When we display a 24 fps movie over a 60 Hz display, from what I know that judder occurs because the unevenness to display every odd frame twice and even frame thrice (11 222 33 444 55 666 ...).
I am not sure how exactly this can be calculated for other combinations.
For example how to display 30 fps over 144 Hz and 24 fps over 165 Hz
Let us assume that we have a display capable of showing some $n$ frames per second which we want to use to watch a movie filmed in $m$ frames per second. I guess you are mostly interested in the case where $m < n$ (the other case, $m > n$ works similarly).
When $m$ divides $n$ evenly, we have no issue as each frame can be repeated exactly $n / m$ times. If however dividing $n$ by $m$ yields a remainder $r$ (that is, $n = km + r$ for some integer $k$) we do not have this option. So to solve this, we try to make the average number of repetitions per frame equal to the ideal $n / m$ by repeating some frames $k = \lfloor n / m \rfloor$ times and others $k + 1 = \lceil n / m \rceil$ times.
Let us study this process for the example you have given. The optimal number of repetitions would be $60/24 = 2 + 12 / 48 = 2.5$, which is not an integer. By using the pattern you mentioned (repeat half of the frames 2 times and the other half 3 times), we achieve an average number of displayed frames per movie frame of $$ \frac{1}{2} \cdot 2 + \frac{1}{2} \cdot 3 = 2.5 $$ which equals the optimum again, cool.
In the general setting, we repeat a fraction $\ell$ of the frames for a longer-than-average amount of times (i.e. $k + 1$ times) and the remaining fraction $1 - s$ for a longer-than-average amount (i.e. $k$ times) to satisfy the generalized equation $$ \frac{n}{m} = (1 - \ell) \cdot k + \ell \cdot (k + 1). $$ Expanding and simplifying the right side of this equation together with using $n = km + r$ yields $$ \frac{n}{m} = k + \frac{r}{m} = k + \ell $$ and we obtain $\ell = r / m$. Hence we can take $m$ frames of the movie and repeat $r$ of those for $k + 1$ times and the remaining ones $k$ times. A practical implementation would use a reduced version of $r / m$ though in order to intersperse the differences in frame repetitions as evenly as possible.
Let us apply this to one of the other scenarios you posed. If we use $m = 30$ and $n = 144$, we find $144 = 4 \cdot 30 + 24$ and thus obtain $k = 4, r = 24$ and $\ell = r / m = 24 / 30 = 4 / 5$. Therefore a suitable pattern would be to repeat each fifth frame 4 times and the other frames 5 times.