<s>[INST] How to find the sum $\sum_{k=1}^{\lfloor n/2\rfloor}\frac{2^{n-2k}\binom{n-2}{2k-2}\binom{2k-2}{k-1}}{k}$ Let $n$ be positive integer, find the value
$$f(n)=\sum_{k=1}^{\lfloor n/2 \rfloor}\dfrac{2^{n-2k}\binom{n-2}{2k-2}\binom{2k-2}{k-1}}{k}. $$
I have found
$$ f(2)=1, \quad f(3)=2, \quad f(4)=5, \quad f(5)=14, \quad f(6)=42, \quad f(7)=132. $$
It seem OEIS (A000108):
$$ f(n)=\dfrac{1}{n}\binom{2n-2}{n-1},$$
but how to prove it?
 [/INST] We have the integral formula  for the Catalan numbers $C_k = \frac{\binom{2k}{k}}{k+1}$ :
$$\ \ \frac{C_k}{2^{2k+1}}=\frac{1}{\pi}\int_{-1}^1 t^{2k}\sqrt{1-t^2} dt \ \ (*)$$
Note also that
$0 = \frac{1}{\pi} \int_{-1}^{1} x^{2k+1} \sqrt{1-x^2} dx$.  Therefore we get
$$(**) \ \ \sum_{k} \binom{n}{2k} \frac{C_{k}}{2^{2k+1}} = \frac{1}{\pi} \int_{-1}^1 (\sum_{k=0}^n \binom{n}{k} x^k) \sqrt{1-x^2} dx =\\ =\frac{1}{\pi} \int_{-1}^1 (1+x)^n \sqrt{1-x^2} dx$$
Now use another formula for the Catalan numbers
$$ C_n = \frac{2^n}{\pi} \int_{-1}^1 (1+x)^{n-1} \sqrt{1-x^2} d x\ \ (***)$$
We are done.
Another possible approach: use the binomial transform for the sequence $\frac{C_{k}}{2^{2k+1}}$ at even index $2k$ and $0$ otherwise. We have the generating function
$$f(x)= \sum_{k\ge 0} \frac{C_{k}}{2^{2k+1}} x^{2k} = \frac{1-\sqrt{1-x^2}}{x^2}$$
Now use that the generating function for the transformed sequence equals $g(x) = \frac{1}{1-x} \cdot f(\pm \frac{x}{1-x})$ ($f$ even so both $\pm$ are OK).  We get
$$g(x) = \frac{1- \sqrt{1- 2 x} - x}{x^2}= \sum_{n\ge 0}\frac {C_{n+1}}{2^{n+1}} x^n$$
Conclusion
$$(**) = \frac{C_{n+1}}{2^{n+1}}$$
$\bf{Added:}$
The formulas $(*),(***)$ are equivalent to the representation of Catalan number as moments of a distribution on $[0,4]$.
$$C_n = \frac{1}{2\pi} \int_{0}^4 t^n  \sqrt{\frac{4-t}{t}} \, dt$$
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