How to solve this conditional recurrence relation?(two variable and conditions)

Question

I am trying to solve the following recurrence relation

$4 \leq n \;\; \; \; \; \;$ and $\; \; \; \; 2\leq i \leq \lfloor{\frac{n}{2}}\rfloor$

$F(2i,n)=$ $\begin{cases} \frac{1}{2(2i)-5}F(2i-2,2i-1),& \text{if } n=2i\text{, } i\geq3\\ \frac{n}{2n-5}F(4,n-1)=\frac{n!*3}{4!(2n-5)!!},& \text{if } i=2 \text{, } n\geq5\\ \frac{2i+n-4}{2n-5}F(2i,n-1)+\frac{n-2i+1}{2n-5}F(2i-2,n-1),& \text{otherwise} \end{cases}$

$F(4,4)=1$

I don't know how to solve a conditional recurrence relation, and I didn't find anything useful about it. any sugestion would be really helpful.

What I did up to now, as the first two conditions and also f(4,4) are initial and special case of the third condition, I only tried to solve the third one without any initial condition, I used generating function. But now I don’t know how to consider the other conditions and even whether the idea of solving the only third one alone was a good idea or not.

Answer 1

Under the assumption that $F(2i,n) = 0$ when $i<4$ or $2i>n$, the first and second cases in the recurrence become partial cases of the third one. Hence, the recurrence reduces to $$F(2i,n) = \begin{cases} 1, & \text{if } i=2,\ n=4;\\ 0, & \text{if } i<2\text{ or } 2i>n;\\ \frac{2i+n-4}{2n-5}F(2i,n-1)+\frac{n-2i+1}{2n-5}F(2i-2,n-1) & \text{otherwise}. \end{cases} $$

Consider the generating function $$f(x,y) := \sum_{i\geq 2} \sum_{n\geq 2i} F(2i,n) x^{n+2i-8} y^{n-2i+1}.$$

Then the first two cases imply $f(x,0) = 0$ and $f(0,y)=y$.

Now, the third case translates into a linear PDE: $$\frac{\partial}{\partial x} (x^2\cdot f(x,y)) + xy\frac{\partial}{\partial y} f(x,y) = 3xy + \frac{y}{x^2}\frac{\partial}{\partial x} (x^5\cdot f(x,y)) + x^4y\frac{\partial}{\partial y} (y^{-1}\cdot f(x,y)),$$ which simplifies to $$(x^2y^2-xy)f_x + (x^3y - y^2)f_y = (2y+x^3-5xy^2)f - 3y^2.$$