Generating function for the sequence $1,1,3,3,5,5,7,7,9,9,\ldots$

Question

The generating function for the sequence $\left\{1,1,1,1,...\right\}$ is $$1 + x + x^2 + x^3 ... = \frac{1}{1-x}$$ What is the generating function for the sequence $\left\{1,1,3,3,5,5,7,7,9,9,\dots \right\}$?

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This is my attempt:

what we want to do is after the initial $\left\{1,1,1,1,...\right\}$ function we want to add two more functions shifted two to the right such as $\left\{0,0,1,1,1,...\right\}$ and keep adding such function 2 at a time and each new two are shifted 2 more to the right then the previous one. so first would be $\left\{1,1,1,1,\dots \right\}$ then add 2 of $\left\{0,0,1,1,1,1,\dots\right\}$ then add 2 more of $\left\{0,0,0,0,1,1,1,1,\dots\right\}$ and keep doing that. this gives us the function

$$\frac{1}{1-x}+\frac{2x^2}{1-x}+\frac{2x^4}{1-x}+...$$ or $$\frac{1}{1-x}+\frac{2x^{2k}}{1-x}$$ however $i$ do not know where to go from here, how do $i$ finish this problem?

Answer 1

You already did the difficult part... You probably mean $$\frac1{1-x}+\sum_{k=1}^\infty\frac{2x^{2k}}{1-x}$$

So you just have to evaluate the geometric series $$\sum_{k=1}^\infty (x^2)^k = \frac1{1-x^2} - 1 = \frac{x^2}{1-x^2}$$

And obtain

$$\frac1{1-x} + \frac2{1-x}\sum_{k=1}^\infty x^{2k} = \frac{1+\frac{2x^2}{1-x^2}}{1-x} = \frac{x^2+1}{(1-x)(1-x^2)}$$

Answer 2

Here is another approach.

The generating function writes $(1+x)(1+3x^2 + 5x^4 + 7x^6 + 9x^8 + \dots)$ where the second term appears as the derivative of $x+x^3 + x^5 + x^7 + x^9 + \dots$

The later being the odd part of $1+x+x^2 + x^3 + x^4 + \dots$, we have: $$ x + x^3 + x^5 + x^7 + x^9 + \dots = \frac{1}{2}\left(\frac{1}{1-x}-\frac{1}{1+x}\right) $$

Hence: $$ 1 + 3x^2 + 5x^4 + 7x^6 + 9x^8 + \dots = \frac{1}{2}\left(\frac{1}{(1-x)^2}+\frac{1}{(1+x)^2}\right) $$

Finnaly: $$ \frac{1+x}{2}\left(\frac{1}{(1-x)^2}+\frac{1}{(1+x)^2}\right) = \frac{x^2+1}{(x-1)^2(1+x)} $$