and you are interested in the "coalescence time" $T$ of this process, i.e. the smallest $t$ for which the composition $f_1\circ\ldots\circ f_t$ of independent (uniform) $N$ to $N$ mappings becomes constant.

View the results of the $N$ throws of the first round as value table $(f_1(1),\ldots,f_1(N))$ of a random mapping $f_1$. Imagine that you write the pairs $(i,f_1(i))$ on the faces of the unlabeled die (instead of just the values $f_1(i)$, as you do). Then the results of N throws of the second round may be viewed as producing the value table $\big((f_2(1),f_1(f_2(1)),\ldots,(f_2(N),f_1(f_2(N))\big)$, (where $f_2$ is a random mapping, and independent of $f_1$), etc,

(I think) this is the simplest (discrete) case of Kingman's coalescence. See .e.g arxiv.org/abs/0809.4233 and the explanations and refernces there.

Another routine proof: observe that ${1 \over \sqrt{1-4x}}=(x\,C(x))^\prime $, and use Bürmann-Lagrange.

What kind of results are you hoping for? In the classic case (I think) the quotient $Q_n:={R_n \over G_n}$ tends a.s to a random variable $Q$ which is distributed as ${Z \over 1-Z}$, with $Z$ being $\mathrm{Beta}(r,g)$ distributed - how do you interpret "tight concentration" here?

Note also that exercise 667 in Whitworth's DDC Exercises precedes Fisher by roughly 30 years.

At the least you should answer your other positivity question (or add a link to the answer there).

For $j=0$ the equality $\sum_{n\geq 0}n^j t^j=\frac{t A_j(t)}{(1-t)^{j+1}}$ uses the convention $0^0:=0$. With $0^0:=1$ you arrive at Fedor Petrov's result.

You can find some results here: projecteuclid.org/euclid.pjm/1102723961

$f$ is just an auxiliary polynomial used to determine the coefficients $\alpha_1,\ldots,\alpha_n$. For $k=0,\ldots,n-1$ $$\big((t\frac{d}{dt})^k f\big) (1)= (-1)^k\frac{(n-1)!}{(n-1-k)!}\big((\frac{d}{dg})^k P\big)(0)=0$$.