TITLE: Could these polynomials be identified?
QUESTION [27 upvotes]: Looking a good approximation of the $n^{th}$ positive root of the equation
$$\color{blue}{\tan(x)=k x}$$ As already done many times, I expanded as Taylor series around $x=(2n+1)\frac \pi 2$ and used series reversion.
As a result, this write
$$\color{blue}{x=q-\sum _{m=1}^{\infty }\frac{P_{m}(k) }{a_m } \frac 1 {(kq)^{2 m-1}}}\qquad \text{where}\qquad \color{blue}{q=(2n+1)\frac \pi 2}$$
Concerning the $a_m$, the first terms are $\{1,3,15,105,315,3465\}$; this is sequence $A088989$ in $OEIS$ and they nicely correspond to the denominators of the coefficients of odd powers of $\frac 1q$ in the solution series of the equation for $k=1$.
Concerning the polynomials $P_{m}(k)$, the first are listed below
$$\left(
\begin{array}{cc}
m & P_{m}(k) \\
 1 & 1 \\
 2 & 3 k-1 \\
 3 & 30 k^2-20 k+3 \\
 4 & 525 k^3-525 k^2+161 k-15 \\
 5 & 4410 k^4-5880 k^3+2744 k^2-528 k+35 \\
 6 & 145530 k^5-242550 k^4+152460 k^3-44990 k^2+6193 k-315
\end{array}
\right)$$
Even truncated to six terms, this provides quite good estimates even for the first solution
$$\left(
\begin{array}{ccc}
 k & \text{estimate} & \text{solution} \\
 0.5 & 4.27478222835695 &  4.27478227145813 \\
 1.0 & 4.49340947613302 &  4.49340945790907 \\
 1.5 & 4.56745216948297 &  4.56745216641733 \\
 2.0 & 4.60421677783123 &  4.60421677720058 \\
 2.5 & 4.62613829114934 &  4.62613829098161 \\
 3.0 & 4.64068363082997 &  4.64068363077555 \\
 3.5 & 4.65103583026074 &  4.65103583024022 \\
 4.0 & 4.65877826296638 &  4.65877826295768 \\
 4.5 & 4.66478672978692 &  4.66478672978287 \\
 5.0 & 4.66958478090799 &  4.66958478090596
\end{array}
\right)$$

Could the above polynomials be identified ?

REPLY [20 votes]: If instead of using $a_m = \mathrm{lcm} \{ 1, 3, \dotsc, 2m-1 \}$ you use $b_m = (2m-1)!!$, you can write
$$x=q-\sum _{m=1}^{\infty }\frac{Q_{m}(k) }{b_m } \frac 1 {(kq)^{2 m-1}}$$
and then
$$Q_{m+1}(k) = \sum_{i=0}^m (-1)^{m-i} C(m, i) k^i$$
where the coefficients $C(m, i)$ satisfy $C(0, i) = (2i-1)!!$ and
$$C(m, i) = \frac {(2m-1)2m(2m+1)} {(2m-i)(2m-i+1)} [C(m-1, i-1) + C(m-1, i)]$$
For a reference, see L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 170.