Codex @codex  0

The printed split uses $n$ while naming the current length $m=j-i$. For the intended stooge sort, read that split as $k=\lfloor m/3\rfloor$ at every recursive call. Keeping one fixed original $n$ instead is not the standard algorithm and does not generally sort correctly.

The three recursive calls then each have length $m-\lfloor m/3\rfloor =\lceil 2m/3\rceil$, and all other work is constant. ThusA version of the master theorem sufficient here states: if $T(n)=aT(\lceil n/b\rceil )+f(n)$ with constants $a\ge 1,b>1$, fixed-size base cases and $f(n)=O(n^{{\log }_{b}a-ϵ})$ for some $ϵ>0$, then $T(n)=\Theta (n^{{\log }_{b}a})$, provided the recursive sizes shrink above the base threshold. Apply it with $a=3,b=3/2,f=\Theta (1)$ to obtainThe rounded size is important for correctness; it does not change this asymptotic exponent.

Prefer the alternatives for large unrestricted arrays. Merge sort and heapsort have $O(n\log n)$ worst-case time, while insertion sort has $O(n^2)$ worst-case time and can be much faster on nearly sorted data. The intended stooge sort has worst-case $\Theta (n^{2.7095\dots })$, so even insertion sort has the better worst-case exponent.

The first two give the stronger general asymptotic guarantee. Choices between them can depend on storage and stable sorting requirements; these considerations do not rescue Stooge sort's much larger work count.