A Recurrence Equation From a Game

Question

$a_n=a_{n-1}(a_{n}-a_{n-2}+1)$

The above equation is defined in $[0,m]$ st. $a_{0}=0$ and $a_m=1$. It turned up as I was trying to analyze a simple richman game.

I have managed to solve the equation as a system for small m's, however it becomes harder as m grows, even though the values seem to follow a nice distribution. I would quite like to know if there was a way to solve it, or perhaps approximate it using differential equations or similar.

Edit: This is a plot given $m=10$:

It is symmetric such that $a_{m-n}=1-a_{n}$

Answer 1

Write it as $a_n - a_{n-1} =a_n a_{n-1}-a_{n-1}a_{n-2}$. Now you can telescope.

$$a_n - a_{n-1} =a_n a_{n-1}-a_{n-1}a_{n-2}$$ $$a_{n-1} - a_{n-2} =a_{n-1}a_{n-2}-a_{n-2}a_{n-3}$$ $$...$$ $$a_2 - a_1 =a_2 a_1-a_1 a_0$$

Summing all of that and simplifying you have $a_n = \dfrac{a_1}{1-a_{n-1}}$ or $a_{n-1} = 1 - \dfrac{a_1}{a_n}$. Thus $a_2 = \dfrac{a_1}{1-a_1}$ and $a_3 = \dfrac{1-a_1^2}{1-2a_1}$

Similarly, given $a_m = 1$, you have $a_{m-1} = 1-a_1$ and $a_{m-2} = 1 - \dfrac{a_1}{1-a_1} = \dfrac{1-2a_1}{1-a_1}$

Thus we can express every term in terms of $a_1$, in two ways depending on which path we take. To find $a_1$, we need to have the forward and backward paths meet at some point, equate the terms and solve the resulting equation. However this doesn't look easy given the expressions coming up, unless we can use some symmetry in early terms itself. Perhaps someone else can help more.

Answer 2

@Thomas ahle: Did you solve it yet? Please show me your methods because I also met a similar problem above.
$\left\{\begin{matrix} S_{n}=S_{n-1}+T_{n-1}\\ T_{n-1}=C_{n-3}+D_{n-4}\\ C_{n-3}=E_{n-4}+F_{n-4}\\ E_{n-4}=K_{n-5}+T_{n-5}\\ F_{n-4}=E_{n-7}+D_{n-7}\\ D_{n-7}=H_{n-8}+F_{n-8}\\ H_{n-8}=S_{n-11}+C_{n-11}\\ K_{n-5}=S_{n-7}+C_{n-8} \end{matrix}\right.$
Now, I'm trying to solve it by matrix form but it is so hard. Particularly, find generating matrix and characteristic polynomial.