Is there any way to know the exact value of
$$\left(\left\lfloor 2^{\frac{n}{2}} \right\rfloor\right)^2$$
for $n$ an integer $n>0$?
When $n$ is even, the solution is trivial, since we do not have to face any fractional part. On the other hand, for $n$ an odd number, it seems difficult to get the exact value.
We could try to get an aporoximation, but using $\lfloor x \rfloor = x + O(1)$ gives a pretty inaccurate result.
Any idea?
Edit: Would it be possible to get it if we knew the value of $$\left(\left\lfloor 2^{\frac{n}{2}-1} \right\rfloor\right)^2$$
? (A recursive fomula)