For which Natural $n\ge2: \phi(n)=n/2$ For which Natural $n\ge2$ does this occur with?: $\phi(n)=n/2$

Suppose $n=2^sd$, where $d$ is odd. If $\phi(n)=n/2$, then $2|n$, forcing $s$ to be greater than zero always. Thus $\phi(n)= \phi(2^s)\phi(d)$.This can happen only if $d=1$, since $\phi(d)\le d-1$ for $d>1$. Therefore, $\phi(n)=n/2$ iff $n=2^s$ for some $s>0$.
