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 A225790 Rubik's Square sequence. 1
 1, 24, 96, 165888, 663552, 165112971264, 660451885056, 23665185138564661248, 94660740554258644992, 488428629217633346355864797184, 1953714516870533385423459188736, 1451626239969468099340993140755597642170368, 5806504959877872397363972563022390568681472 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS The n-th term of this sequence is by definition the order of the group of admissible positions of the n X n Rubik's Square. The group is generated by 2n elements of order 2: 180-degree rotations of n columns, and 180-degree rotations of n rows. It is naturally a subgroup of the symmetric group on n^2 symbols. The sequence was discovered by Carlos Aguirre during a semester project supervised by Sean D. Lawton. Proof of the formula: for any n>0, we split the square of 2n X 2n to four parts. It is easy to show that an element (x, y) has only four places to arrive: ((x, y), (2n-x+1,y), (x,2n-y+1) and (2n-x+1, 2n-y+1)), we call them a, b, c, d. By some rotation we can make any even permutation of abcd, without changing other elements. So we can pair rows i and (2n-i+1), and enumerate which of them are reversed by an odd number; and we can pair columns j and (2n-j-1), enumerate which of them are reversed by an odd number. Then each quad tuple (a, b, c, d) is determined to have an odd permutation, or even permutation; so they have 12 permutations to choose. This gives 12^(n^2) * 2^(2n) permutations, but reversing all columns and rows gives the same permutations, so there are 12^(n^2) * 2^(2n-1) permutations in total. - Qingyu Ren, Aug 12 2019 LINKS Qingyu Ren, Table of n, a(n) for n = 1..60 (first 25 terms from Eric M. Schmidt) FORMULA For any n>0, a(2*n+1) = 4*a(2*n). - Eric M. Schmidt, May 24 2013 Conjecture: a(2*n) = 2^A142463(n) * 3^(n^2) = 2^A142463(n) * 3^A000290(n). - Eric M. Schmidt, Nov 05 2013 For any n>0, a(2*n) = 12^(n^2) * 2^(2n-1) = 2^A142463(n) * 3^(n^2). - Qingyu Ren, Aug 12 2019 EXAMPLE Draw an square n X n array of squares (one face of an n X n X n Rubik's cube). Starting with 1, number the n^2 squares of the array from left to right and from top to bottom. One is allowed to permute this labeling by a finite succession of 180-degree rotations of rows or columns. To compute the terms of the sequence, compute the order of the group of allowed positions. The 1 X 1 case corresponds to the trivial group and so its order is 1: the first term. Here are computations for the next three terms of this sequence using the computer program GAP: gap> G2:=Group((1,2),(3,4),(1,3),(2,4)); gap> Order(G2); 24 gap> G3:=Group((1,3),(4,6),(7,9),(1,7),(2,8),(3,9)); gap> Order(G3); 96 gap> G4:=Group((1,4)(2,3), (5,8)(6,7), (9,12)(10,11), (13,16)(14,15), (1,13)(5,9), (2,14)(6,10), (3,15)(7,11), (4,16)(8,12)); gap> Order(G4); 165888 PROG (GAP) A225790 := n -> Size(grp(n)); grp := n -> Group(Concatenation(List([1, n+1..n^2-n+1], s->flip(s, n, 1)), List([1..n], s->flip(s, n, n)))); flip := function(start, nterms, skip) return Product([1..Int(nterms/2)], m->(start + skip*(m - 1), start + skip*(nterms - m)), ()); end; # Eric M. Schmidt, Nov 05 2013 (Haskell) a225790 1 = 1 a225790 n = 12 ^ (n1 * n1) * 2 ^ (2 * n1 - 1) * k   where     n1 = div n 2     k = if odd n then 4 else 1 -- Qingyu Ren, Aug 12 2019 (Python) a1, n = 1, 1 print(n, a1) while n < 12:     n = n+1     if n%2 == 0:         nn = n//2         a = 2**(2*nn*nn+2*nn-1)*3**(nn*nn)         a1 = a     else:         a = 4*a1     print(n, a) # A.H.M. Smeets, Aug 15 2019 CROSSREFS Sequence in context: A272871 A319577 A277563 * A042122 A042124 A042126 Adjacent sequences:  A225787 A225788 A225789 * A225791 A225792 A225793 KEYWORD nonn AUTHOR Sean D Lawton, May 16 2013 EXTENSIONS More terms from Eric M. Schmidt, Nov 05 2013 STATUS approved

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Last modified August 17 23:13 EDT 2022. Contains 356204 sequences. (Running on oeis4.)