jmc

We rewrite the given equation as $$5^{a_{n+1}-a_n}=1+\frac{1}{n+\frac{2}{3}}=\frac{3n+5}{3n+2}.$$Then, we observe a telescoping product: $$\begin{array}{rl}{5}^{a_n-a_1}& =5^{a_2-a_1}\cdot 5^{a_3-a_2}\cdots 5^{a_n-a_{n-1}}\\ & =\frac{8}{5}\cdot \frac{11}{8}\cdots \frac{3n+2}{3n-1}\\ & =\frac{3n+2}{5}.\end{array}$$Since $a_1=1$, we have $$5^{a_n}=3n+2$$for all $n\ge 1$. Thus, $a_k$ is an integer if and only if $3k+2$ is a power of $5$. The next power of $5$ which is of the form $3k+2$ is $5^3=125$, which is $3(41)+2$. Thus $k=\boxed{41}$.