An explicit representation for polynomials generated by a power of $x/\sin(x)$ 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.
 A: Faà di Bruno's formula implies that
$$d_k(n) = 
\sum_{m_1,\dots,m_k\geq 0\atop 1\cdot m_1+\cdots+k\cdot m_k=k} \frac{(2k)!}{m_1!\,2!^{m_1}\,m_2!\,4!^{m_2}\,\cdots\,m_k!\,(2k)!^{m_k}}\cdot (n)_{m_1+\cdots+m_k}\cdot \prod_{j=1}^k d_j(1)^{m_j},$$
where $(n)_t=n(n-1)\cdots(n-t+1)$ is the falling factorial.
Correspondingly, the coefficient of $n^\ell$ in $d_k(n)$ is given by
$$\sum_{m_1,\dots,m_k\geq 0\atop 1\cdot m_1+\cdots+k\cdot m_k=k} \frac{(2k)!}{m_1!\,2!^{m_1}\,m_2!\,4!^{m_2}\,\cdots\,m_k!\,(2k)!^{m_k}}\cdot s(m_1+\cdots+m_k,\ell)\cdot \prod_{j=1}^k d_j(1)^{m_j},$$
where $s(z,t)$ are the (signed) Stirling numbers of first kind.
It remains to notice that  and $d_k(1)=(-1)^{k-1}(2^{2k}-2)B_{2k}$.
ADDED. Equivalently, in terms of Bell polynomials $\mathcal{B}_{n,k}()$, we have
$$[n^\ell]\ d_k(n) = \sum_{m=1}^k s(m,\ell)\cdot \mathcal{B}_{2k,m}(0,d_1(1),0,d_2(1),\dots).$$
A: 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*}
A: Experiment suggests that 
$$ d_k(k)= \prod_{i=1}^k \frac{2i-1}{3}$$
whereas $(4^k-1)d_k(1)$ gives the sequence A000182.
