Floor function of powers of $2$ 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)

Using identity:
$$\lfloor x\rfloor =x-\frac{1}{2}+\frac{\sum _{k=1}^{\infty } \frac{\sin (2 k \pi  x)}{k}}{\pi }$$
Solve sum ,substitute x=2^(n/2) and raise to the power 2 we have:
$$\left\lfloor 2^{n/2}\right\rfloor ^2=\frac{\left(\left(-1+2^{1+\frac{n}{2}}\right) \pi -i \log \left(1-e^{-i
   2^{1+\frac{n}{2}} \pi }\right)+i \log \left(1-e^{i 2^{1+\frac{n}{2}} \pi }\right)\right)^2}{4 \pi ^2} = 1/4\,{\frac { \left( {2}^{1+n/2}\pi-2\,{\rm arccot} \left(\cot \left( 
\pi\,{2}^{n/2} \right) \right) \right) ^{2}}{{\pi}^{2}}}
$$
and:
$$\left\lfloor 2^{\frac{n}{2}-1}\right\rfloor ^2=\frac{\left(\left(-1+2^{n/2}\right) \pi -i \log \left(1-e^{-i 2^{n/2}
   \pi }\right)+i \log \left(1-e^{i 2^{n/2} \pi }\right)\right)^2}{4 \pi ^2} = 1/4\,{\frac {1}{{\pi}^{2}} \left( \pi\,{2}^{n/2}-\pi-2\,\arctan
 \left( {\frac {\sin \left( \pi\,{2}^{n/2} \right) }{-1+\cos \left( 
\pi\,{2}^{n/2} \right) }} \right)  \right) ^{2}}
$$
