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Given that: $$n={k^2 \choose k}$$ what is $k$ as a function of $n$?

So far, I have found the following approximation:

$$ n \approx (k^2)^k = (k^k)^2 $$ $$ \sqrt{n} \approx k^k $$

If we take this approximation as an equality, then according to this answer:

$$ k = {\ln{\sqrt{n}} \over W(\ln{\sqrt{n}})} $$ where $W$ is Lambert's function.

Asymptotically, $W(z) = \Theta(\ln z)$, so we get:

$$ k = \Theta\left({\ln n \over \ln \ln n}\right) $$

Is this approximation correct (asymptotically)? Is there a better approximation?

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Erel Segal-Halevi
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    You can arrive at the same approximation more easily from $n \approx k^k$ (follows from Stirling approximation), so it seems to be correct in that sense. – Sil Apr 10 '16 at 08:24

1 Answers1

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Let us start from basics (or "basics" with two huge quotation marks): $\binom{k^2}{k} = \frac{(k^2)!}{k!(k^2-k)!}$, which allows you to directly invoke (and enjoy) the use of Stirling's approximation. Specifically, it is easy to see that $n\to \infty$ implies $k\to\infty$. This will get you the following (slightly more general statement):

For any fixed $a> 0$, $$ \binom{ak^2}{k} = c_a e^{k\ln k + k\ln(ae) - \frac{1}{2}\ln k + o(1)}, $$ with $c_a \stackrel{\rm def}{=} \frac{e^{-\frac{1}{2a}}}{2\pi}$.

In your case, $a=1$, so $$ \binom{k^2}{k} = \frac{e^{-\frac{1}{2}}}{2\pi} e^{k\ln k + k - \frac{1}{2}\ln k + o(1)} = e^{k\ln k + k - \frac{1}{2}\ln k + O(1)}. $$

From there, you can try to get lower-order terms, but note that first this implies that $ e^{\ln n} = e^{(1+o(1))k\ln k} $ and then $$ \ln n = (1+o(1))k\ln k $$ from which you get that $k= \Theta(\frac{\ln n}{\ln\ln n})$.

Clement C.
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