TITLE: An explicit representation for polynomials generated by a power of $x/\sin(x)$
QUESTION [9 upvotes]: The coefficients $d_{k}(n)$ given by the power series 
$$\left(\frac{x}{\sin x}\right)^{n}=\sum_{k=0}^{\infty}d_{k}(n)\frac{x^{2k}}{(2k)!}$$
are polynomials in $n$ of degree $k$. First few examples:
$$d_{0}(n)=1,\quad d_{1}(n)=\frac{n}{3}, \quad d_{2}(n)=\frac{2 n}{15}+\frac{n^2}{3}, \quad d_{3}(n)=\frac{16n}{63}+\frac{2 n^{2}}{3}+\frac{5n^3}{9}.$$
Question: Is there an explicit formula for the coefficients of polynomials $d_{k}(n)$?
Remark: I am aware of their connection with the Bernoulli polynomials of higher order $B_{n}^{(a)}(x)$. Namely, one has $d_{k}(n)=(-4)^{k}B_{2k}^{(n)}(n/2)$. This formula and several other alternative expressions can be found in the book of N. E. Norlund (Springer, 1924) but none of them seems to be very helpful.

REPLY [2 votes]: This is really a comment rather than an answer, but it's too long for a comment.
Here's a simple proof that the coefficients of the polynomials $d_k(n)$ are positive.
We have
$$\left(\frac{x}{\sin x}\right)^n =
  \exp\left(n\log\left(\frac{x}{\sin x}\right)\right)$$
so the coefficient of $n^l$ in $d_k(n)$ is 
the coefficient of $x^{2k}/(2k)!$ in 
$\bigl(\log(x/\sin x)\bigr)^l/l!$. So it suffices to show that 
$\log(x/\sin x)$ has positive coefficients. But
$$\frac{d\ }{dx} \log \left(\frac{x}{\sin x}\right) = \frac{1}{x} -\cot x$$
in which the coefficients are easily seen to be positive by expressing  the 
coefficients of $\cot x$ in terms of Bernoulli numbers:
\begin{align*}\cot x&= \sum_{n=0}^\infty (-1)^n 2^{2n} B_{2n} \frac{x^{2n-1}}{(2n)!}\\
  &=\frac{1}{x}- \sum_{n=1}^\infty  2^{2n} |B_{2n}| \frac{x^{2n-1}}{(2n)!}.
\end{align*}