10. (14 points) The sequence $\left\{a_{n}\right\}$ is defined as follows: $a_{1}=1$, and for $n \geqslant 2$,
$a_{n}=\left\{\begin{array}{ll}a_{\frac{n}{2}}+1, & \text { when } n \text { is even; } \\ \frac{1}{a_{n-1}}, & \text { when } n \text { is odd. }\end{array}\right.$
It is known that $a_{n}=\frac{30}{19}$. Find the positive integer $n$.
10. From the given conditions, it is easy to know that $a_{n}>0(n=1,2, \cdots)$.

From $a_{1}=1$, we know that:
when $n$ is even, $a_{n}>1$;
when $n(n>1)$ is odd, $a_{n}=\frac{1}{a_{n-1}}1$, so $n$ is even.
Thus, $a_{\frac{n}{2}}=\frac{30}{19}-1=\frac{11}{19}1, \frac{n}{2}-1$ is even;
$a_{\frac{n-2}{4}}=\frac{19}{11}-1=\frac{8}{11}1, \frac{n-6}{4}$ is even;
$a_{\frac{n-6}{8}}=\frac{11}{8}-1=\frac{3}{8}1, \frac{n-14}{8}$ is even;
$a_{\frac{n-14}{16}}=\frac{8}{3}-1=\frac{5}{3}>1, \frac{n-14}{16}$ is even;
$a_{\frac{n-14}{32}}=\frac{5}{3}-1=\frac{2}{3}1, \frac{n-46}{32}$ is even;
$a_{\frac{n-46}{64}}=\frac{3}{2}-1=\frac{1}{2}1, \frac{n-110}{64}$ is even;
$a_{\frac{n-110}{128}}=2-1=1$.
Therefore, $\frac{n-110}{128}=1$. Solving for $n$ gives $n=238$.