OFFSET
1,2
COMMENTS
More precisely, order of group of n X n X n Rubik cube, under assumptions not-s, not-m, not-i.
The three possible assumptions considered here are the following:
s (for n odd) indicates that we are working in the "supergroup" and so take account of twists of the face centers.
m (for n > 3) indicates that the pieces are marked so that we take account of the permutation of the identically-colored pieces on a face.
i (for n > 3) indicates that we are working in the theoretical invisible group and solve the pieces on the interior of the cube as well as the exterior. It is assumed that the M and S traits apply to the interior pieces as if they were on the exterior of a smaller cube.
REFERENCES
Dan Hoey, posting to Cube Lovers List, Jun 24, 1987.
Rowley, Chris, The group of the Hungarian magic cube, in Algebraic structures and applications (Nedlands, 1980), pp. 33-43, Lecture Notes in Pure and Appl. Math., 74, Dekker, New York, 1982.
LINKS
Robert Munafo, Table of n, a(n) for n = 1..27 (first 10 terms from Robert G. Wilson v)
Answers.com, Rubik's Cube.
Isaiah Bowers, How To Solve A Rubik's Cube.
Cube Lovers, Discussions on the mathematics of the cube
Cube Lovers Archive, Mailing List
Cube20.org, God's Number is 20
Christophe Goudey, Information
Jaap Scherphuis, Puzzle Pages
Eric Weisstein's World of Mathematics, Rubik's Cube
Wikipedia, Rubik's Cube
Wikipedia, Professor's Cube
FORMULA
a(1)=1; a(2)=7!*3^6; a(3)=8!*3^7*12!*2^10; a(n)=a(n-2)*24^6*(24!/24^6)^(n-2). - Herbert Kociemba, Dec 08 2016
a(n) = ceiling(3674160*11771943321600^(n mod 2)*620448401733239439360000^floor((n - 2)/2)*3246670537110000^floor(((n - 2)/2)^2)). - Davis Smith, Mar 20 2020
MAPLE
f := proc(n) local A, B, C, D, E, F, G; if n mod 2 = 1 then A := (n-1)/2; F := 0; B := 1; C := 1; D := 0; E := (n+1)*(n-3)/4; G := (n-1)*(n-3)/4; else A := n/2; F := 1; B := 1; C := 0; D := 0; E := n*(n-2)/4; G := (n-2)^2/4; fi; (2^A*((8!/2)*3^7)^B*((12!/2)*2^11)^C*((4^6)/2)^D*(24!/2)^E)/(24^F*((24^6)/2)^G); end;
MATHEMATICA
f[n_] := Block[{a, b, c, d, e, f, g}, If[OddQ@ n, a = (n - 1)/2; b = c = 1; d = f = 0; e = (n + 1) (n - 3)/4; g = (n - 1) (n - 3)/4, a = n/2; b = f = 1; c = d = 0; e = n (n - 2)/4; g = (n - 2)^2/4]; Ceiling[(2^a*((8!/2)*3^7)^b*((12!/2)*2^11)^c*((4^6)/2)^d*(24!/2)^e)/(24^f*((24^6)/2)^g)]]; Array[f, 10] (* Robert G. Wilson v, May 23 2009 *)
f[1]=1; f[2]=7!3^6; f[3]=8!3^7 12!2^10; f[n_]:=f[n-2]*24^6*(24!/24^6)^(n-2); Table[f[n], {n, 1, 10}] (* Herbert Kociemba, Dec 08 2016 *)
f[1]=1; f[n_]:=7!3^6(6*24!!)^(s=Mod[n, 2])24!^(r=(n-s)/2-1)(24!/4!^6)^(r(r+s)); Array[f, 5] (*Herbert Kociemba, Jul 03 2022*)
PROG
(Maxima) A075152(n) := block( if n = 1 then return (1), [a:1, b:1, c:1, d:1, e:1, f:1, g:1], if mod(n, 2) = 1 then ( a : (n-1)/2, f : 0, b : 1, c : 1, d : 0, e : (n+1)*(n-3)/4, g : (n-1)*(n-3)/4 ) else ( a : n/2, f : 1, b : 1, c : 0, d : 0, e : n*(n-2)/4, g : (n-2)^2/4 ), return ( (2^a * ((factorial(8)/2)*3^7)^b * ((factorial(12)/2)*2^11)^c * ((4^6)/2)^d * (factorial(24)/2)^e) / (24^f * ((24^6)/2)^g) ) )$ for i:1 thru 27 step 1 do ( sprint(i, A075152(i)), newline() )$ // Robert Munafo, Nov 12 2014
(PARI) A075152(n)=ceil(3674160*(11771943321600)^(n%2)*620448401733239439360000^floor((n-2)/2)*(3246670537110000)^floor(((n-2)/2)^2)) \\ Davis Smith, Mar 20 2020
CROSSREFS
KEYWORD
nonn
AUTHOR
Warren Power, Sep 05 2002
EXTENSIONS
Entry revised by N. J. A. Sloane, Apr 01 2006
Offset changed to 1 by N. J. A. Sloane, Sep 02 2009
STATUS
editing