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.

A287650 Number of doubly symmetric diagonal Latin squares of order 4n with the first row in ascending order.

Original entry on oeis.org

2, 12288, 81217160478720, 6101215007109090122576072540160
Offset: 1

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Author

Eduard I. Vatutin, May 29 2017

Keywords

Comments

A doubly symmetric square has symmetries in both the horizontal and vertical planes.
The plane symmetry requires one-to-one correspondence between the values of elements a[i,j] and a[N+1-i,j] in a vertical plane, and between the values of elements a[i,j] and a[i,N+1-j] in a horizontal plane for 1 <= i,j <= N. - Eduard I. Vatutin, Alexey D. Belyshev, Oct 09 2017
Belyshev (2017) proved that doubly symmetric diagonal Latin squares exist only for orders N == 0 (mod 4).
Every doubly symmetric diagonal Latin square also has central symmetry. The converse is not true in general. It follows that a(n) <= A293777(4n). - Eduard I. Vatutin, May 26 2021
a(n)/(A001147(n)*2^(n*(4*n-3))) is the number of 2n X 2n grids with two instances of each of 1..n on the main diagonal and in each row and column with the first row in nondescreasing order. - Andrew Howroyd, May 30 2021

Examples

			Doubly symmetric diagonal Latin square example:
  0 1 2 3 4 5 6 7
  3 2 7 6 1 0 5 4
  2 3 1 0 7 6 4 5
  6 7 5 4 3 2 0 1
  7 6 3 2 5 4 1 0
  4 5 0 1 6 7 2 3
  5 4 6 7 0 1 3 2
  1 0 4 5 2 3 7 6
Reflection of all rows is equivalent to the exchange of elements 0 and 7, 1 and 6, 2 and 5, 3 and 4; hence, this square is horizontally symmetric. Reflection of all columns is equivalent to the exchange of elements 0 and 1, 2 and 4, 3 and 5, 6 and 7; hence, the square is also vertically symmetric.
From _Andrew Howroyd_, May 30 2021: (Start)
a(2) = 4*3*1024 = 12288. The 4 base quarter square arrangements are:
  1 1 2 2  1 1 2 2  1 1 2 2  1 1 2 2
  1 2 1 2  1 2 2 1  2 2 1 1  2 2 1 1
  2 1 2 1  2 2 1 1  1 1 2 2  2 2 1 1
  2 2 1 1  2 1 1 2  2 2 1 1  1 1 2 2
(End)
		

Crossrefs

Formula

a(n) = A292517(n) / (4n)!.

Extensions

a(2) corrected by Eduard I. Vatutin, Alexey D. Belyshev, Oct 09 2017
Edited and a(3) from Alexey D. Belyshev added by Max Alekseyev, Aug 23 2018, Sep 07 2018
a(4) from Andrew Howroyd, May 31 2021