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I'm trying to find the closed form of the ordinary generating function for the following sequences:

(1) $2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0,\dotsc.$

(2) $0, 0, 1, 3, 9, 27, 81, 243,\dotsc$

Here is my work. I think I've figured out the sequences. My problem is, how do I get these into closed form? From class, I'm used to seeing closed form in other ways; for example, the geometric series would be $1/(1-x)$; binomial theorem would be $(1+z)^n$. But how would I write these in closed form? Also, am I correct in thinking that these numbers are meant to represent coefficients?

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Your second sequence as a generating function is $$ z^3+3z^4+9z^5+\dotsb=z^3(1+3z+9z^2+\dotsb)=\frac{z^3}{1-3z}. $$ while your first is $$ 2z+2z^5+2z^9+\dotsb=2z(1+z^4+z^8+\dotsb)=\frac{2z}{1-z^4}. $$ These answers are based on the fact that your indexing starts at $1$.

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$0,\; 0,\; 1,\; 3,\; 9,\; 27,\; 81,\; 243,\;\ldots$

its generating function is

$0+0x+1x^2+3x^3+9x^4+27x^5+\ldots$

$x^2(1+3x+9x^2+27x^3+\ldots)$

$$x^2 \sum_{n=0}^\infty (3x)^n=x^2\frac{1}{1-3x}=\color{red}{\frac{x^2}{1-3x}}$$

The first sequence has generating function

$$2+2x^4+2x^8+2x^{12}+\ldots=2\sum_{n=0}^\infty (x^4)^n=\color{red}{\frac{2}{1-x^4}}$$

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  • $\begingroup$ Quick question: Could you explain where (x4)^2 came from in the summation? Would it not just be x^4? $\endgroup$ – gbm0102 Oct 29 '17 at 16:56
  • $\begingroup$ @GarrettMcClure It was a typo. It was $(x^4)^n$ I have fixed it. Thank you... $\endgroup$ – Raffaele Oct 29 '17 at 17:15

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