |
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| Feb 11, 2021 at 14:53 | comment | added | T. Amdeberhan | @SamHopkins: I now understand what you meant. Yes, you are right there. Corrected. | |
| Feb 11, 2021 at 14:51 | history | edited | T. Amdeberhan | CC BY-SA 4.0 |
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| Feb 11, 2021 at 14:19 | history | edited | T. Amdeberhan | CC BY-SA 4.0 |
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| Feb 11, 2021 at 14:01 | history | edited | T. Amdeberhan | CC BY-SA 4.0 |
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| Feb 11, 2021 at 14:00 | comment | added | Sam Hopkins | In your previous question, you write the same equality but with the terms in the sum squared. This seems strange to me. | |
| S Feb 11, 2021 at 2:01 | history | suggested | markvs | CC BY-SA 4.0 |
Fixed misprint in the title
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| Feb 11, 2021 at 0:22 | review | Suggested edits | |||
| S Feb 11, 2021 at 2:01 | |||||
| Feb 10, 2021 at 23:44 | comment | added | Sam Hopkins | Should the terms in the first sum be squared? | |
| Feb 10, 2021 at 22:49 | history | asked | T. Amdeberhan | CC BY-SA 4.0 |