Find the smallest nomial of this sequence that $a_1=1993^{1994^{1995}}$ and 
\[ a_{n+1}=\begin{cases}\frac{a_n}{2}&\text{if $n$ is even}\\a_n+7 &\text{if $n$ is odd.} \end{cases} \]
1. **Clarify the sequence definition:**
   The sequence is defined as follows:
   \[
   a_{n+1} = \begin{cases}
   \frac{a_n}{2} & \text{if } a_n \text{ is even}, \\
   a_n + 7 & \text{if } a_n \text{ is odd}.
   \end{cases}
   \]
   The initial term is \( a_1 = 1993^{1994^{1995}} \).

2. **Determine the parity of \( a_1 \):**
   Since \( 1993 \) is an odd number, any power of \( 1993 \) will also be odd. Therefore, \( a_1 \) is odd.

3. **Behavior of the sequence:**
   - If \( a_n \) is odd, then \( a_{n+1} = a_n + 7 \), which is also odd.
   - If \( a_n \) is even, then \( a_{n+1} = \frac{a_n}{2} \).

4. **Reduction to smaller values:**
   - When \( a_n \) is odd, it will eventually become even after adding 7 enough times.
   - Once \( a_n \) is even, it will be repeatedly halved until it becomes odd again.

5. **Modulo 7 analysis:**
   - Since \( a_1 = 1993^{1994^{1995}} \), we need to find \( 1993 \mod 7 \).
   - \( 1993 \equiv 4 \mod 7 \) (since \( 1993 = 7 \times 284 + 4 \)).

6. **Quadratic residues modulo 7:**
   - The quadratic residues modulo 7 are \( 0, 1, 2, 4 \).
   - Since \( a_1 \equiv 4 \mod 7 \), all terms in the sequence will be quadratic residues modulo 7.

7. **Possible minimum values:**
   - The possible minimum values are \( 1 \) and \( 4 \) (since \( 2 \) and \( 0 \) are not achievable by the sequence rules starting from \( 4 \mod 7 \)).
   - We need to check if the sequence can reach 1.

8. **Sequence behavior towards 1:**
   - If \( a_n \equiv 4 \mod 7 \), then \( a_n \) can be reduced to 1 by repeated halving (since \( 4 \) is even).
   - For example, \( 4 \rightarrow 2 \rightarrow 1 \).

Therefore, the smallest nominal value in the sequence is \( 1 \).

The final answer is \( \boxed{1} \).