For $k = \lfloor \log_{2}(n+1) \rfloor - 1$ evaluate $n \times \frac{n-1}{2} \times\frac{n-3}{4} \times \frac{n-7}{8} \times \dots \times \frac{n-(2^{k}-1)}{2^k}$
So the product goes up to $k$ and I want to sub in $k = \lfloor \log_{2}(n+1) \rfloor - 1$ to evaluate: $n \times \frac{n-1}{2} \times \dots \times \frac{n-(2^{( \lfloor \log_{2}(n+1) \rfloor -1)}-1}{2^{ \lfloor \log_{2}(n+1) \rfloor -1 } }$
To make computations less disgusting, I think it would be okay to assume that n is such that log(n+1)-1 is an integer.
Any help appreciated!