With regard to the general question, I defer to Joel Hamkins's answer. For your specific function, though, the first few cases give
$$\begin{align} w(2) &= 2^n - 2 \\ w(3) &= 3^n -3\cdot 2^n + 3\\ w(4) &= 4^n - 4\cdot 3^n + 6\cdot 2^n - 4\\ w(5) &= 5^n - 5\cdot 4^n + 10\cdot 3^n - 10\cdot 2^n + 5 \end{align}$$
This suggests the explicit formula (where by "explicit" I just mean that the function $w$ doesn't appear on the right hand side)
$$w(i) = \sum_{j=0}^{i-1} (-1)^j(i-j)^n{i \choose j}$$
which I suspect (without taking time to verify) can be proved by straightforward induction.
A minor additional note: The OP's second example does not have chaotic trajectories. Because the coefficient 5 is greater than 4, the typical trajectory quickly escapes the unit interval and henceforth heads for $-\infty$.
Added 12/19/11: An elementary mathoverflow question that was opened and closed yesterday made me realize there's an interpretation for the function $w(i)$ that makes it easy to see the equivalence of the OP's recursive formula and the "explicit" formula I came up with: The function $w$ counts the number of maps from a set of $n$ objects onto a set of $i$ objects (with $n$ assumed greater than or equal to $i$). The OP's recursive formula does the count by starting with the number of all maps ($i^n$) and subtracting the number that are mappings onto subsets of size $j$ for $j=1$ to $i-1$. My explicit formula is an inclusion-exclusion count of all maps (not necessarily onto) from a set of size $n$ to sets of size $i-j$. The only little mystery from this point of view is why the equivalence works for all $n$, not just $n\ge i$.
