Timeline for Why is this alternating sum involving Catalan numbers $\sum_{i=0}^{\lfloor t/2 \rfloor} (-1)^{i+1} \binom{t-i}{i} C_{t-i-1} = 0$ for all $t$?
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
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| Jan 22, 2023 at 17:24 | comment | added | Alexander Burstein | Denoting the second sum as $xC$, where $C=C(x)=1+xC^2$, another way to put it is that $(xC)\circ(x-x^2)=x$ because $(x-x^2)\circ(xC)=xC-x^2C^2=x$. | |
| Jan 19, 2023 at 19:22 | comment | added | Aaron Li | Ah, thank you! My colleague had a very nice combinatorial argument, but this is much more straightforward. | |
| Jan 19, 2023 at 19:22 | vote | accept | Aaron Li | ||
| Jan 18, 2023 at 4:35 | history | answered | Ira Gessel | CC BY-SA 4.0 |