Computing problems and generating functions 
Q. Find the generating function for the sequence $\{a_n\}$, where $a_n$ is the number of solutions to the equation: $a+b+c=k$ when $a, b, c$ are non-negative integers such that $a\ge2, 0\le b\le3$ and $2\le c\le5.$

As far as I have understood, here, $a_n$ must be the coefficient of $x^k$ in the expansion of $(x^2+x^3+x^4+\cdots)(1+x+x^2+x^3)(x^2+x^3+x^4+x^5).$
This follows because we obtain the term $x^k$ in the product by picking a term $x^a$ in the first sum, $x^b$ in the second and $x^c$ in the third, where the exponents $a, b, c$ satisfy the equation $a+b+c=k$ along with the given constraints.
But since the positive constant $k$ is not specified here so I'm not aware of how to obtain the desired coefficient in the above product. Does there exist any closed form for the same? Please suggest.. Give me some insights. It would be so helpful. Thanks in advance...

Let's do a few simplifications:

*

*$(x^{2} + x^{3} + \cdots) = x^{2}(1 + x + x^{2} + \cdots) = x^{2} \cdots \dfrac{1}{1-x}$


*$1 + x + x^{2} + x^{3} = \dfrac{1-x^{4}}{1-x}$ (by the sum of a geometric series)


*$x^{2} + x^{3} + x^{4} + x^{5} = x^{2}(1 + x + x^{2} + x^{3}) = x^{2} \cdot \dfrac{1-x^{4}}{1-x}$
So really, you have the expression:
$$x^{4} \cdot (1-x^{4})^{2} \cdot \frac{1}{(1-x)^{2}}$$
It suffices to look for the coefficient of $x^{k-4}$ in the expansion of:
$$(1-x^{4})^{2} \cdot \frac{1}{(1-x)^{2}}$$
Now using the Binomial and (Derivative of a) Geometric Series Expansions, we have that:
\begin{align*}
(1-x^{4})^{2} \cdot \frac{1}{(1-x)^{2}} &= \left( 1 -2x^{4} + x^{8} \right) \cdot \left( \sum_{i=0}^{\infty} \binom{i+2-1}{i} x^{i} \right)
\end{align*}
By expanding things out, we have that the coefficient of $x^{k-4}$ is:
\begin{align*}
\binom{(k-4)+2-1}{k-4} - 2\binom{(k-8)+2-1}{k-8} + \binom{(k-12) + 2 - 1}{k-8}
\end{align*}
