Squarefree-part bound for smooth numbers
= Squarefree-part bound for smooth numbers
If $S$ is a finite set of <prime numbers>, every <S-smooth number> has a unique expression $n=ab^2$, where $a$ is <squarefree> and its <prime factors> lie in $S$. There are at most $2^{|S|}$ choices for $a$ and at most $\lfloor\sqrt x\rfloor$ choices for $b$ when $n\le x$. Consequently $f_S(x)\le2^{|S|}\sqrt x$.