Counting the number of ways three dice can sum to $n$ I want to count in how many ways three dice can sum to a given number without brute forcing it. In fact, I would like to do it using generating functions and without having to expand out the product.

To do this, I have thought of making a differential equation out of $y=(x+x^2+x^3+x^4+x^5+x^6)^3$ and then solving the equation through the use of power series, or getting the $n$th taylor coefficient from the equation. 

What I attempted was:
$$y=(x+x^2+x^3+x^4+x^5+x^6)^3$$
$$y'=3y^{2/3}(1+2x+3x^2+4x^3+5x^4+6x^5)$$
Then, by setting $x=0$ we see $y(0)=0$ and $y'(0)=0$, which is what I want, however, when trying to obtain $y''(0)$ by differentiating both sides, I get division by $0$ on the RHS. 

Am I making a mistake in the set up or is calculating the expression through the use of a differenrial equation impossible?
 A: Here is an alternate approach. We could at first  transform  $y(x)$ so that an expansion after that becomes less cumbersome. Maybe this variant is also useful for your needs.
It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series. We also use Iverson brackets 
\begin{align*}
[[P(x)]]=\begin{cases}
1&\qquad P(x) \ \text{  true}\\
0&\qquad P(x) \ \text{ false}
\end{cases}
\end{align*}
This way we can treat multiple cases in one expression.

We obtain
  \begin{align*}
[x^n]y(x)&=[x^n](x+x^2+x^3+x^4+x^5+x^6)^3\\
&=[x^n]x^3(1+x+x^2+x^3+x^4+x^5)^3\\
&=[x^{n-3}][[n\geq 3]]\left(\frac{1-x^6}{1-x}\right)^3\tag{1}\\
&=[x^{n-3}][[n\geq 3]](1-3x^6+3x^{12}-x^{18})\sum_{j=0}^\infty\binom{-3}{j}(-x)^j\tag{2}\\
&=\left([x^{n-3}][[n\geq 3]]-3[x^{n-9}][[n\geq 9]]\right.\\
&\qquad\quad\left.+3[x^{n-15}][[n\geq 15]-[x^{n-21}][[n\geq 21]\right)
\sum_{j=0}^\infty\binom{j+2}{2}x^j\tag{3}\\
&=\binom{n-1}{2}[[n\geq 3]]-3\binom{n-7}{2}[[n\geq 9]]\\
&\qquad\quad+3\binom{n-13}{2}[[n\geq 15]]-\binom{n-19}{2}[[n\geq 21]]\tag{4}
\end{align*}

Comment:


*

*In (1) we apply the coefficient of rule 
\begin{align*}
[x^p]x^qA(x)=[x^{p-q}]A(x)
\end{align*}
and we use the formula for the finite geometric series. Since there is no contribution to the coefficient of $x^n$ if $n<3$ we respect this by using $[[n\geq 3]]$.

*In (2) we expand the binomial and we also expand $\frac{1}{(1-x)^3}$ using the binomial series expansion.

*In (3) we use the linearity of the coefficient of operator and we use the binomial identity
\begin{align*}
\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q=\binom{p+q-1}{p-1}(-1)^q
\end{align*}

*In (4) we select the coefficient of $x^{n-k}, k\in\{3,9,15,21\}$.
Note: The usage of the Iverson brackets covers the general case. If we need to calculate a specific case only, the calculation becomes even more straight forward.

Example: $[x^{10}]y(x)$
We obtain
  \begin{align*}
[x^{10}]y(x)&=[x^{10}](x+x^2+x^3+x^4+x^5+x^6)^3\\
&=[x^7]\left(\frac{1-x^6}{1-x}\right)^3\\\
&=[x^7](1-3x^6)\sum_{j=0}^\infty\binom{-3}{j}(-x)^j\\
&=\left([x^7]-3[x]\right)\sum_{j=0}^\infty\binom{j+2}{2}x^j\\
&=\binom{9}{2}-3\binom{3}{2}\\
&=27
\end{align*}

