Algorithm for fastest division below threshold
The optimal strategy is the greedy one: repeatedly split chunks of size $j$. This strategy results in $\lceil \frac{i}{j} \rceil - 1$ splits. We can prove that this strategy is optimal by induction on $i$. The strategy is clearly optimal for $i \leq j$. Now suppose that we split $i &gt; j$ as $i = i_1 + i_2$. In total, this choice will use up at least this many splits:
$$
1 + \lceil \tfrac{i_1}{j} \rceil - 1 + \lceil \tfrac{i_2}{j} \rceil - 1 = \lceil \tfrac{i_1}{j} \rceil + \lceil \tfrac{i_2}{j} \rceil - 1.
$$
Let $\alpha = i/j$, $\alpha_1 = i_1/j$, $\alpha_2 = i_2/j$, so that $\alpha = \alpha_1 + \alpha_2$. To complete the proof, notice that $\lceil \alpha_1 \rceil + \lceil \alpha_2 \rceil \geq \lceil \alpha \rceil$, and so the quantity above is always at least $\lceil \tfrac{i}{j} \rceil - 1$, as claimed.

I have described one optimal strategy. Using the proof above, we can actually describe all optimal strategies. Splitting $i$ to $i_1+i_2$ is optimal iff $\lceil \frac{i_1}{j} \rceil + \lceil \frac{i_2}{j} \rceil = \lceil \frac{i}{j} \rceil$. When does that happen? Define
$$
i = aj-b, \quad i_1 = a_1j-b_1, \quad i_2 = a_2j-b_2,
$$
where $0 \leq b,b_1,b_2 &lt; j$. A splitting is optimal iff $a_1 + a_2 = a$. Notice that
$$
i_2 + i_2 = (a_1+a_2)j - b_1 - b_2.
$$
This shows that the splitting is optimal iff $b_1 + b_2 = b$ (rather than $b + j$).