Finding a recurrence for $c_n=a_{n+1}-a_n$ provided that $a_n = 3a_{n-1}-3a_{n-2}+a_{n-3}$

Question

Suppose $a_0,a_1,a_2$ satisfy the recurrence $a_n = 3a_{n-1}-3a_{n-2}+a_{n-3}$ for $n\ge3$.

Let $c_n=a_{n+1}-a_n$ for $n\ge1$ and $c_0=0$.

Find a linear recurrence of degree at most $2$ for the sequence $c_0,c_1,c_2,\dots$.

I'm not entirely sure what they are asking for but I think it is asking for a recurrence relation for $c_n$ where the characteristic polynomial has degree at most $2$.

I tried finding a closed form for $a_n$ to try and get rid of the $a_n$ in the $c_n$ equation, but I can't figure it out when the initial conditions aren't specified ($a_0,a_1,a_2$).

Any help would be appreciated!

Answer 1

Here is a start. First shift the index, then rearrange as $$ a_n = 3a_{n-1}-3a_{n-2}+a_{n-3} \implies a_{n+1} = 3a_{n}-3a_{n-1}+a_{n-2} $$

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

Now, use $c_n= a_{n+1}-a_n $ in the last equation to get a recurrence relation of degree $2$ in terms of $c_n$. I think you can finish it.

Answer 2

Hint

More empirically than Mhenni Benghorbal, you can assume that you know the first terms of $a(n)$ (just name them $a_0$,$a_1$ and $a_2$). The condition $c(0)=0$ shows that $a_0=a_1$. Compute a very few terms for $a(n)$ and $c(n)$. The pattern becomes obvious.