\item You are given a function F : $\mathbb{Z} -> \mathbb{Z}^+$\\ 
where F ( n ) =$$\begin{cases} 1 & \text{if n is 1} \\
F$(n/2)$ & \text{if n is even }\\ 1+ F(3n-8)& \text{if n is odd and n$ > $1 }
\end{cases}$$\\
for each integer n $\ge 1$. \\
\item If F well defined?

To determine if the function $ F: \mathbb{Z} \rightarrow \mathbb{Z}^+ $ is well-defined, we need to check whether the function produces a unique output for each input $ n \geq 1 $.

The function is defined as follows:

\[
F(n) =
\begin{cases} 
1 & \text{if } n = 1 \\
F\left(\frac{n}{2}\right) & \text{if } n \text{ is even} \\
1 + F(3n - 8) & \text{if } n \text{ is odd and } n > 1
\end{cases}
\]

### Analysis of Each Case

1. **Base Case ($ n = 1 $):**
   - When $ n = 1 $, $ F(1) = 1 $. This is a fixed value and poses no issue.

2. **Even $ n $:**
   - If $ n $ is even, $ F(n) = F\left(\frac{n}{2}\right) $.
   - This means that the function value for even $ n $ is determined by the function value at $ \frac{n}{2} $.
   - Since $ n $ is halved each time, eventually $ n $ will reach 1, where $ F(1) = 1 $ is already defined. Thus, for even numbers, the function is well-defined and finite because the halving process will always reach 1.

3. **Odd $ n > 1 $:**
   - If $ n $ is odd and greater than 1, $ F(n) = 1 + F(3n - 8) $.
   - Here, we need to consider the sequence produced by repeatedly applying this function.
   - The sequence $ 3n - 8 $ for odd $ n $ can either increase or decrease depending on the value of $ n $. The critical point is whether or not this sequence will eventually reach an even number or 1, which would then lead the sequence to eventually terminate.

### Behavior of $ 3n - 8 $ for Odd $ n > 1 $

- **If $ n = 3 $:** 
  - $ F(3) = 1 + F(3(3) - 8) = 1 + F(9 - 8) = 1 + F(1) = 1 + 1 = 2 $.
  
- **If $ n = 5 $:**
  - $ F(5) = 1 + F(3(5) - 8) = 1 + F(15 - 8) = 1 + F(7) $.
  - Continuing this process for $ n = 7 $, $ n = 9 $, etc., it can be complex. But in each case, we need to check if the function eventually leads to an even $ n $ or reaches 1.

### Conclusion

The function $ F(n) $ will be well-defined if the sequences generated by the rules for odd $ n > 1 $ either eventually reach 1 or reach an even $ n $ which then leads to 1. The behavior of these sequences can be intricate, but for most small values of $ n $, the sequences do reach 1. However, without a general proof that $ 3n - 8 $ always reduces to an even number or 1 for all $ n $, we cannot guarantee $ F(n) $ is well-defined for all $ n \geq 1 $.

Therefore, **$ F $ is well-defined for many cases, but it requires a deeper analysis to prove definitively for all $ n $.** The general behavior of the sequence $ 3n - 8 $ for arbitrary odd $ n $ would need to be thoroughly explored to ensure that it does not produce an infinite loop without reaching an even number or 1.

---

Would you like to explore this behavior for specific values or go deeper into proving the general case? Here are some related questions:

1. What happens to $ F(n) $ when $ n $ is a large odd number?
2. Can you show that $ F(n) $ reaches 1 for all small values of $ n $?
3. How does the behavior of $ 3n - 8 $ differ for different values of $ n $?
4. What would be the function values for specific $ n $, such as $ F(7) $ or $ F(9) $?
5. How can we construct a proof to determine if the function is well-defined for all integers $ n \geq 1 $?

**Tip:** Analyzing recursive functions often involves understanding the behavior of sequences they generate. Working through small examples can offer valuable insights.

Explore the function F(n) defined recursively for integers n >= 1. This analysis covers the behavior of F(n) for even and odd integers, examining sequences and termination conditions. Suitable for advanced high school students and college-level learners interested in recursive functions and number theory.

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