The sequence $ (a_n)$ satisfies $ a_1 \equal{} 1$ and $ \displaystyle 5^{(a_{n\plus{}1}\minus{}a_n)} \minus{} 1 \equal{} \frac{1}{n\plus{}\frac{2}{3}}$ for $ n \geq 1$. Let $ k$ be the least integer greater than $ 1$ for which $ a_k$ is an integer. Find $ k$.
1. Given the sequence \( (a_n) \) with \( a_1 = 1 \) and the recurrence relation:
   \[
   5^{(a_{n+1} - a_n)} - 1 = \frac{1}{n + \frac{2}{3}}
   \]
   for \( n \geq 1 \).

2. We start by rewriting the recurrence relation:
   \[
   5^{(a_{n+1} - a_n)} = 1 + \frac{1}{n + \frac{2}{3}}
   \]
   Let \( b_n = a_{n+1} - a_n \). Then:
   \[
   5^{b_n} = 1 + \frac{1}{n + \frac{2}{3}}
   \]

3. Taking the natural logarithm on both sides:
   \[
   b_n \ln 5 = \ln \left(1 + \frac{1}{n + \frac{2}{3}}\right)
   \]
   For small \( x \), \( \ln(1 + x) \approx x \). Thus:
   \[
   b_n \ln 5 \approx \frac{1}{n + \frac{2}{3}}
   \]
   \[
   b_n \approx \frac{1}{\ln 5} \cdot \frac{1}{n + \frac{2}{3}}
   \]

4. Summing \( b_n \) from \( n = 1 \) to \( n = k-1 \):
   \[
   a_k - a_1 = \sum_{n=1}^{k-1} b_n
   \]
   \[
   a_k - 1 \approx \frac{1}{\ln 5} \sum_{n=1}^{k-1} \frac{1}{n + \frac{2}{3}}
   \]

5. The sum \( \sum_{n=1}^{k-1} \frac{1}{n + \frac{2}{3}} \) can be approximated by the harmonic series:
   \[
   \sum_{n=1}^{k-1} \frac{1}{n + \frac{2}{3}} \approx \sum_{n=1}^{k-1} \frac{1}{n} = H_{k-1}
   \]
   where \( H_{k-1} \) is the \((k-1)\)-th harmonic number.

6. Using the approximation \( H_{k-1} \approx \ln(k-1) + \gamma \) (where \( \gamma \) is the Euler-Mascheroni constant):
   \[
   a_k - 1 \approx \frac{1}{\ln 5} (\ln(k-1) + \gamma)
   \]

7. We need \( a_k \) to be an integer. Since \( a_1 = 1 \), we need:
   \[
   a_k \approx 1 + \frac{1}{\ln 5} (\ln(k-1) + \gamma)
   \]
   to be an integer.

8. We hypothesize that \( a_n = \log_5(3n + 2) \). Then:
   \[
   a_{n+1} = \log_5(3(n+1) + 2) = \log_5(3n + 5)
   \]
   \[
   a_{n+1} - a_n = \log_5(3n + 5) - \log_5(3n + 2) = \log_5 \left( \frac{3n + 5}{3n + 2} \right)
   \]

9. Given \( 5^{(a_{n+1} - a_n)} = \frac{3n + 5}{3n + 2} \):
   \[
   5^{\log_5 \left( \frac{3n + 5}{3n + 2} \right)} = \frac{3n + 5}{3n + 2}
   \]
   which is consistent with our hypothesis.

10. We need \( 3k + 2 \) to be a power of 5. The smallest power of 5 greater than 1 is 125:
    \[
    3k + 2 = 125
    \]
    \[
    3k = 123
    \]
    \[
    k = 41
    \]

The final answer is \( \boxed{41} \)