Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.