cp's OEIS Frontend

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.

A343867 Number of semicyclic pandiagonal Latin squares of order 2*n+1 with the first row in ascending order.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1560, 0, 34000, 175104, 0, 22417824, 313235960, 0, 83574857328, 1729671003296
Offset: 0

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Author

Andrew Howroyd, May 08 2021

Keywords

Comments

Pandiagonal Latin squares exist only for odd orders not divisible by 3. All pandiagonal Latin squares for orders less than 13 are cyclic which are not counted by this sequence.
Semicyclic Latin squares are defined in the Atkin reference where the first nonzero term of this sequence is given. They are cyclic in a single direction. The direction can be horizontal or vertical or any other step such as a knights move.
Each symbol in a semicyclic Latin square occupies the same pattern of squares up to translation on the torus which in the case of a pandiagonal square is a solution to the toroidal n-queens problem.
For prime 2n+1, a(n) is a multiple of 2n+1.

Examples

			The following is an example of an order 13 semicyclic square with a step of (1,4). This means moving down one row and across by 4 columns increases the cell value by 1 modulo 13. Symbols can be relabeled to give a square with the first row in ascending order.
   0 11  1  7  5  9  3 10  4  8  6 12  2
   9  7  0  3  1 12  2  8  6 10  4 11  5
  11  5 12  6 10  8  1  4  2  0  3  9  7
   1  4 10  8 12  6  0  7 11  9  2  5  3
  10  3  6  4  2  5 11  9  0  7  1  8 12
   8  2  9  0 11  4  7  5  3  6 12 10  1
   7  0 11  2  9  3 10  1 12  5  8  6  4
   6  9  7  5  8  1 12  3 10  4 11  2  0
   5 12  3  1  7 10  8  6  9  2  0  4 11
   3  1  5 12  6  0  4  2  8 11  9  7 10
  12 10  8 11  4  2  6  0  7  1  5  3  9
   2  6  4 10  0 11  9 12  5  3  7  1  8
   4  8  2  9  3  7  5 11  1 12 10  0  6
...
a(12) = 4*(A071607(12) - A123565(25)) + 11240. - _Jim White_, Jul 22 2021
a(14) = 4*(A071607(14) - A123565(29)) + 91176. - _Jim White_, Jul 24 2021
a(15) = 4*(A071607(15) - A123565(31)) + 334800. - _Jim White_, Aug 03 2021
		

Crossrefs

Cf. A071607, A123565 (cyclic), A338620, A343868.

Programs

  • PARI
    \\ See Links

Formula

a(n) >= 4*(A071607(n) - A123565(2*n+1)).

Extensions

a(12)-a(15) from Jim White, Aug 03 2021