A probability problem by Shivam Jadhav

First, evaluate $a_{0}$ , $a_{1} = 2a_{0} \Rightarrow a_{0} = \frac{3}{2}$

Then, use Z-transform: $a_{n+1} - 2a_{n} + n^2 - n = 0$

$z(\mathbf{A}(z)-a_{0}) - 2\mathbf{A}(z) + \dfrac{z(z+1)}{(z-1)^3} - \dfrac{z}{(z-1)^2} = 0$

$\Rightarrow \mathbf{A}(z) = \dfrac{z(3z^3 -9z^2 + 9z - 7)}{2(z-2)(z-1)^3}$

$\Rightarrow \mathbf{A}(z) = \dfrac{z(3z^3 -9z^2 + 9z - 7)}{2(z-2)(z-1)^3}$

$\Rightarrow \mathbf{A}(z) = \dfrac{2z}{z-1} + \dfrac{z}{(z-1)^2} + \dfrac{z(z+1)}{(z-1)^3} - \dfrac{z}{2(z-2)}$

Working in the inverse of the Z-transform: $\boxed{a_{n} = 2 + n + n^2 - 2^{n-1}}$

Now: $a_{20} = 422-2^{19}$ ; $a_{15} = 242-2^{14}$ ;

$\dfrac{\left\vert a_{20} - a_{15} \right\vert}{18133} = \dfrac{2^{19} - 2^{14} + 180}{18133} = \boxed{\boxed{28}}$