Question about generating functions about an exercise I have a question about generating functions,
I've understood that generating functions are also representing some series of numbers, but is this series could be formed by 2 different functions?
For example, an exercise that I've encountered goes like this:
Find the generating function that represents the number of solutions to the equation:
$$ x_1 + x_2 + \dots + x_k=n$$
Where the variables are even numbers that are not divisible by 3,
So I assume we should find a function that represents the series:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 ...
0 1 0 1 0 0 0 1 0  1  0  0  0  1  0  1  0  0 ..
The zero's and one's stands for the coefficient of $x^n$
So to get to that series we have few options the first one i've thought about is to take the generating function: $1 + x + x^2 + x^3 + \dots$
$$ 1 + x + x^2 + x^3 + ... = \frac{1}{1-x} $$
Which represents the series of 1 1 1 1 1 ...
and then to subtract those (subtract odd numbers and multiples of 6):
$$ f(x) = ((1 + x + x^2 + x^3 + ...) - x(1 + x^2 + x^4 + ...) - (1 + x^6 + x^{12} +...))^k $$
Which gives us the series we wanted.
But the solution to this exercise shows different answer and the generating function goes like this:
$$ g(x) = ((x^2 + x^4) + (x^8 + x^{10}) + (x^{14} + x^{16}) + ... )^k $$
which represents the same series as well but looks differently, does those 2 functions coefficients represent the same number of solutions to the equation?
Thanks alot !
 A: Your approach is fine. We can show the generating functions are equal: $f=g$.

We have
\begin{align*}
\color{blue}{g(x)}&=\left(\left(x^2 + x^4\right) + \left(x^8 + x^{10}\right) + \left(x^{14} + x^{16}\right) + ... \right)^k \\
&=\left(x^2\left(1+x^2\right)+x^8\left(1+x^2\right)+x^{14}\left(1+x^2\right)+\cdots\right)^k\\
&=\left(1+x^2\right)^k\left(x^2+x^8+x^{14}+\cdots\right)^k\\
&=\left(1+x^2\right)^kx^{2k}\left(1+x^6+x^{12}+\cdots\right)^k\\
&\,\,\color{blue}{=\left(\frac{x^2\left(1+x^2\right)}{1-x^6}\right)^k}
\end{align*}

On the other hand we obtain

\begin{align*}
\color{blue}{f(x)}&= \left(\left(1 + x + x^2 + x^3 +\cdots\right) - x\left(1 + x^2 + x^4 + \cdots\right)
- \left(1 + x^6 + x^{12} +\cdots\right)\right)^k\\
&=\left(\frac{1}{1-x}-\frac{x}{1-x^2}-\frac{1}{1-x^6}\right)^k\\
&=\left(\frac{1+x}{1-x^2}-\frac{x}{1-x^2}-\frac{1}{1-x^6}\right)^k\\
&=\left(\frac{1}{1-x^2}-\frac{1}{1-x^6}\right)^k\\
&=\left(\frac{\left(1-x^6\right)-\left(1-x^2\right)}{\left(1-x^2\right)\left(1-x^6\right)}\right)^k\\
&=\left(\frac{x^2\left(1-x^4\right)}{\left(1-x^2\right)\left(1-x^6\right)}\right)^k\\
&\,\,\color{blue}{=\left(\frac{x^2\left(1+x^2\right)}{1-x^6}\right)^k}
\end{align*}
and the claim follows.

A: The functions $f$ and $g$ are identical.
You have taken all the powers then deleted all odd ones and then the multiples of $6$.
The book has started with the even powers and then deleted the multiples of $6$.
It's just the same!
