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What would be an asymptotic (big-O) notation for ${n \choose k}$ which puts it into perspective with $2^n$?

EDIT: for $k$ constant, this post states ${n \choose k} = \Theta(n^k)$. I'm searching for a more general asymptotic notation, not limited to $k$ constant.

EDIT 2: and for $k = \frac{n}{2}$, this post obtains ${n \choose k} = \Theta(\frac{2^n}{\sqrt{n}})$.

J. Schmidt
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    Note $\binom{n}{k}\le\binom{n}{\lfloor n/2\rfloor}\sim\frac{2^n}{\sqrt{n\pi/2}}$ by Stirling. – J.G. Nov 22 '22 at 11:33
  • @J.G. this implies that ${n \choose k} = O(\frac{2^n}{\sqrt{n}}$ which is a nice bound I was looking for! However, we lose all information on $k$ which is unfortunate. – J. Schmidt Nov 22 '22 at 12:56

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Two bounds from information theory might be helpful. If you let the function $H(p) = -p\log_2 p-(1-p)\log_2(1-p)$ denote the binary entropy function for $0\leq p\leq 1$, then you have

$$\frac{1}{n+1}2^{n H(k/n)}\leq \binom{n}{k}\leq 2^{n H(k/n)}.$$

It can be further tightened, but the connection with $2^n$ is already there.

adrien_vdb
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If you consider$${n \choose k}={n \choose a n}=\frac{\Gamma (n+1)}{\Gamma ((1-a) n+1)\, \Gamma (a n+1)}$$ and use Stirling approximation $$\log \Bigg[\binom{n}{a n}\Bigg]=-((1-a) \log (1-a)+a \log (a))\,n-$$ $$\frac 12\log (2 \pi (1-a) a)-\frac 12 \log(n)+O\left(\frac{1}{n}\right)$$

If you make $a=\frac 12$, the leading coefficient is $\log(2)$ and then the $2^n$

Claude Leibovici
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