In the sequence of integers whose sine values approach $0$, does $a_n/a_{n-1}$ have an upper bound? What about $a_n/na_{n-1}$?

Question

Consider the sequence of all positive increasing integers whose sine values monotonically approach $0$.

The first few terms are shown in red:

$|\sin{\color{red}{1}}|\approx0.842$
$|\sin{\color{red}{3}}|\approx0.141$
$|\sin{\color{red}{22}}|\approx8.85\times 10^{-3}$
$|\sin{\color{red}{333}}|\approx8.82\times 10^{-3}$
$|\sin{\color{red}{355}}|\approx3.01\times 10^{-5}$
$|\sin{\color{red}{103993}}|\approx1.91\times 10^{-5}$
$|\sin{\color{red}{104348}}|\approx1.10\times 10^{-5}$
$|\sin{\color{red}{208341}}|\approx8.11\times 10^{-6}$

Here is a graph of $\dfrac{a_n}{a_{n-1}}$ against $n$.

And here is a graph of $\dfrac{a_n}{na_{n-1}}$ against $n$.

Does $\dfrac{a_n}{a_{n-1}}$ have an upper bound, and if so, what is it? If not, does $\dfrac{a_n}{na_{n-1}}$ have an upper bound, and if so, what is it?

I suspect that $\dfrac{a_n}{a_{n-1}}$ has no upper bound, but $\dfrac{a_n}{na_{n-1}}$ has an upper bound of $\dfrac{a_6}{6\times a_5}=\dfrac{103993}{6\times 355}\approx48.8$. If so, then I guess the proof might be related to the (supposed) sameness of the sequence in this question with the sequence of numerators in the simple continued fraction of $\pi$.