How to show $p^3 \approx e^{-p{k \choose 2}}$ implies $p \approx \frac{12\ln(k)}{k^2}$? Just a step in a proof I'm reading that I don't get. Answer or hint would be appreciated.  
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Edit: This is in the context of an application of the Lovász Local Lemma to find a bound on the Ramsey number $R(3,k)$.


 A: 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
$$p \sim {12 \over k^2} \ln k$$
A: 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}
$
