<s>[INST] Expressing $\frac{1+x+x^2}{1-x^7}$ as a closed form for the $n^{th}$ term of the associated sequence First of all I'm a little confused about the wording, am I suppose to find the sequence associated, lets say for example $(1,1,1,1,\ldots)$, and come up with an expression like $a_n=1^n$? I've gotten this far with this question
\begin{align}
\frac{1+x+x^2}{1-x^7}=&\ \frac{1}{1-x^7}+\frac{x}{1-x^7}+\frac{x^2}{1-x^7}\\
=&\ \sum^{\infty}_{k=0}x^{7k} +x\sum^{\infty}_{k=0}x^{7k}+x^2\sum^{\infty}_{k=0}x^{7k}\\
=& \sum^{\infty}_{k=0}x^{7k} +\sum^{\infty}_{k=0}x^{7k+1}+\sum^{\infty}_{k=0}x^{7k+2}
\end{align}
But I'm unsure what to do next, am I supposed to reindex each sum s.t. the exponents are all the same?
eg: $k=k-\frac{1}{7}$ for the second sum, and  $k=k-\frac{2}{7}$ for the third sum? But if I do that, then the lower limit of the sums wouldn't be integers and I'm assuming that isn't allowed. Or am I overcomplicated it and I can just do:
\begin{align}
 \sum^{\infty}_{k=0}x^{7k} +\sum^{\infty}_{k=0}x^{7k+1}+\sum^{\infty}_{k=0}x^{7k+2}=& \sum^{\infty}_{k=0}x^{7k} +\sum^{\infty}_{k=0}x\cdot x^{7k}+\sum^{\infty}_{k=0}x^2\cdot x^{7k}\\
=&\ \sum^{\infty}_{k=0}(1+x+x^2)x^{7k} 
\end{align}
If so, would the closed form of the associated sequence be $a_n = 1+n+n^2$?
 [/INST] From expanding $$\sum_{k=0}^\infty (1+x+x^2)x^{7k} = 1 + x + x^2+x^7+x^8+x^9+x^{14}+x^{15}+x^{16}+\dots$$
the sequence associated with this G.F. is $(1,1,1,0,0,0,0,1,1,1,0,0,0,0, \dots)$, which can be written as:
$$a_n=\begin{cases}1 &\text{for } n \equiv 0,1,2 &\pmod 7\\0 &\text{for } n \equiv 3,4,5,6 &\pmod 7\end{cases}$$
where $1+x+x^2$ contributes to the $n\equiv 0,1,2$ and $x^{7k}$ contributes to$\pmod 7$.
We can also see this from the original $$\sum_{k=0}^\infty x^{7k} + \sum_{k=0}^\infty x^{7k+1} + \sum_{k=0}^\infty x^{7k+2}$$
so the coefficient of $x^n$ is nonzero and equal to $1$ if and only if $n = 7k,7k+1$ or $7k+2$.

As a side note, the G.F. for $a_n=1+n+n^2$ would look like:
$$1 + 3x + 7x^2 + 13x^3 + 21 x^4 + 31 x^5 + 43 x^6 + \cdots=\frac{1+x^2}{(1-x)^3}.$$
</s>