# Number of maneuvers for Rubik's Cube

Submitted by Herbert Kociemba on Sat, 05/19/2007 - 10:35.

By accident I just ran across a formula I developed many years ago for the number of maneuvers in FTM which cannot be shortened in a trivial way. I did not see it anywhere else, so maybe it is of some interest.

Let r = Sqrt(6) and k the maneuver length, then we have

N(k) = [(3+r)(6+3r)^n + (3-r)(6-3r)^n]/4

which gives 1, 18, 243, 3240, 43254, ...

Round[(3+r)(6+3r)^n] is a good approximation even for small n. and we see that 6+3r = 13.348... is the asymtotic branching factor.

Let r = Sqrt(6) and k the maneuver length, then we have

N(k) = [(3+r)(6+3r)^n + (3-r)(6-3r)^n]/4

which gives 1, 18, 243, 3240, 43254, ...

Round[(3+r)(6+3r)^n] is a good approximation even for small n. and we see that 6+3r = 13.348... is the asymtotic branching factor.