Initial value problem for a recurrence

Remarks

Every such problem has exactly one solution. The terms $u_k, u_{k+1}, \dots$ are computed one after the other from the recurrence, and induction on $n$ shows that no other choice is possible. Example: $u_{n+2} = u_{n+1} + u_n$ with $u_0 = 0$ and $u_1 = 1$ has the solution $0, 1, 1, 2, 3, 5, \dots$, the Fibonacci numbers. Non-example: giving only $u_0 = 0$ for this order $2$ recurrence is not an initial value problem in this sense. Both $0, 0, 0, \dots$ and $0, 1, 1, 2, \dots$ satisfy the recurrence with $u_0 = 0$, so one initial value is too few. Initial conditions may also be placed at another starting index $n_0$ instead of $0$.