Number of solutions of $x_1+2\cdot x_2+2\cdot x_3 = n$ I have to find number of solutions of $x_1+2\cdot x_2+2\cdot x_3 = n$. I guess it would be $[x^n](1+x+x^2 \dots)(1 + x^2 + x^4 \dots)^2$, but how to compute it? I know only that $\frac{1}{1-x} = 1+x+x^2 \dots$.
 A: Generating Function
The generating function is
$$
\begin{align}
\frac1{(1-x)\left(1-x^2\right)^2}
&=\frac1{(1-x)^3(1+x)^2}\\
&=\sum_{j=0}^\infty\binom{-3}{j}(-x)^j\sum_{k=0}^\infty\binom{-2}{k}x^k\\
&=\sum_{j=0}^\infty\binom{j+2}{2}x^j\sum_{k=0}^\infty\binom{k+1}{1}(-x)^k\\
&=\sum_{n=0}^\infty\sum_{k=0}^n(-1)^k\binom{k+1}{1}\binom{n-k+2}{2}x^n\tag{1}
\end{align}
$$
Thus, the number of solutions is
$$
\bbox[5px,border:2px solid #C0A000]{\sum_{k=0}^n(-1)^k\binom{k+1}{1}\binom{n-k+2}{2}=\frac{2n^2+10n+11+(-1)^n(2n+5)}{16}}\tag{2}
$$

Alternating Sums of Powers
To compute the sum in $(2)$ we have used the following
$$
\begin{align}
\sum_{k=0}^n(-1)^k1&=\frac{(-1)^n+1}2\tag{3}\\
\sum_{k=0}^n(-1)^kk&=\frac{(-1)^n(2n+1)-1}4\tag{4}\\
\sum_{k=0}^n(-1)^kk^2&=\frac{(-1)^n\left(n^2+n\right)}2\tag{5}\\
\sum_{k=0}^n(-1)^kk^3&=\frac{(-1)^n\left(4n^3+6n^2-1\right)+1}8\tag{6}
\end{align}
$$

Recursion for the Alternating Sums of Powers
The formulas in $(3)$-$(6)$ were derived by induction and $(9)$.
Combining the following two sums by reindexing the first gives
$$
\begin{align}
\sum_{k=0}^n(-1)^{k+1}(k+1)^m+\sum_{k=0}^n(-1)^kk^m
&=\sum_{k=1}^{n+1}(-1)^kk^m+\sum_{k=0}^n(-1)^kk^m\\
&=(-1)^{n+1}(n+1)^m+2\sum_{k=0}^n(-1)^kk^m\tag{7}\\
\end{align}
$$
Combining the same two sums without reindexing gives
$$
\sum_{k=0}^n(-1)^{k+1}(k+1)^m+\sum_{k=0}^n(-1)^kk^m
=\sum_{k=0}^n(-1)^{k+1}\left((k+1)^m-k^m\right)\tag{8}
$$
Equating the right sides of $(7)$ and $(8)$ gives
$$
\sum_{k=0}^n(-1)^kk^m
=\frac12\left[(-1)^n(n+1)^m-\sum_{k=0}^n(-1)^k\left((k+1)^m-k^m\right)\right]\tag{9}
$$
A: If $n$ is odd, then $x_1$ must be odd, and if $n$ is even, then $x_1$ must be even. Thus there is an integer $y$ such that $x_1=2y+1$ for former and $x_1=2y$ for latter. Thus
$$
y+x_2+x_3=\frac{n-1}{2}\text{ or }y+x_2+x_3=\frac{n}{2}.
$$
Therefore, the answer is
\begin{align}
\binom{\frac{n+3}{2}}{\frac{n-1}{2}} &= \binom{\frac{n+3}{2}}{2}\\
&=\frac{n+3}{2}\cdot\frac{n+1}{2}\cdot\frac{1}{2}
\end{align}
for odd $n$ and
\begin{align}
\binom{\frac{n+4}{2}}{\frac{n}{2}}&=\binom{\frac{n+4}{2}}{2}\\
&=\frac{n+4}{2}\cdot\frac{n+2}{2}\cdot\frac{1}{2}
\end{align}
for even $n$.
