$F_n = \frac{1}{n+3/2} \left ( F_{n+2} - c F_{n+1} \right )$
where $c$ is a constant parameter.
or, more general:
$F_n = g(n) \left ( F_{n+2} - c F_{n+1} \right )$
On
Use generating functions: Define $f(z) = \sum_{n \ge 0} F_n z^n$, multiply your recurrence by $z^n$ and sum over $n \ge 0$:
$$ \sum_{n \ge 0} F_{n + 2} z^n = c \sum_{n \ge 0} F_{n + 1} z^n + \sum_{n \ge 0} n F_n z^n + \frac{3}{2} \sum_{n \ge 0} F_n z^n $$
Recognize some of the sums:
$$ \frac{f(z) - F_0 - F_1 z}{z^2} = c \frac{f(z) - F_0}{z} + z f'(z) + \frac{3}{2} f(z) $$
Now you've got a first order, linear differential equation for $f$. Solve that one (yes, chances are you get a mess), and pull the coefficients out of that one.
For more general forms of the equation you'll probably be out of luck.
The "backward" recurrence relation: $$ F_n = \frac{1}{n+3/2}\left(F_{n+2}-c F_{n+1}\right) $$ is equivalent to the standard recurrence relation: $$ F_{n+2}= c F_{n+1} + (n+3/2) F_n$$ so any choice of $F_0,F_1$ leads to a solution, and the situation is the same for the second recurrence.