<s>[INST] Solving the recursion $3a_{n+1}=2(n+1)a_n+5(n+1)!$ via generating functions I have been trying to solve the recurrence: 
\begin{align*}
a_{n+1}=\frac{2(n+1)a_n+5((n+1)!)}{3},
\end{align*}
where $a_0=5$, via generating functions with little success. My progress until now is this:
Let $A(x)=\sum_{n=0} ^{\infty} a_nx^n$. By multiplying both sides of our recurrence relation by $x^n$ and summing over $n$ from $0$ to $\infty$, we see that 
\begin{align}
\sum_{n=0} ^{\infty} a_{n+1} x^n = \frac{2}{3}\sum_{n=0} ^{\infty} (n+1)a_nx^n + \sum_{n=0} ^{\infty} (n+1)!x^n.
\end{align}
Using our definition of $A(x)$ we can rewrite the left hand side as
\begin{align*}
\sum_{n=0} ^{\infty} a_{n+1} x^n=\frac{A(x)-a_0}{x}.
\end{align*}
Such manipulations of the right hand side have been difficult because of the coefficients of the power series. 
Is there anyway to proceed from here, or are generating functions not suited to solve such a recurrence?
 [/INST] Exponential generating function of sequence $\{a_n\}$ is $f(x) = \sum_{n=0}^\infty a_n \frac{x^n}{n!}$. Rewriting the recurrence equation as 
$$
   \frac{a_{n+1}}{n+1} = \frac{2}{3} a_n + \frac{5}{3} n!
$$
Now multiplying both sides by $\frac{x^n}{n!}$ and using recurrence relation for factorial:
$$
   a_{n+1} \frac{x^n}{(n+1)!} = \frac{2}{3} a_n \frac{x^n}{n!} + \frac{5}{3} x^n
$$
Summing from $n=0$ to infinity:
$$
  \frac{1}{x} \sum_{n=1}^\infty a_n \frac{x^n}{n!} = \frac{2}{3} \sum_{n=0}^\infty a_n \frac{x_n}{n!} + \frac{5}{3} \sum_{n=0}^\infty x^n
$$
or
$$
    \frac{1}{x} \left( f(x) - a_0 \right) = \frac{2}{3} f(x) + \frac{5}{3} \frac{1}{1-x}
$$
Solving for $f(x)$ we readily get:
$$
  f(x) = \frac{5}{1-x} = \sum_{n=0}^\infty (5 \cdot n!) \frac{x^n}{n!}
$$
Thus the solution is $a_n = 5 \cdot n!$.
</s>