Explicit Form for Coefficients of Extended Hyperbolic Secant Function Consider the function:
$$\frac{3}{e^x+e^{{w_3}x}+e^{{w_3^2}x}}=\sum_{n=0}^\infty{E_{3,n}\frac{x^n}{n!}}$$
Note here that $w_3=e^{\frac{2i\pi}{3}}$
I am trying to get an explicit formula for the $E_{3,i}$.  So I try and rewrite the LHS:
$$\frac{3}{e^x+e^{{w_3}x}+e^{{w_3^2}x}}=\frac{3e^{-x}}{1-\left[-e^{(w_3-1)x}-e^{(w_3^2-1)x}\right]}$$
$$=3e^{-x}\sum_{n=0}^\infty(-1)^n\left[e^{(w_3-1)x}+e^{(w_3^2-1)x}\right]^n$$
$$=3e^{-x}\sum_{n=0}^\infty(-1)^n\left\{e^{(w_3-1)x}\left[1+e^{(w_3^2-w_3)x}\right]\right\}^n$$
$$=3e^{-x}\sum_{n=0}^\infty(-1)^ne^{(w_3-1)nx}\sum_{k=0}^n\binom{n}{k}e^{(w_3^2-w_3)kx}$$
$$=3\sum_{n=0}^\infty\sum_{k=0}^\infty\binom{n}{k}(-1)^ne^{[(w_3-1)n+(w_3^2-w_3)k-1]x}$$
$$=3\sum_{n=0}^\infty\sum_{k=0}^\infty\sum_{j=0}^\infty\binom{n}{k}(-1)^n[(w_3-1)n+(w_3^2-w_3)k-1]^j\frac{x^j}{j!}$$
It is here I get confused how to finish.  How do i get it simpler and finalize so that I can extract the coefficient and get my formula?
EDIT:
Is better approach this?  Consider Coefficient extraction:
$$E_{3,n}=n![x^n]\frac{3e^{-x}}{1-\left[-e^{(w_3-1)x}-e^{(w_3^2-1)x}\right]}$$
$$=n![x^n]3e^{-x}\sum_{r=0}^n(-1)^r\left[e^{(w_3-1)x}+e^{(w_3^2-1)x}\right]^r$$
$$...$$
$$=n![x^n]3\sum_{r=0}^n\sum_{k=0}^r\binom{r}{k}(-1)^re^{[(w_3-1)r+(w_3^2-w_3)k-1]x}$$
$$=n![x^n]3\sum_{r=0}^n\sum_{k=0}^r\sum_{j=0}^\infty\binom{r}{k}(-1)^r[(w_3-1)r+(w_3^2-w_3)k-1]^j\frac{x^j}{j!}$$
And since I"m considering the $n$-th coefficient, then i need the case when $j=n$?
$$3\sum_{r=0}^n\sum_{k=0}^r\binom{r}{k}(-1)^r[(w_3-1)r+(w_3^2-w_3)k-1]^n$$
Then I can simplify at least one of the expressions in the trinomial above, since $w_3^2-w_3=-i\sqrt{3}$.  So
$$E_{3,n}=\sum_{r=0}^n\sum_{k=0}^r\binom{r}{k}(-3)^r[(w_3-1)r-i\sqrt{3}k-1]^n$$
I know this is not the right answer because I've put the formula into Mathematica and am not getting the correct coefficients.  The coefficients are:
$$1,0,0,-1,0,0,19,...$$
 A: Take a look at this sum: $e^x+e^{{w_3}x}+e^{{w_3^2}x}$. By the property of $w_3$ you can easily compute the power series. So you have:
$$\frac{3}{e^x+e^{{w_3}x}+e^{{w_3^2}x}} = \frac{1}{\sum_{n=0}^\infty \frac{x^{3n}}{(3n)!}} = \frac{1}{1 + \sum_{n=1}^\infty \frac{x^{3n}}{(3n)!}}$$
We can compute this using the geometric series formula
$$\frac{1}{1 + \sum_{n=1}^\infty \frac{x^{3n}}{(3n)!}} = 1 + (-\sum_{n=1}^\infty \frac{x^{3n}}{(3n)!}) + (-\sum_{n=1}^\infty \frac{x^{3n}}{(3n)!})^2 + (-\sum_{n=1}^\infty \frac{x^{3n}}{(3n)!})^3 + \cdots $$
It is clear that the only nonzero $E_{3,n}$ must have $n$ a multiple of three. Moreover:
$$\frac{3}{e^x+e^{{w_3}x}+e^{{w_3^2}x}} = 1 - \frac{x^3}{3!} - \frac{x^6}{6!} + \frac{x^6}{3! 3!} + \cdots = 1 - \frac{x^3}{3!} - \frac{19 x^6}{6!} + \cdots $$
Which is what you got with Mathematica.
I hope this helps, or at least shows you how to repeat your Mathematica result.
A: Here are some hints which could be helpful.

