 Integer rounding in a divide-and-conquer recurrence

An integer-valued problem size requires integer subproblem sizes and a base case. For the displayed unequal split with $n\ge 2$, both children have positive size less than $n$, and $1/4\le m/n\le 1/2$. The entropy potential for a divide-and-conquer recurrence therefore supplies a uniform positive potential decrease. The squared sizes satisfy $m^2+(n-m{)}^2\le 5n^2/8$, which similarly closes a quadratic inductive hypothesis when the work at the root is $n^2$.