# 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.

• What sort of formula are you looking for? There might be a formula as a double sum, but probably nothing simpler. Would that do any good? – Ira Gessel Aug 23 '17 at 6:44
• What is wrong with $B_n^{(a)}$? If this is not explicit enough for you, then you have to explain in more detail what exactly you want. Otherwise, as others already suggested, this is too vague a question to answer. – Alex Gavrilov Aug 23 '17 at 10:01
• @Ira Gessel: My motivation comes from an asymptotic expansion of a certain special function where these polynomials appear (in more complicated way, unfortunately). The desired thing is to be able to deduce, for example, sign or monotonicity of those complicated expressions. As a first phase, I try to have a better intuition of $d_{k}(n)$ as function of $n$. I was able to collect some of their properties mainly from Norlund's book. But a sort of an explicit representation of these polynomials is missing (if there is any). A double sum would be much better then $k$ sums. – Twi Aug 23 '17 at 13:00
• @Alex Gavrilov: Nothing is wrong with Norlund's polynomials. It's just a different notation; see also the comment to Ira. – Twi Aug 23 '17 at 13:03

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

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*}

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.