Finding the number of solutions to $x+2y+4z=400$ My question is how to find the easiest way to find the number of non-negative integer solutions to $$x+2y+4z=400$$
I know that I can use generating functions, and think of it as partitioning $400$ with $x$ $1$'s, $y$ $2$'s, and $z$ $4$'s. The overall generating function is: $$(1+x+x^2 + \cdots x^{400})(1+x^2+x^4+\cdots x^{200})(1+x^4+x^8+\cdots x^{100})$$
And then from this I have to calculate the coefficient of $x^{400}$, which I don't know how to do. If there's an easier way to do, I'd love to know.

You were on the right track, but instead of truncating the factors, just consider the coefficient of $x^{400}$ in:
$$(1+x+x^2+x^3+\ldots)(1+x^2+x^4+x^6+\ldots)(1+x^4+x^8+x^{12}+\ldots)=\frac{1}{(1-x)(1-x^2)(1-x^4)},\tag{1}$$
then write the RHS of $1$ as a sum of terms like $\frac{A}{(1-\xi x)^k}$, with $\xi\in\{1,-1,i,-i\}$, and exploit the identities:
$$\frac{1}{1-\xi x}=\sum_{k=0}^{+\infty}(\xi x)^k,$$
$$\frac{1}{(1-\xi x)^2}=\sum_{k=0}^{+\infty}(k+1)(\xi x)^k,$$
$$\frac{1}{(1-\xi x)^3}=\sum_{k=0}^{+\infty}\frac{(k+2)(k+1)}{2}(\xi x)^k$$
to recover the final expression:

$$[x^n]\frac{1}{(1-x)(1-x^2)(1-x^4)}=\frac{2n^2+14n+21}{32}+\frac{(7+2n) (-1)^n}{32}+\frac{1}{8} \cos\left(\frac{n \pi }{2}\right)+\frac{1}{8} \sin\left(\frac{n \pi }{2}\right)$$

that if $8\mid n$ becomes:

$$[x^n]\frac{1}{(1-x)(1-x^2)(1-x^4)}=\frac{1}{16}(n+4)^2.$$

