 Squaring recurrence under integer halving

For $T(n)=1$ at $n\le 3$ and $T(n)=2T(\lfloor n/2\rfloor {)}^2$ otherwise, put $d=\max (0,\lfloor {\log }_{2}n\rfloor -1)$ for positive integer $n$. Taking logarithms gives $S(n)=1+2S(\lfloor n/2\rfloor )$, hence $S(n)=2^d-1$ and $T(n)=2^{2^d-1}$. These constant-value plateaux have exponential envelopes $2^{\Theta (n)}$, but do not justify $\Theta (2^{n/2})$ for all integer inputs.