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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

User: Sean D Lawton

Sean D Lawton's wiki page.

Sean D Lawton has authored 2 sequences.

A237623 Two-Special Pairs in a Free Group.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 2, 6, 14, 30, 62, 129, 252, 499, 984, 1884, 3624, 7025, 13358, 25694, 49316
Offset: 1

Views

Author

Sean D Lawton, Feb 10 2014

Keywords

Comments

a(n) is the number of pairs of positive words of length n in a free group of rank 2 such that the words are not conjugate and they have the same SL(2,C) trace functions. This sequence was discovered by Julian Caballero, Ruben Espinoza, Carlos Salinas under the supervision of Sean D Lawton.

Crossrefs

Cf. A059076.

A225790 Rubik's Square sequence.

Original entry on oeis.org

1, 24, 96, 165888, 663552, 165112971264, 660451885056, 23665185138564661248, 94660740554258644992, 488428629217633346355864797184, 1953714516870533385423459188736, 1451626239969468099340993140755597642170368, 5806504959877872397363972563022390568681472
Offset: 1

Views

Author

Sean D Lawton, May 16 2013

Keywords

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

Examples

			Draw a 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

Links

Programs

  • 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

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

Extensions

More terms from Eric M. Schmidt, Nov 05 2013

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