Solving Recurrences with Difference Equations

Here's an example from Robert Sedgewick's course on analytic combinatorics.

Solve the recurrence

$a_n = 3 a_{n−1} − 3 a_{n−2} + a_{n−3}$ for

$n>2$ with

$a_0=a_1=0$ and

$a_2=1$ .

Let

$f(n) = a_n$ ; in the language of difference equations the above becomes simply

$\frac{{\Delta^3} f}{{\Delta n}^3 } = 0 .$ Immediately,

$f(n) = c_2 n^2 + c_1 n + c_0 .$ Applying the initial conditions we get

$c_0 = 0, c_2 + c_1 = 0, 4 c_2 + 2 c_1 = 1,$ and so the solution is

$a_n = \frac{1}{2} n^2 - \frac{1}{2} n$ .

Now what if the initial conditions are changed so

$a_1 = 1$ ?

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