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Date: Sun, 07 Sep 1997 00:20:36 +0200
From: Herbert Kociemba <kociemba@hrz1.hrz.th-darmstadt.de>
To: cube-lovers@ai.mit.edu
Subject: Number of maneuvers with n face turns

The number of maneuvers with 1, 2, 3,.. face turns for Rubik's cube are
of course well known and are 18, 243, 3240... But I did not see a closed
formula for these numbers before, so maybe you find the following
formula interesting:

Let r:= sqrt(6), then you have with n face turns

P(n) = [(3+r)*(6+3r)^n + (3-r)*(6-3r)^n]/4

maneuvers.  Because the second part in brackets is much smaller than the
first, asymptotically you have

(3+r)*(6+3r)^n /4 maneuvers.

Even for small n, this approximation is very good. So for n=3 you get
3240.33 instead of 3240. The asymptotic branching factor P(n+1)/P(n) is
therefore (6+3r), which is about 13.348469 .

Herbert