Is there a closed form solution to $$a_{n+2}=\dfrac{(n+1)(n-2)a_{n+1} + (4n+3)a_n - a_{n-1}}{(n+2)(n+1)}$$ that can be written in terms of $a_0$ and $a_1$ given the fact that that $$a_2 = \dfrac{2a_1 - 3a_0}{-2}?$$
I'm doing series solutions to differential equations and I was solving for the coefficients and came to this. I generally can not find a closed form solution in terms of $a_1$ and $a_0$. However, I have seen examples that seem to have cleverly come up with a way to do this. Usually that involves the gamma function somehow by using $\Gamma(n+1) = n\Gamma(n)$ (but they tend to do more that just use the factorial of an integer).
Also, I know that in this case the denominator is $n!$ but beyond that, it just gets messy when I do it.
Any suggestions? I'm hoping that any insight into being able to do this somehow would generalize to a method (or at least tricks) to helping me do this in the future.
Thanks for any suggestions.
While this does not fully answer the question, one can indeed simplify the problem as per my comment above. Let $c_n:=n!\,a_n$. Then we can deduce a recurrence relation for $c_n$ from $a_n$:
\begin{align} c_0&=a_0,\quad c_1=a_1,\quad c_2=2a_2=-2a_1+3a_0=-2c_1+3c_0,\\\\ c_{n+2}&=(n+2)!a_{n+2}\\ &=(n+2)!\dfrac{(n+1)(n-2)a_{n+1} + (4n+3)a_n - a_{n-1}}{(n+2)(n+1)}\\ &=(n-2)(n+1)!\,a_{n+1} + (4n+3)n!\,a_n - n!\,a_{n-1}\\ &=(n-2)c_{n+1}+(4n+3)c_n-n c_{n-1} \end{align} The coefficients of this recurrence relation still depend on $n$, but as polynomials rather than rational functions in $n$.