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 $\tan x$.
Best Answer
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 $k^{\text{th}}$ 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 $n^{\text{th}}$ 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 $1$.