Constant term in $\cot^n(x)$

Question

The Laurent expansion for $\cot x$ is $$\cot x =\frac{1}{x} +\sum_{k=1}^\infty \frac{(-1)^k 4^k B_{2k}}{(2k)!}x^{2k-1}$$ where $B_k$ are the Bernoulli numbers. I would like to have a formula for the constant term of $\cot^n x$. I mean something more explicit that writing down the sum of all products of $n$ terms whose exponents add to zero. From numerical experimentation, it seems that the numerators are sequence A216254 and the denominators are A225149 but I have no proof and in any case there are no formulas or recurrence relations on OEIS for those sequences.

Answer 1

This follows from Cauchy's formula after the change of variables $ z = \arctan t$: $$ \frac{1}{{2\pi i}}\oint_{(0 + )} {\cot ^n z\frac{{dz}}{z}} = \frac{1}{{2\pi i}}\oint_{(0 + )} {\frac{1}{{\tan ^n z}}\frac{{dz}}{z}} = \frac{1}{{2\pi i}}\oint_{(0 + )} {\frac{t}{{(t^2 + 1)\arctan t}}\frac{{dt}}{{t^{n+1} }}} . $$ Thus, the constant term in the Laurent expansion of $\cot^n z$ is precisely the $n$th coefficient in the Maclaurin expansion of $$ \frac{z}{{(z^2 + 1)\arctan z}}. $$ Now if we write $$ \frac{z}{{(z^2 + 1)\arctan z}} = \sum\limits_{n = 0}^\infty {a_n z^{2n} } $$ and multiply both sides by $$ \frac{{\arctan z}}{z} = \sum\limits_{n = 0}^\infty {\frac{{( - 1)^{n} }}{{2n + 1}}z^{2n} }, $$ use the geometric series on the left-hand side, and compare like powers of $z$, we obtain the recurrence relation $$ ( - 1)^n = \sum\limits_{k = 0}^n {\frac{{( - 1)^k }}{{2k + 1}}a_{n - k} } \Longleftrightarrow a_n = ( - 1)^n - \sum\limits_{k = 1}^n {\frac{{( - 1)^k }}{{2k + 1}}a_{n - k} } $$ with $a_0=1$.

To obtain an asymptotic formula for the coefficients $a_n$, we can proceed as follows. The singularities of the generating function are located at $z=\pm i$ and $$ \frac{z}{{(z^2 + 1)\arctan z}} \sim - \frac{1}{{(1 \pm iz)\log (1 \pm iz)}} $$ as $z\to \pm i$. Using Theorem VI.2 in Analytic Combinatorics by Flajolet and Sedgewick, we derive $$ a_n \sim ( - 1)^n \frac{2}{{\log (2n + 1)}}\left( {1 - \frac{\gamma }{{\log (2n + 1)}} - \frac{{\pi ^2 - 6\gamma ^2 }}{{6\log^2 (2n + 1)}} + \cdots } \right) $$ as $n\to +\infty$, where $\gamma$ is the Euler–Mascheroni constant.