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Here's an example from Robert Sedgewick's course on analytic combinatorics . Solve the recurrence an=3an−1−3an−2+an−3 for n>2 with a0=a1=0 and a2=1. Let f(n)=an; in the language of difference equations the above becomes simply Δ3fΔn3=0. Immediately, f(n)=c2n2+c1n+c0. Applying the initial conditions we get c0=0,c2+c1=0,4c2+2c1=1, and so the solution is an=12n2−12n. Now what if the initial conditions are changed so a1=1?