Is there a closed expression for this one? Has anyone ever seen a closed expression (and its sequence) for the following one:
$$\partial_x^{n} \, \text{arctanh}^k(x)|_{x=0}\,\,\,\,(1)$$
Provided $k,n \in \mathbb{N_0}$ and $x \in \mathbb{R}$.
When $k=1$, things become easier:
$$d_x^{n} \, \displaystyle\sum_{n=0}^\infty \frac{x^{2n+1}}{2n+1}|_{x=0}, |x| < 1 $$
When $k=1$, it gives the following sequence:
$1,0,2,0,24,0,720,0,40320,0, \ldots$
Which clearly is:
$0!, 0,2!,0,4!,0,6!,0,8!,0 \ldots$
However, when $k=2$:
$0,0,2,0,16,0,368,0,16896,0,1297152, \ldots$
Which becomes hard to find a pattern.
But, for any natural $k$, is there another way to rewrite (1) and then find its sequence? Any suggestion on how to proceed? I don't think raising the McLaurin series to the $k$-th power and then expanding it is a good idea.
Thanks
 A: First approach. If $k>n$, the answer is $0$ since the Taylor series about $0$ begins with $x^k$. Suppose that $n\geq k$. By the Cauchy formula, the quantity in question is
\begin{align*}
\frac{{n!}}{{2\pi i}}\oint_{(0 + )} {\frac{{(\tanh ^{ - 1} z)^k }}{{z^{n + 1} }}dz}&  = \frac{{n!}}{{2\pi i}}\oint_{(0 + )} {\frac{{s^k }}{{(\tanh s)^{n + 1} }}\frac{1}{{\cosh ^2 s}}ds} 
\\ & = \frac{{n!}}{{2\pi i}}\oint_{(0 + )} {\left( {\frac{{2s}}{{e^{2s}  - 1}}} \right)^{n + 1} e^{2s} \left( {\frac{{e^{2s}  + 1}}{2}} \right)^{n - 1} \frac{{ds}}{{s^{n - k + 1} }}} .
\end{align*}
Now, in terms of the generalised Bernoulli- and Euler polynomials,
$$
\left( {\frac{{2s}}{{e^{2s}  - 1}}} \right)^{n + 1} e^{2s}  = \sum\limits_{j = 0}^\infty  {2^j B_j^{(n + 1)} (1)\frac{{s^j }}{{j!}}} 
$$
and
$$
\left( {\frac{{e^{2s}  + 1}}{2}} \right)^{n - 1}  = \sum\limits_{j = 0}^\infty  {2^j E_j^{(1 - n)} (0)\frac{{s^j }}{{j!}}} ,
$$
near $s=0$. Thus, our integral becomes
$$
\frac{{n!}}{{2\pi i}}\sum\limits_{j = 0}^\infty  {\left[ {\frac{{2^j }}{{j!}}\sum\limits_{m = 0}^j {\binom{j}{m}B_m^{(n + 1)} (1)E_{j - m}^{(1 - n)} (0)} } \right]\oint_{(0 + )} {\frac{{ds}}{{s^{n - k - j + 1} }}} } \\ = \frac{{2^{n - k} n!}}{{(n - k)!}}\sum\limits_{m = 0}^{n - k} {\binom{n-k}{m}B_m^{(n + 1)} (1)E_{n - k - m}^{(1 - n)} (0)} .
$$
In terms of the Stirling numbers of the first kind,
$$
B_m^{(n + 1)} (1) = \frac{{n - m}}{n}B_m^{(n)}  = \binom{n}{m}^{ - 1} s(n,n - m).
$$
The generalised Euler numbers may be expressed in terms of the Stirling numbers of the second kind as follows.
\begin{align*}
&\left( {\frac{{e^{2s}  + 1}}{2}} \right)^{n - 1}  = \left( {1 + \frac{{e^{2s}  - 1}}{2}} \right)^{n - 1}  = \sum\limits_{k = 0}^{n - 1} {\binom{n-1}{k}\frac{1}{{2^k }}(e^{2s}  - 1)^k } 
\\ & = \sum\limits_{j = 0}^{n - 1} {\binom{n-1}{k}\frac{{k!}}{{2^k }}\sum\limits_{i = k - 1}^\infty  {2^j S(j,k)\frac{{s^j }}{{j!}}} }  = \sum\limits_{j = 0}^\infty  {\left[ {2^j \sum\limits_{k = 0}^j {\binom{n-1}{k}\frac{{k!}}{{2^k }}S(j,k)} } \right]\frac{{s^j }}{{j!}}} ,
\end{align*}
i.e.,
$$
E_j^{(1 - n)} (0) = \sum\limits_{k = 0}^j {
   \binom{n - 1}{k} \frac{{k!}}{{2^k }}S(j,k)} .
$$
Thus, after some simplification, the final result is that for $n\geq k$,
$$\boxed{
\left[ {\frac{{d^n }}{{dx^n }}(\tanh ^{ - 1} x)^k } \right]_{x = 0} = \\ =  2^{n - k} k!\sum\limits_{m = 0}^{n - k} {\sum\limits_{r = 0}^{n - k - m} {\binom{n - m}{k} \binom{n - 1}{r}\frac{{r!}}{{2^r }}s(n,n - m)S(n - k - m,r)} } .}
$$
Second approach. We expand $(\tanh ^{ - 1} x)^k$ about $x=0$ using the Stirling numbers of the first kind:
\begin{align*}
(\tanh ^{ - 1} x)^k & = \frac{1}{{2^k }}(\log (1 + x) - \log (1 - x))^k 
\\ & = ( - 1)^k \frac{{k!}}{{2^k }}\sum\limits_{m = 0}^k {( - 1)^m \frac{{\log ^m (1 + x)}}{{m!}}\frac{{\log ^{k - m} (1 - x)}}{{(k - m)!}}} 
\\ & = ( - 1)^k \frac{{k!}}{{2^k }}\sum\limits_{m = 0}^k {( - 1)^m \left( {\sum\limits_{j = 0}^\infty  {\frac{{s(j,m)}}{{j!}}x^j } } \right)\left( {\sum\limits_{j = 0}^\infty  {( - 1)^j \frac{{s(j,k - m)}}{{j!}}x^j } } \right)} 
\\ &
 = ( - 1)^k \frac{{k!}}{{2^k }}\sum\limits_{m = 0}^k {( - 1)^m \sum\limits_{j = 0}^\infty  {\left[ {\sum\limits_{r = 0}^j {( - 1)^{j - r} \binom{j}{r}s(r,m)s(j - r,k - m)} } \right]\frac{{x^j }}{{j!}}} } 
\\ & = \sum\limits_{j = 0}^\infty  {\left[ {\frac{{k!}}{{2^k }}\sum\limits_{r = 0}^j {\sum\limits_{m = 0}^r {( - 1)^{m + r} \binom{j}{r}\left| {s(r,m)s(j - r,k - m)} \right|} } } \right]\frac{{x^j }}{{j!}}} .
\end{align*}
Therefore, by the Taylor formula,
$$\boxed{
\left[ {\frac{{d^n }}{{dx^n }}(\tanh ^{ - 1} x)^k } \right]_{x = 0}  = \frac{{k!}}{{2^k }}\sum\limits_{r = 0}^n {\sum\limits_{m = 0}^r {( - 1)^{m + r} \binom{n}{r}\left| {s(r,m)s(n - r,k - m)} \right|} } .}
$$
