Maclaurin series of $\frac{x^2}{1- x \cot x}$ I wonder if there is an explicit formula for the Maclaurin expansion of $\frac{x^2}{1 - x \cot x}$. We know an explicit formula for $1- x \cot x$.
Due to the continued fraction formula for $\tan x$, we know that all of the coefficients after the first are negative.
$\bf{Added:}$ I was lead to this question in trying to prove that in the Maclaurin expansion of $ \frac{x^2}{1 - x \cot x} + \frac{3}{5} ( 1 - x \cot x) - 2 $ all of the coefficients are positive. Since we have formulas for the expansion of the second term, we are interested in explicit formulas for the expansion of the first term.
$\bf{Added:}$ We have the continued fraction
$$\frac{x^2}{1 - x \cot x} = 3 - \frac{x^2}{ 5 - \frac{x^2}{ 7 - \cdots} } $$
following easily from the continued fraction for $\tan x$.

Using Bessel functions, we find
\begin{align*}
\frac{{x^3 }}{{1 - x\cot x}} & = 3 - x\frac{{J_{5/2} (x)}}{{J_{3/2} (x)}} = \frac{3}{2} + x\frac{{J'_{3/2} (x)}}{{J_{3/2} (x)}} = 3 - 2x^2 \sum\limits_{k = 1}^\infty  {\frac{1}{{1 - (x/j_{3/2,k}^2 )^2 }}} 
\\ & = 3 - 2x^2 \sum\limits_{k = 1}^\infty  {\sum\limits_{n = 0}^\infty  {\frac{1}{{j_{3/2,k}^{2n} }}x^{2n} } }  = 3 - 2x^2 \sum\limits_{n = 0}^\infty  {\left( {\sum\limits_{k = 1}^\infty  {\frac{1}{{j_{3/2,k}^{2n} }}} } \right)x^{2n} } \\& = 3 - 2x^2 \sum\limits_{n = 0}^\infty  {\sigma _{2n} \!\left( \tfrac{3}{2} \right)x^{2n} }, 
\end{align*}
where $j_{3/2,k}$ denotes the $k$th positive zero of $J_{3/2}$ and $\sigma_n$ is the Rayleigh function of order $n$. If $n\geq 1$, we can write
$$
\sigma _{2n} \!\left( {\tfrac{3}{2}} \right) = ( - 1)^{n - 1} 3 \cdot 2^{2n - 1} \frac{{V_{2n} }}{{(2n)!}},
$$
where $V_n$ is the $n$th van der Pol number. See this paper for properties of these numbers, including recurrence relations. In particular, your generating function is equation $(d)$ in Section $1$.
