Jonathan has a magical coin machine which takes coins in amounts of $7, 8$, and $9$. If he puts in $7$ coins, he gets $3$ coins back; if he puts in $8$, he gets $11$ back; and if he puts in $9$, he gets $4$ back. The coin machine does not allow two entries of the same amount to happen consecutively. Starting with $15$ coins, what is the minimum number of entries he can make to end up with $4$ coins?
1. **Understanding the Problem:**
   - Jonathan starts with 15 coins.
   - He can make entries of 7, 8, or 9 coins.
   - Each entry has a specific outcome:
     - Entry of 7 coins: loses 4 coins (15 - 7 + 3 = 11 coins).
     - Entry of 8 coins: gains 3 coins (15 - 8 + 11 = 18 coins).
     - Entry of 9 coins: loses 5 coins (15 - 9 + 4 = 10 coins).
   - The goal is to reach exactly 4 coins with the minimum number of entries.
   - Consecutive entries of the same amount are not allowed.

2. **Working Backwards:**
   - We need to find a sequence of entries that leads to exactly 4 coins.
   - Let's consider the possible states before reaching 4 coins:
     - If he has 4 coins, he could have:
       - Gained 3 coins from having 1 coin (not possible since he can't have less than 7 coins to make an entry).
       - Gained 4 coins from having 0 coins (not possible since he can't have less than 7 coins to make an entry).
       - Lost 4 coins from having 8 coins.
       - Lost 5 coins from having 9 coins.

3. **Case Analysis:**
   - **Case 1: Starting from 8 coins:**
     - To reach 8 coins, he could have:
       - Gained 3 coins from having 5 coins (not possible since he can't have less than 7 coins to make an entry).
       - Gained 4 coins from having 4 coins (not possible since he can't have less than 7 coins to make an entry).
       - Lost 4 coins from having 12 coins.
       - Lost 5 coins from having 13 coins.
   - **Case 2: Starting from 9 coins:**
     - To reach 9 coins, he could have:
       - Gained 3 coins from having 6 coins (not possible since he can't have less than 7 coins to make an entry).
       - Gained 4 coins from having 5 coins (not possible since he can't have less than 7 coins to make an entry).
       - Lost 4 coins from having 13 coins.
       - Lost 5 coins from having 14 coins.

4. **Finding the Minimum Moves:**
   - **From 15 coins:**
     - Entry of 7 coins: 15 - 7 + 3 = 11 coins.
     - Entry of 8 coins: 15 - 8 + 11 = 18 coins.
     - Entry of 9 coins: 15 - 9 + 4 = 10 coins.
   - **From 11 coins:**
     - Entry of 7 coins: 11 - 7 + 3 = 7 coins.
     - Entry of 8 coins: 11 - 8 + 11 = 14 coins.
     - Entry of 9 coins: 11 - 9 + 4 = 6 coins.
   - **From 10 coins:**
     - Entry of 7 coins: 10 - 7 + 3 = 6 coins.
     - Entry of 8 coins: 10 - 8 + 11 = 13 coins.
     - Entry of 9 coins: 10 - 9 + 4 = 5 coins.
   - **From 13 coins:**
     - Entry of 7 coins: 13 - 7 + 3 = 9 coins.
     - Entry of 8 coins: 13 - 8 + 11 = 16 coins.
     - Entry of 9 coins: 13 - 9 + 4 = 8 coins.
   - **From 9 coins:**
     - Entry of 7 coins: 9 - 7 + 3 = 5 coins.
     - Entry of 8 coins: 9 - 8 + 11 = 12 coins.
     - Entry of 9 coins: 9 - 9 + 4 = 4 coins.

5. **Conclusion:**
   - The minimum number of entries to reach 4 coins from 15 coins is 4 moves:
     - 15 coins -> 10 coins (entry of 9 coins)
     - 10 coins -> 13 coins (entry of 8 coins)
     - 13 coins -> 9 coins (entry of 7 coins)
     - 9 coins -> 4 coins (entry of 9 coins)

The final answer is $\boxed{4}$.