Justification behind this heuristic in SMA*+

I'm reading the paper Enhanced Simplified Memory-bounded A Star (SMA*+). In short, the authors propose a limited-memory version of A* where we cull nodes if we are out of memory, to re-expand them later on.

To do this they choose the worst node to cull. And they propose a heuristic to maximize to choose what is considered the worst node:

For example, a culling heuristic based on the ratio between f-cost and depth, given as $c(n) = f(n) / \ln(d(n) + e)$, where $e$ is Euler’s number, can improve performance on problems with many similar f-costs, if we can reasonably assume that the goal has a high depth.

A quick reminder, $f(n) = g(n) + h(n)$ where $g(n)$ is the cost to reach node $n$ and $h(n)$ is the heuristic to estimate the distance from $n$ to the goal node. $d(n)$ is the depth of the node.

The authors don't further elaborate on this. Is there some logical justification I'm missing for the formula $c(n)$, in particular the natural logarithm and adding $e$ to $d(n)$?

From what I can tell the function $c(n)$ was just chosen as a reasonably simple trade-off between f-cost and depth.
It appears that $e$ is added to simply have the denominator $\ln(d(n) + e)$ always be $\geq 1$ and prevent $\ln(0)$.