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Trying to prove some uncorrelated things, I came across the following identity: $$\binom{m}{n}=\sum_{k=0}^{\lfloor n/2 \rfloor} 2^{1-\delta_{k,n-k}} \binom{m/2}{k} \binom{m/2}{n-k}, $$ where $\delta_{i,j}$ is the Kronecker delta, equal to 1 if $i=j$ and vanishing otherwise. This identity seems to hold for every $m$ and $n$ (I checked it with Mathematica for each pair of integers $n, m$ from 1 up to 100). I've never seen such an identity, and it doesn't seem straightforward to prove.

Is this some known identity? And how could I go in proving (or disproving) it?

Pichi Wuana
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glS
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2 Answers2

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It's true. The number of ways we can pick $n$ things from $m$ is:

Divide the $m$ things into two half-sized chunks. Then we need to do one of the following:

  • pick at most $n/2$ things out of the first chunk (say we pick $k$ from the first chunk), and the remaining $n-k$ things out of the second chunk;
  • pick at most $n/2$ things out of the second chunk, and the remaining $n-k$ things out of the first chunk.

By symmetry, we take care of the second case by simply doubling all terms from the first case for which $k \not = n-k$.

Patrick Stevens
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Vandermonde Identity gives $$\sum_{k=0}^nu_k=\sum_{k=0}^n\binom {m/2}k\binom {m/2}{n-k}=\binom mn\qquad\qquad (1) $$ If $n$ is even:
then number of terms is odd, with the middle term being the $n/2$-th, with symmetrical terms on both sides of this, as $k, n-k$ are symmetrical about $n/2$, i.e. $$\begin{align} u_0&=u_n\\ u_1&=u_{n-1}\\ u_2&=u_{n-2}\\ &\vdots\\ u_{\frac n2-1}&=u_{\frac n2+1}\\ &u_{\frac n2} \end{align}$$ Hence

$$\sum_{k=0}^n\binom {m/2}k\binom {m/2}{n-k}=\left[2^\color{lightgrey}{1-0}\sum_{k=0}^{\frac n2-1}\binom {m/2}k\binom {m/2}{n-k}\right]+\color{lightgrey}{2^{1-1}}\binom {m/2}{n/2}\binom{m/2}{n/2} \qquad\qquad(2)$$

If $n$ is odd:
then number of terms is even and there is no middle term as $n/2$ is not an integer, i.e. $$\begin{align} u_0&=u_n\\ u_1&=u_{n-1}\\ u_2&=u_{n-2}\\ &\vdots\\ u_{ \big\lfloor \frac n2 \big\rfloor}&=u_{\big\lfloor \frac n2\big\rfloor +1} \end{align}$$

Hence

$$\sum_{k=0}^n\binom {m/2}k\binom {m/2}{n-k} =2^\color{lightgrey}{1-0}\sum_{k=0}^{\big\lfloor \frac n2 \big\rfloor}\binom {m/2}k\binom {m/2}{n-k} \qquad\qquad(3) $$

To cater for both odd and even $n$, the RHS of equations $(2),(3)$ can be expressed as

$$\sum_{k=0}^{\lfloor n/2 \rfloor} 2^{1-\delta_{k,n-k}} \binom{m/2}{k} \binom{m/2}{n-k}$$

As this is equal to $(1)$, we have

$$\binom{m}{n}=\sum_{k=0}^{\lfloor n/2 \rfloor} 2^{1-\delta_{k,n-k}} \binom{m/2}{k} \binom{m/2}{n-k}\qquad\blacksquare$$

Hypergeometricx
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