mystery regarding power series of $\frac{1}{\sqrt{1+x^{x}}}$ In the course of playing around with $\sum_{n=1}^{\infty} \frac{1}{\sqrt{1+n^{n}}}$, I used w|α to obtain the power series for $f(x)=\frac{1}{\sqrt{1+x^{x}}}$
which is 
\begin{align*}
\frac{1}{\sqrt{1+x^{x}}} =& \frac{1}{\sqrt{2}} - \frac{x\log(x)}{4\sqrt{2}} -\frac{x^{2}\log^{2}(x)}{32\sqrt{2}}+ \frac{5x^{3}\log^{3}(x)}{384\sqrt{2}}\\\
 \\\
&+ \frac{17x^{4}\log^{4}(x)}{6144\sqrt{2}} - \frac{121x^{5}\log^{5}(x)}{122880\sqrt{2}} - \frac{721x^{6}\log^{6}(x)}{2949120\sqrt{2}} \ldots
\end{align*}
Before I realized that I couldn't really use this to help me with the sum, I found that the denominators (ignoring the $\sqrt{2}$, because all of them have it in common) correspond to $4^{n}n!$, what is baffling is that the numerators appear to correspond to the coefficients in the exponential generating function for $f(x)=e^{\tanh^{-1}(\tan(x))}$ (I believe that the 7th entry should be 1369 and not 6845), and I'm curious what the explanation is, because $f(x)=e^{\tanh^{-1}(\tan(x))}$ is a mighty weird looking function.
 A: It's simple: $\rm\quad\quad  f(x)\ =\  \cos x - \sin x \ =\ \frac{1+i}2 e^{ix} + \frac{1-i}2 e^{-ix}$
$\rm\quad\displaystyle \Rightarrow\quad\quad\quad\quad\ f(\tan^{-1} x)\ =\ \frac{1-x}{\sqrt{x^2+1}}\quad $ via $\rm\displaystyle\quad e^{\:i\:tan^{-1} x}\ =\ \frac{1+ x\: i}{\sqrt{x^2 + 1}}$
$\rm\quad\displaystyle \Rightarrow\quad f(\tan^{-1}\tanh x)\ =\ \ \frac{\sqrt 2}{e^{4x}+1}\quad\ $ via $\rm\displaystyle\quad\ \tanh{x}\ =\ 1 - \frac{2}{e^{2x}+1}$
A: Well,
$$\frac1{\sqrt{1+x^x}}=\frac1{\sqrt{1+e^y}}$$
where $y=x\log x$ so it's unsurprising you get a series in terms of $y=x\log x$.
Then
$$\frac1{\sqrt{1+e^y}}=\frac1{\sqrt2}\frac1{\sqrt{1+u}}$$
where $u=\frac12(e^y-1)=o(y)$. This explains the fact that the coefficients
are rationals over $\sqrt2$.
Now consider $f(x)=\exp(\tanh^{-1}(\tan x))$. Now
$$\tanh z=\frac{e^{2z}-1}{e^{2z}+1}$$
so that
$$e^z=\sqrt{\frac{1+\tanh z}{1-\tanh z}}.$$
Hence,
$$\exp(\tanh^{-1} t)=\sqrt{\frac{1+t}{1-t}}.$$
Putting $t=\tan x$ gives
$$f(x)=\sqrt{\frac{1+\tan x}{1-\tan x}}=\sum_{n=0}^\infty a_n x^n.$$
Then
\begin{eqnarray*}
&&(1+i)f(ix)+(1-i)f(-ix)\\
&=&
2\sum_{m=0}^\infty
(a_{4m} x^{4m}-a_{4m+1} x^{4m+1}-a_{4m+2} x^{4m+2}+a_{4m+3} x^{4m+3}).
\end{eqnarray*}
But
$$f(ix)=\sqrt{\frac{1+i\tanh x}{1-i\tanh x}}
=\frac{1+i\tanh x}{\sqrt{1+\tanh^2x}}$$
and so
\begin{eqnarray*}
&&(1+i)f(ix)+(1-i)f(-ix)\\
&=&
\frac{2-2\tanh x}{\sqrt{1+\tanh^2 x}}
=\frac{2\cosh x-2\sinh x}{\sqrt{\cosh^2x+\sinh^2x}}\\
&=&\frac{2e^{-x}}{\sqrt{(e^{2x}+e^{-2x})/2}}=\frac{2\sqrt2}{\sqrt{1+e^{4x}}}
\end{eqnarray*}
which explains why the coefficients in the two series are the same
up to signs and powers of 4.
