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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 $n\ge i$.

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 $-\infty$.

2 Align is better.

With regard to the general question, I defer to Joel Hamkins's answer. For your specific function, though, the first few cases give

w(2) = \begin{align} w(2) &= 2^n - 2 $$w(3) = 2 \\ w(3) &= 3^n -3\cdot 2^n + 3$$ $$w(4) = 3\\ w(4) &= 4^n - 4\cdot 3^n + 6\cdot 2^n - 4$$ $$w(5) = 4\\ w(5) &= 5^n - 5\cdot 4^n + 10\cdot 3^n - 10\cdot 2^n + 5$$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.

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