<s>[INST] How would you find the number of solutions to $m_1+2m_2+3m_3=n$? Assuming $m_i$ are nonnegative integers
I understand that we need to use $C(n+k-1,n-1)$ here but I am not sure how the coefficients of $m_i$ affect the equation?
For example to find the number of solutions to $m_1+m_2+m_2=n$, we could use the equation above to find that it is $C(n+2,2)=\frac{1}{2}(n+2)(n+1)$ 
 [/INST] I'm presuming $n$ is given.  The answer (call it $f(n)$) is the coefficient of $x^n$ in the Taylor expansion of the generating function
$$ g(x) = (1+x+x^2 + \ldots)(1+x^2 + x^4 +\ldots)(1 + x^3 + x^6 + \ldots) = \frac{1}{(1-x)(1-x^2)(1-x^3)}$$
Using partial fractions, 
$$ g(x) = \frac{1}{6(1-x)^3} + \frac{1}{4(1-x)^2} + \frac{1}{8(1+x)} + \frac{17}{72(1-x)} + \frac{2+x}{9(1+x+x^2)}$$
and we can write
$$ f(n) = \frac{47}{72} + \frac{n}{2} + \frac{n^2}{12} + \frac{(-1)^n}{8} + \frac{\omega^n + \overline{\omega}^n}{9} $$
where $\omega = (-1 + i \sqrt{3})/2$.
Hmm, seems to be OEIS sequence A001399
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