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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