Floor-sum rearrangement
= Floor-sum rearrangement
For an <arithmetic function> $a$, changing the order of the finite <summation> gives $\sum_{d=1}^n a(d)\lfloor n/d\rfloor=\sum_{t=1}^n\sum_{d\le\lfloor n/t\rfloor}a(d)$. Both sides count pairs of <positive integers> $(d,t)$ with $dt\le n$, assigning weight $a(d)$ to each pair. This expresses a weighted <floor sum> in terms of a <summatory function>.