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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!

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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 $n\ge 1$.

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  • $\begingroup$ Thank you @Brian M. Scott Could you explain why we don't need to count the $\frac{1}{4}$ for $n\geq 1$ please? $\endgroup$ – cheeseboardqueen Aug 2 '20 at 2:32
  • $\begingroup$ @cheeseboardqueen: Because it really is just a constant. If you write out your series ‘longhand’, you get $$\frac14+\frac34+\frac{2x}{1!}+\frac{6x^2}{2!}+\ldots\;,$$ and you can simply combine the two constant terms into one. $\endgroup$ – Brian M. Scott Aug 2 '20 at 2:36
  • $\begingroup$ This helps a lot! Appreciate it :) @Brian M. Scott $\endgroup$ – cheeseboardqueen Aug 2 '20 at 2:37
  • $\begingroup$ @cheeseboardqueen: You’re welcome! $\endgroup$ – Brian M. Scott Aug 2 '20 at 2:38

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