Linear recursion with constant coefficients that fullfil $a_n = 3^n + 7^n$ for $n \in \mathbb{N}_0$

Question

Let $a_n = 3^n + 7^n$ for $n \in \mathbb{N}_0$

I know that the generating function of $a_n= 3^n+7^n = A(x) = \frac{1}{1 - 3x} + \frac{1}{1 - 7x}.$

The exponential generating function

$$A(x) = \sum_{n=0}^{\infty}\frac{a_n}{n!}x^n$$

of the sequence $(a_n)_{n \in \mathbb{N}}$ is

$A(x)= \sum_{n=0}^{\infty}\frac{3^n}{n!}x^n+ \sum_{n=0}^{\infty}\frac{7^n}{n!}x^n= \sum_{n=0}^{\infty}\frac{(3x)^n}{n!}+ \sum_{n=0}^{\infty}\frac{(7x)^n}{n!}=e^{3x}+e^{7x}.$

I want to find out a linear recursion with constant coefficients, that fullfil $a_n$, but I don't know how its done

Answer 1

$$a_n-3^n = 7^n$$

so $$ a_{n+1}-3^{n+1}= 7^{n+1}$$

so $$a_{n+1}-3^{n+1}=7\cdot 7^n = 7(a_n-3^n )$$

so $$a_{n+1}-7a_n = -4\cdot 3^n$$

so $$a_{n+2}-7a_{n+1} = -4\cdot 3^{n+1}$$ $$=-12\cdot 3^n$$ $$= 3(-4\cdot 3^n)$$ $$ =3 (a_{n+1}-7a_n )$$

So $$a_{n+2}-10a_{n+1} +21a_n =0$$