How many solutions the equation $x_1 + x_2 +x_3 = m$ for $m\in\Bbb{N}, x_{1,3}\in \Bbb{N_{even}}, x_2\in\Bbb{N}$ ?
Show that $i\in\Bbb{N}$, $2i$ and $2i+1$ has the same amount of solutions.
My Work -
So we can define $x_i$ as - $x_{1,3} = 1+x^2 +x^4...$ and $x^2 = 1+x+x^2+x^3+...$ and define our generating function $f$ as -
$f(x) = \lambda x\in\Bbb{R}. x_1\cdot x_2\cdot x_3 \,\, \Rightarrow$ $f(x) = \lambda x\in\Bbb{R}. (1+x^2+x^4+...)^2(1+x+x^2+x^3+..)$
$f(x) = \lambda x\in\Bbb{R}. \dfrac{1}{1-x^2} \cdot \dfrac{1}{1-x^2} \cdot \dfrac{1}{1-x} \Rightarrow $ $f(x) = \lambda x\in\Bbb{R}.\dfrac{1}{(1-x)^3}\cdot \dfrac{1}{(1+x)^2}$
$\iff f(x) = \sum_\limits{n=0}^\infty \binom{n+2}{2} x^n \cdot \Big(\sum_\limits{n=0}^\infty(-1)^k\cdot x^n\Big)^2$ $\iff f(x) = \sum_\limits{n=0}^\infty \binom{n+2}{2} x^n \cdot \sum_\limits{n=0}^\infty x^{2n}$
So we want to find the coefficient of $m$, which if i am right is $\binom{m+2}{2} + 1$ for $m\in\Bbb{N_{even}}$ and $\binom{m+2}{2}$ for $m\in\Bbb{N_{odd}}$ ? Is everything legit ? I think i miss something because $\binom{m+2}{2}$ $\not=$ $\binom{m+2}{2} +1$ (?)