Zelaron Gaming Forum - View Single Post - [C#,Math] Anyone think they can do this better?

Goodlookinguy · 07:07 AM

goodlookinGuy said:

I'm trying to see if there's an easier way, mathematically [...]

Not enough math in this "math" thread. Anyway, I found you a closed-form function for your problem:

$\reverse <br /> \mathbf{f}(n) = \sum_{p=0}^7 $ \sum_{k=\frac{2^p + H[p-1] +1}{2}}^{2^p} \(2^{(8-p) H[n-k]} + H[n-2k] + H[n-2k+1] -1$ \)<br />$

where $\reverse H[t]$ is the discrete Heaviside step function defined as

$\reverse H[t]=\begin{cases} 0, & t < 0, \\ 1, & t \ge 0. \end{cases}$