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We know from the extended binomial theorem that the OGF corresponding to the concise expression is $\sum_{k \geqslant 0}{1/2 \choose k}(-8x)^k$. And we need to find the coefficient $x^n$, which is ${1/2 \choose n}(-8)^n$. Then,

\begin{align} {1/2 \choose n}(-8)^n &= \frac{(\frac{1}{2})(\frac{1}{2} - 1)(\frac{1}{2} - 2)\cdots(\frac{1}{2} - n + 1)(-8)^n}{n!}\\ &= \frac{(\frac{1}{2})(-\frac{1}{2})(-\frac{3}{2})\cdots(\frac{3 - 2n}{2})(-8)^n}{n!}\\ &= \frac{(-1)(1)(3)\cdots(2n - 3)(-1)^n(-8)^n}{2^nn!}\\ &= -\frac{(1)(3)(5)\cdots(2n - 3)8^n}{2^nn!}\\ &= -\frac{(1)(3)(5)\cdots(2n - 3)4^n}{n!}\\ &= -\frac{(1)(3)(5)\cdots(2n - 3)\cdot 4^n}{n!} \cdot \frac{(2)(4)(6)\cdots(2n - 2)}{(2)(4)(6)\cdots(2n - 2)}\\ &= -\frac{(1)(2)(3)(4)(5)\cdots(2n - 3)(2n - 2)\cdot 4^n}{n!\cdot(2)(4)(6)\cdots(2n - 2)}\\ &= -\frac{(2n - 2)! \cdot 2^n \cdot 2^n}{n! \cdot 2^{n - 1} \cdot (1)(2)(3)\cdots(n - 1)}\\ &= -\frac{(2n - 2)! \cdot 2^{n + 1}}{n! \cdot (n - 1)!}\\\ \end{align}

And then I've hit several dead ends. Any hints? Have I done something wrong to get to where I am?

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You already got it. From Catalan numbers we know that: $$ \sum_{n\geq 0}\frac{z^n}{n+1}\binom{2n}{n} = \frac{1-\sqrt{1-4z}}{2z} \tag{1}$$ hence: $$ \sum_{n\geq 0}\frac{2^{n+2}\,z^{n+1}}{n+1}\binom{2n}{n} = 1-\sqrt{1-8z}\tag{2}$$ and for every $n\geq 2$, $$ [x^n]\sqrt{1-8x} = -\frac{2^{n+1}}{n}\binom{2n-2}{n-1}. \tag{3}$$

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Your work and the resulting expression is accurate. The only thing left is to perhaps write it as $$-\binom{2n-1}{n} \frac{2^{n+1}}{2n-1},$$ or something of that sort.

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