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3 votes

A polynomial identity related to Catalan numbers

These assertions can be proved using (formal) generating functions. Using that for $j\geq 0, k\geq 1$ \begin{align*} \sum_{n\geq 0} {n-j+kj \choose kj} t^n &=\frac{t^j}{(1-t)^{kj+1} }\;\;\mbox{ ...
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4 votes

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 $t$?

The formula (which holds for $t>1$ but not for $t=1$), is equivalent to $$\sum_{t=1}^\infty C_{t-1}\bigl(x(1-x)\bigr)^t = x,$$ which follows directly from the generating function $$\sum_{t=1}^\...
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