If $F(n) = 2F(\lfloor\frac{n}{2}\rfloor)$, how do we prove that $F(n) = 2^{\lfloor\log_2(n)\rfloor}$? Let $F(n) = 2F(\lfloor\frac{n}{2}\rfloor)$
The final case is $F(1) = 1$.
How do we prove that $F(n) = 2^{\lfloor\log_2(n)\rfloor}$?
If $n$ happens to be a power of $2$ then $F(n) = n$ but when it's not a power of $2$ I am not sure what the standard method is for handling floor cases because it doesn't seem cleanly simple anywhere. I can show that:
$F(n) = 2^1F(\lfloor\frac{n}{2^1}\rfloor) = 2^2F(\lfloor\frac{n}{2^2}\rfloor) = 2^3F(\lfloor\frac{n}{2^3}\rfloor) = 2^kF(\lfloor\frac{n}{2^k}\rfloor)$
and $k$ ends when $\lfloor\frac{n}{2^k}\rfloor = 1$ which makes $F(1) = 1$ so the answer is clearly $2^{k}$ for some special $k$ but I don't know how to get the $k$.
So maybe my real question is $\lfloor\frac{n}{2^k}\rfloor = 1$, how do we solve for $k$?

Hint: It might be useful to think of $n$ in binary. Check this example:
Let $n = 110101100$, then
\begin{align*}
F(n) &= 10 F(11010110) \\ &= 100 F(1101011) \\ &= 1000F(110101) \\ &= 10000F(11010) \\ &= 100000F(1101) \\ &= 1000000F(110) \\ &= 10000000F(11) \\&= 100000000F(1) \\ &= 100000000. 
\end{align*}
$F$ converts every $1$ after the leftmost one, to a $0$.
