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Let $$C_n=\frac{1}{2n+1}\binom{2n+1}{n}$$ be a Catalan number. It is well-known that $$(\sum_{n\ge{0}}C_n x^n)^k=\sum_{n\ge{0}}C(n,k)x^n$$ with $$C(n,k)=\frac{k}{2n+k}\binom{2n+k}{n}.$$ It is also ...