4 ms·
yes, but note that those two are the same when C (2k) is applied as well. C (2k + 1) = C (3k + 2) saves you one step of computation, which can save you a lot of
by mattashii 5y ago
yes, but note that those two are the same when C (2k) is applied as well. C (2k + 1) = C (3k + 2) saves you one step of computation, which can save you a lot of time if you're applying the steps.