An explicit representation for polynomials generated by a power of $x/\sin(x)$

Question

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.

Answer 1

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 $$[n^\ell]\ d_k(n) = \sum_{m=1}^k s(m,\ell)\cdot \mathcal{B}_{2k,m}(0,d_1(1),0,d_2(1),\dots).$$

Answer 2

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*}

Answer 3

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.