Find the smallest term of the sequence $a_1, a_2, a_3, \ldots$ defined by $a_1=2014^{2015^{2016}}$ and
$$
a_{n+1}= 
\begin{cases}
\frac{a_n}{2} & \text{ if } a_n \text{ is even} \\
a_n + 7 & \text{ if } a_n \text{ is odd} \\
\end{cases}
$$
1. **Initial Term Analysis**:
   The sequence starts with \( a_1 = 2014^{2015^{2016}} \). We need to determine the smallest term in the sequence defined by the recurrence relation:
   \[
   a_{n+1} = 
   \begin{cases}
   \frac{a_n}{2} & \text{if } a_n \text{ is even} \\
   a_n + 7 & \text{if } a_n \text{ is odd}
   \end{cases}
   \]

2. **Behavior of the Sequence**:
   - If \( a_n \) is even, the next term \( a_{n+1} \) is \( \frac{a_n}{2} \).
   - If \( a_n \) is odd, the next term \( a_{n+1} \) is \( a_n + 7 \).

3. **Modulo Analysis**:
   To understand the behavior of the sequence, we analyze \( 2014^{2015^{2016}} \mod 7 \):
   - Note that \( 2014 \equiv 5 \pmod{7} \).
   - Therefore, \( 2014^{2015^{2016}} \equiv 5^{2015^{2016}} \pmod{7} \).

4. **Cycle of Powers Modulo 7**:
   The powers of 5 modulo 7 cycle every 6 terms:
   \[
   \begin{aligned}
   5^1 &\equiv 5 \pmod{7}, \\
   5^2 &\equiv 25 \equiv 4 \pmod{7}, \\
   5^3 &\equiv 20 \equiv 6 \pmod{7}, \\
   5^4 &\equiv 30 \equiv 2 \pmod{7}, \\
   5^5 &\equiv 10 \equiv 3 \pmod{7}, \\
   5^6 &\equiv 15 \equiv 1 \pmod{7}.
   \end{aligned}
   \]
   Since \( 2015^{2016} \mod 6 \) determines the exponent modulo 7, we need to find \( 2015^{2016} \mod 6 \):
   - \( 2015 \equiv -1 \pmod{6} \).
   - \( (-1)^{2016} \equiv 1 \pmod{6} \).

5. **Final Modulo Calculation**:
   Thus, \( 5^{2015^{2016}} \equiv 5^1 \equiv 5 \pmod{7} \).

6. **Sequence Modulo 7**:
   Since \( a_1 \equiv 5 \pmod{7} \), all terms in the sequence will be congruent to 5 modulo 7. This is because:
   - If \( a_n \equiv 5 \pmod{7} \) and \( a_n \) is even, \( a_{n+1} = \frac{a_n}{2} \equiv \frac{5}{2} \pmod{7} \), which is not possible since 5 is odd.
   - If \( a_n \equiv 5 \pmod{7} \) and \( a_n \) is odd, \( a_{n+1} = a_n + 7 \equiv 5 + 7 \equiv 5 \pmod{7} \).

7. **Smallest Term Analysis**:
   The smallest term in the sequence must be among the numbers congruent to 5 modulo 7. The smallest positive integer congruent to 5 modulo 7 is 5 itself. However, since the sequence is defined by the given recurrence relations, we need to check if 3 can appear in the sequence.

8. **Checking for 3**:
   - If \( a_n = 3 \), then \( a_{n+1} = 3 + 7 = 10 \), and \( a_{n+2} = \frac{10}{2} = 5 \).
   - Since 5 is congruent to 5 modulo 7, and 3 is not congruent to 5 modulo 7, 3 cannot appear in the sequence.

Therefore, the smallest term in the sequence is \( \boxed{5} \).