Composition of formal power series
Let's consider the following part of OPs calculation
\begin{align*}
\frac{1}{1-\left[-e^{(w_3-1)x}-e^{(w_3^2-1)x}\right]}=\sum_{n=0}^{\infty}(-1)^n
\left[e^{(w_3-1)x}+e^{(w_3^2-1)x}\right]^n\tag{1}
\end{align*}
This representation is not valid when considering the ring of formal power series. The expression above is in fact the composition of the power series
\begin{align*}
G(x)=\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n
\end{align*}
  with
  \begin{align*}
F(x)=-\left[e^{(w_3-1)x}+e^{(w_3^2-1)x}\right]
\end{align*}
Note, that a composition
\begin{align*}
G(F(x))=\sum_{n=0}^{\infty}\left(F(x)\right)^n
\end{align*}
  is only valid, iff $F(0)=0$. If $F$ has no constant term it is assured, that a finite number of summands is used as contribution for the coefficients of the powers of $x$. (You might also look at this section of the referred Wiki page.)
Since $F(x)=-e^{(w_3-1)x}-e^{(w_3^2-1)x}=-2+\cdots$ has a constant term $-2\neq 0$, it does not fulfill this requirement and the representation (1) is not valid.

Observe, that on the other hand the correct representation provided in the answer of @amcalde
\begin{align*}
\frac{1}{1+\sum_{n=1}^{\infty}\frac{x^{3n}}{(3n)!}}
=\sum_{t=0}^{\infty}\left(-\sum_{n=1}^{\infty}\frac{x^{3n}}{(3n)!}\right)^t\tag{2}
\end{align*}
is a valid composition of formal power series $G(x)=\frac{1}{1+x}$ and $H(x)=-\sum_{n=1}^{\infty}\frac{x^{3n}}{(3n)!}$ with $H(0)=0$.

Calculation of reciprocals of power series
The paper Composita and its properties by V.V. Kruchinin and D.V. Kruchinin presents in section 5 techniques to obtain the coefficients of reciprocals of formal power series. In order to obtain a valid composition of power series the authors consider also $xF(x)$ instead of $F(x)$.

But note, that it is not easy to find simple representations. Before trying to obtain the coefficients of the reciprocal of the rather complex expression
\begin{align*}
e^x+e^{{w_3}x}+e^{{w_3^2}x}
\end{align*}
you could try to start with the easier expression
\begin{align*}
e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!},
\end{align*}
Here we know the solution $e^{-x}=\sum_{n=0}^{\infty}(-1)^n\frac{x^n}{n!}$ and we can better see, how Kruchinins approach could be applied.

OEIS: A002115
Using the representation (2) to obtain the coefficients of small powers of $x$ we get
  \begin{align*}
\frac{1}{1+\sum_{n=1}^{\infty}\frac{x^{3n}}{(3n)!}}=1-\frac{x^3}{3!}+\frac{19x^6}{6!}-\frac{1513x^9}{9!}+\cdots
\end{align*}
  We can find this sequence  in OEIS as A002115.
Since there is no simple representation of the coefficients of the series stated, this could indicate that no one is available.

