How to show $p^3 \approx e^{-p{k \choose 2}}$ implies $p \approx \frac{12\ln(k)}{k^2}$?

Question

Just a step in a proof I'm reading that I don't get. Answer or hint would be appreciated.

======
Edit: This is in the context of an application of the Lovász Local Lemma to find a bound on the Ramsey number $R(3,k)$.

Answer 1

Here's a more elementary way, without Lambert the sheepish function.

If $p^3 \approx e^{-p{k \choose 2}} $ then $3\ln(p) \approx -p{k \choose 2} =-pk(k-1)/2 $.

Letting $r = 1/p$, $3\ln(r) =k(k-1)/(2r) $ so $r\ln(r) \approx k(k-1)/6 $.

Since the inverse of $r\ln(r)=x$ is about $r = x/\ln(x) $,

$\begin{array}\\ r &\approx (k(k-1)/6)/\ln(k(k-1)/6)\\ &\approx (k^2/6)/\ln(k^2/6)\\ &\approx k^2/(6(\ln(k^2)-\ln(6)))\\ &\approx k^2/(12\ln(k))\\ \text{so}\\ p &\approx 12\ln(k)/k^2\\ \end{array} $

Answer 2

This can be seen as follows:

Suppose $p^3 = Ce^{-p{k(k-1) \over 2}}$ for some value of $C$ which is specified to be in a bounded range. Then taking cube roots of both sides we get the following, where $C_1 = C^{1 \over 3}$. $$p = C_1 e^{-p{k(k-1) \over 6}}$$ This gives $$p {k(k-1) \over 6} = {k(k-1) \over 6}C_1 e^{-p{k(k-1) \over 6}}$$ We rewrite this as $$p {k(k-1) \over 6}e^{p{k(k-1) \over 6}} = C_1{k(k-1) \over 6}$$ In terms of the Lambert $W$ function, we have $$p{k(k-1) \over 6} = W\bigg(C_1{k(k-1) \over 6}\bigg)$$ This is solved for $p$ by $$p = \bigg({k(k-1) \over 6}\bigg)^{-1} W\bigg(C_1{k(k-1) \over 6}\bigg)$$ Since the main term of $W(x)$ as $x \rightarrow \infty$ is $\ln x$, we get the asymptotics $$p \sim \bigg({k(k-1) \over 6}\bigg)^{-1}\ln\bigg({k(k-1) \over 6}\bigg)$$ Since $k \sim k - 1$ for large $k$, this simplifies to $$p \sim {12 \over k^2} \ln\bigg({k \over 6}\bigg)$$ Since $\ln k \sim \ln k - \ln 6 = \ln({k \over 6})$, this then simplifies to $$p \sim {12 \over k^2} \ln k$$