Solution:
If $a_{1}$ is odd, then we can see by induction that $a_{j}=(j+1) a_{1}$ when $j$ is even and $a_{j}=a_{1}$ when $j$ is odd (using the fact that no even $j$ can divide $a_{1}$). So we have infinitely many $j$'s for which $a_{j}=a_{1}$.

If $a_{1}>2$ is even, then $a_{2}$ is odd, since $a_{2}=a_{1} / 2$, and $a_{1}$ may have only one factor of $2$. Now, in general, let $p=\min (\{p_{1}, \ldots, p_{n}\} \setminus\{2\})$. Suppose $1<j<p$. By induction, we have $a_{j}=(j+1) a_{1} / 2$ when $j$ is odd, and $a_{j}=a_{1} / 2$ when $j$ is even. So $a_{i} \neq a_{1}$ for all $1<j<p$. It follows that $a_{p}=a_{1} / 2 p$. Then, again using induction, we get for all nonnegative integers $k$ that $a_{p+k}=a_{p}$ if $k$ is even, and $a_{p+k}=(p+k+1) a_{p}$ if $k$ is odd. Clearly, $a_{p} \neq a_{1}$ and $p+k+1 \neq 2 p$ when $k$ is odd (the left side is odd, and the right side even). It follows that $a_{j}=a_{1}$ for no $j>1$.

Finally, when $a_{1}=2$, we can check inductively that $a_{j}=j+1$ for $j$ odd and $a_{j}=1$ for $j$ even.

So our answer is just the number of odd elements in $S$. There are $9$ odd prime numbers smaller than $30$, so the answer is $2^{9}=512$.