Determine the number of ways to color a 1-by-n chessboard with the colors red, blue, green and orange. Determine the number of ways to color a 1-by-n chessboard, using the colors red, blue, green and orange if an even number of squares is to be colored red and an even number is to be colored green.
Current result:
I will use exponential generating functions.
\begin{align*}
g^{(e)}(x)&=
\left(1+\frac{x^{2}}{2!}+\frac{x^{4}}{4!}+\dots\right)^2\left(1+\frac{x^{1}}{1!}+\frac{x^{2}}{2!}+\dots\right)^2\\
&=\left(\frac{e^x+e^{-x}}{2}\right)^2e^{2x}
\\&=\frac{1}{4}\left(e^{2x}+2e^xe^{-x}+e^{-2x}\right)e^{2x}\\
&=\frac{1}{4}(e^{4x}+2e^{2x}+1)\\
&=\frac{1}{4}\left(\sum_{n=0}^{\infty}4^n\frac{x^n}{n!}+2\sum_{n=0}^{\infty}2^n\frac{x^n}{n!}+1\right)\\
&=\sum_{n=0}^{\infty}\left(\frac{4^n+2\times 2^n}{4}\right)\frac{x^n}{n!}+\frac{1}{4}
\end{align*}
I'm wondering how I can move $1$ into the sum? I've been suggested to use $x^0$ but I'm not certain how to incorporate that as a sum since I want to find $h_n$ which will be the coefficient of $\frac{x^n}{n!}$.
Thanks in advance!

You can simply read off the coefficients $h_n$ from the sum
$$\sum_{n=0}^\infty\left(\frac{4^n+2\cdot 2^n}4\right)\frac{x^n}{n!}+\frac14\;:$$
$h_0=\frac{1+2}4+\frac14=1$, and
$$h_n=\frac{4^n+2\cdot 2^n}4=4^{n-1}+2^{n-1}=2^{n-1}\left(2^{n-1}-1\right)$$
for $n\ge 1$.
