I claim, for $\operatorname{artanh}(\rho) = \frac{1}{2} \ln\left(\frac{1+\rho}{1-\rho}\right)$, i.e., the inverse hyperbolic tangent function, the following holds **approximately** under assumptions given below:
\begin{align}
y & = 1 + b_1\operatorname{artanh}(\rho) + b_2\operatorname{artanh}(\rho)^2 \\[5pt]
{} & = 1+a_1\rho+a_2\rho^2+a_3\rho^3 +\cdots+a_j\rho^j+ \cdots
\end{align}

for $a_j$ in $[0, 1]$. In other words, the given quadratic may be approximated as the sum of a power series. The power series converges to a non-negative constant, as proven by Goel and Ramalingam (1989, pp. 38-39). 

Specifically, let $(X, Y)$ be a random sample of size $n$ from a bivariate normal population with correlation parameter $\rho$. Let $R(\cdot)$ be a ranking function that takes a real number sequence and replaces each real number with a positive integer indicating its position in the sorted sequence. Then G&R showed the expected number of fixed points between $R(X)$ and its permutation $R(Y)$ is the sum of the above power series.

The polynomial is a regression model fitted to data generated via Monte Carlo simulation. For my simulation parameters $(n = 30$ and $60, \rho = .1, .2, .3, .4, .5, .6, .7, .8, .9)$, fit is near perfect. It can be shown, however, that the approximation weakens for increasing $n$ and fails completely as $n$ goes to infinity, where all coefficients of the power series equal 1. [I assume the asymptotics are responsible for the inconsistencies identified by Carlo Beenakker.]

Now, given that context, my question is: can I relate $(b_1, b_2)$ to $a_j$, so that knowing the one allows me to compute the other? If this is not solvable for the given information, is it solvable in the special case of the power series being a geometric series, so that all $a_j$ are equal? I'm aware there is a relationship between Taylor series and power series that may be exploitable to express a polynomial as a power series, but not sure how or whether it relates here.

[Edit: I have corrected the post to say the expression is approximate and definitely fails under certain conditions. I've also added detail to the context to explain the distributional assumptions.]