# Binomial transform of a scaled version of the catalan numbers.

I was looking at the mathworld entry for Catalan Numbers http://mathworld.wolfram.com/CatalanNumber.html and was surprised to find formula (11) there:

(1) $C_n= \sum_{k=0}^n (-1)^k 2^{n-k}\binom{n}{k}\binom{k}{\lfloor k/2\rfloor}$

where $C_n=\frac{1}{n+1}\binom{2n}{n}$ is the $n$-th Catalan number.

I tried to prove that formula but failed. One thing that I tried is the following. Define $D_n = \binom{n}{\lfloor n/2 \rfloor}$, then formula (1) states that $(C_n/2^n)_n$ is the (alternating) binomial transform of $(D_k/2^k)_k$. Since the alternating binomial transform is self-inverse. Proving (1) is equivalent to proving the following

(2) $\binom{n}{\lfloor n/2 \rfloor}=D_n = \sum_{k=0}^n (-1)^k 2^{n-k} \binom{n}{k}C_k$

I haven't been able to prove this formula either.

Could anyone help me find a proof for them?

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