I'm reading this paper which says

Let $d=d(n)$ be the positive real number for which

$$ \binom n d p^{\binom d 2} = 1 $$

where $ 0 < p \le 1$, then

$$ d(n) = 2 \log_bn - 2 \log_b (\log_b n) + 2 \log_b\left(\frac 1 2 e\right) + 1 + O(1) \\= 2\log_bn+ O( \log_b \log_b n) $$

where $b = \frac 1 p$.

As the authors skimmed the proof, I've completely no idea how they reached the conclusion.

  • 1
    $\begingroup$ I suppose you could try to use stirling and then just plug in $d$ and see what happens? $\endgroup$ – Maximilian Janisch Jun 30 at 22:07
  • $\begingroup$ @MaximilianJanisch what you said probably works, but I think OP might want a motivated proof. $\endgroup$ – mathworker21 Jul 1 at 0:34

Setting $b=1/p$, the equation can be written as




Applying the Stirling approximation, the LHS becomes

$$\log_b \left(\sqrt{\frac{n}{2\pi d(n-d)}} \frac{ n^n}{d^d(n-d)^{n-d}}\right)\\ =\left(n+\frac{1}{2}\right)\log_b n- \left(d+\frac{1}{2}\right)\log_b d-\left(n-d+\frac{1}{2}\right)\log_b(n-d)+O(1) $$

We can consider the expansion for $n=\infty$ of the term $\log_b(n-d)$, which is

$$\log_b n-\frac{d}{n\log b}+O(n^{-2})$$

Substituting we get

$${ d}\,{\log_b n} +\frac{d}{\log b} -\frac{d^2}{n \log b} -\left(d+\frac{1}{2}\right)\log_b d +O(1) $$

Since this quantity has to be equal to $d(d-1)/2$, grouping the terms according to $d$ we have

$$d^2\left(\frac{1}{2}+\frac{1}{n\log b }\right)-d\left(\log_b n+\frac{1}{2}+\frac{1}{\log b} \right) +\left(d+\frac{1}{2}\right)\log_b d =O(1)$$

From this equation, we get that, as $n \rightarrow \infty$, the main term of the asymptotic expansion of $d$ is $ 2\log_b n$.

To determine the successive term of the expansion, we can set $d=2\log_b n +f(n)$. Taking the last equation, dividing by $d$, and making the substitution we get

$$\bigg[2\log_b n+f(n)\bigg]\left(\frac{1}{2}+\frac{1}{n\log b }\right)- \left(\log_b n+\frac{1}{2}+\frac{1}{\log b } \right) +\left(1+\frac{1}{2[2\log_b n +f(n)]}\right) \\ \log_b \bigg[2\log_b n+f(n)\bigg] =O(1)$$

Taking into account that the order of $f(n)$ is lower than $\log n$, incorporating some terms into the $O(1)$ error, the last equation becomes

$$\frac{1}{2}\left[2\log_b n+{f(n)}\right]- \log_b n + \log_b [2\log_b n+f(n)] =O(1)$$

from which

$$f(n)=-2\log_b (\log_b n) +O(1)$$

To further go into the expansion, we can set

$$d=2\log_b n -2\log_b (\log_b n ) +g(n)$$

Proceding as above by making the substitution, taking into account that the order of $g(n)$ is $O(1)$, and incorporating some terms into a residual $O(1)$ term, we get

$$\frac{1}{2}\left[2\log_b n -2\log_b (\log_b n)+{g(n)}\right]\\- \log_b n - \frac{1}{2} -\frac{1}{\log b}+\log_b 2\\+ \log_b [\log_b n+2\log_b(\log_b n) +f(n)] =O(1)$$

from which

$$g(n)=1+ \frac{2}{\log b}-2 \left(\log_b 2 \right) +O(1)\\ = 1+ 2\log_b e +2 \left(\log_b \frac{1}{2}\right) +O(1) \\ =1+ 2\log_b \left(\frac{1}{2}e \right) +O(1)$$

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  • 2
    $\begingroup$ Nice answer. But the exponent in the OP's question is $d\choose2$, not $n\choose2$, which changes your starting equation. Maybe it's just a typo in your answer, though. $\endgroup$ – cjferes Jul 3 at 22:27
  • 2
    $\begingroup$ Thank you for your note. I just edited my answer. $\endgroup$ – Anatoly Jul 3 at 22:32
  • $\begingroup$ Thank you for the clear and understandable answer! $\endgroup$ – UmbQbify -Key20- Jul 4 at 10:57

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