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Mail-from: From kociemba@hrz1.hrz.th-darmstadt.de Sun Sep 7 17:51:08 1997
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Date: Sun, 07 Sep 1997 00:20:36 +0200
From: Herbert Kociemba
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