Recurrences seem lovely!

Thanks to Danesh Chavan for his solution to the first problem of this series. Else, I would not be able to provide a solution here.

We have to change the form of the recurrent relation as follows:

$a_n = 7a_{n-1}-10a_{n-2} + 7 \\ \Rightarrow a_n + b = a_n = 7(a_{n-1}+b)-10(a_{n-2} +b) \\ \Rightarrow a_n = a_n = 7(a_{n-1}+b)-10(a_{n-2} +b) - b \\ \Rightarrow 7b-10b -b = 7 \quad \Rightarrow b = - \frac{7}{4}$

Therefore, $a_n - \frac{7}{4} = 7\left( a_{n-1} - \frac{7}{4} \right) - 10\left( a_{n-2} - \frac{7}{4} \right)$ and the characteristic polynomial is $x^2 = 7x - 10 \quad \Rightarrow x^2-7x+10 = 0\quad \Rightarrow (x-2)(x-5)=0$

Then we have $a_n$ is of the following form:

$a_{n} - \frac{7}{4} = c_1(2^n) + c_2(5^n) \\ \Rightarrow \begin{cases} a_{0} - \frac{7}{4} = 0 - \frac{7}{4} = c_1 + c_2 & \Rightarrow 2c_1+2c_2 = -\frac{14}{4} &...(1) \\ a_{1} - \frac{7}{4} = 1 - \frac{7}{4} = 2c_1 + 5c_2 & \Rightarrow 2c_1 + 5c_2 = - \frac{3}{4} & ...(2) \end{cases} \\ Eq.2-Eq.1: \space 3c_2 = \frac {11}{4} \quad \Rightarrow c_2 = \frac{11}{12} \quad \Rightarrow c_1 = - \frac{8}{3}$

Therefore, we have $a_n = \frac{11}{12}(5^n) - \frac{8}{3}(2^n)+ \frac{7}{4}$

$\Rightarrow a_{20} = \frac{11}{12}(5^{20}) - \frac{8}{3}(2^{20})+ \frac{7}{4} = 87420142874\boxed{372}$