OFFSET
0,3
COMMENTS
A pandiagonal Latin square is a Latin square in which the diagonal, antidiagonal and all broken diagonals and antidiagonals are transversals.
For orders n = 5, 7 and 11 all pandiagonal Latin squares are cyclic, so a(n) = A123565(2n+1) for n < 6. For n=6 (order 13), this is not true and there are 12386 inequivalent squares; of these 10 are cyclic (in all directions) and 1560 are semi-cyclic (A343867).
Pandiagonal Latin squares exist only for odd orders not divisible by 3. This is because the positions of each symbol are a solution to the toroidal n-queens problem which only has solutions for these sizes. - Andrew Howroyd, May 26 2021
The number of semicyclic and cyclic pandiagonal Latin squares given by A123565 and A343867 give the following lower bounds: a(8) >= 34014, a(9) >= 175120, a(11) >= 22417844, a(12) >= 313235970. - Eduard I. Vatutin, Oct 02 2025
LINKS
A. O. L. Atkin, L. Hay, and R. G. Larson, Enumeration and construction of pandiagonal Latin squares of prime order, Computers & Mathematics with Applications, Volume. 9, Iss. 2, 1983, pp. 267-292.
Vahid Dabbaghian and Tiankuang Wu, Constructing non-cyclic pandiagonal Latin squares of prime orders, Journal of Discrete Algorithms 30, 2015.
Vahid Dabbaghian and Tiankuang Wu, Constructing Pandiagonal Latin Squares from Linear Cellular Automaton on Elementary Abelian Groups, Journal of Combinatorial Designs 23(5).
Eduard I. Vatutin, About getting Dabbaghian-Wu pandiagonal Latin squares of orders 13, 19 and 25 (in Russian).
Eduard I. Vatutin, About getting Dabbaghian-Wu pandiagonal Latin squares of orders N=p1*p2 (in Russian).
FORMULA
a(n) = A342306(n) / (2n+1)!. - Eduard I. Vatutin, Jun 13 2021
EXAMPLE
Example of a cyclic pandiagonal Latin square of order 5:
0 1 2 3 4
2 3 4 0 1
4 0 1 2 3
1 2 3 4 0
3 4 0 1 2
Example of a cyclic pandiagonal Latin square of order 7:
0 1 2 3 4 5 6
2 3 4 5 6 0 1
4 5 6 0 1 2 3
6 0 1 2 3 4 5
1 2 3 4 5 6 0
3 4 5 6 0 1 2
5 6 0 1 2 3 4
Example of a cyclic pandiagonal Latin square of order 11:
0 1 2 3 4 5 6 7 8 9 10
2 3 4 5 6 7 8 9 10 0 1
4 5 6 7 8 9 10 0 1 2 3
6 7 8 9 10 0 1 2 3 4 5
8 9 10 0 1 2 3 4 5 6 7
10 0 1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9 10 0
3 4 5 6 7 8 9 10 0 1 2
5 6 7 8 9 10 0 1 2 3 4
7 8 9 10 0 1 2 3 4 5 6
9 10 0 1 2 3 4 5 6 7 8
For order 13 there is a square
7 1 0 3 6 5 12 2 8 9 10 11 4
2 3 4 10 0 7 6 9 12 11 5 8 1
4 11 1 7 8 9 10 3 6 0 12 2 5
6 5 8 11 10 4 7 0 1 2 3 9 12
8 9 2 5 12 11 1 4 3 10 0 6 7
3 6 12 0 1 2 8 11 5 4 7 10 9
10 0 3 2 9 12 5 6 7 8 1 4 11
1 7 10 4 3 6 9 8 2 5 11 12 0
11 4 5 6 7 0 3 10 9 12 2 1 8
5 8 7 1 4 10 11 12 0 6 9 3 2
12 2 9 8 11 1 0 7 10 3 4 5 6
9 10 11 12 5 8 2 1 4 7 6 0 3
0 12 6 9 2 3 4 5 11 1 8 7 10
that is pandiagonal but not cyclic (Dabbaghian and Wu).
CROSSREFS
KEYWORD
nonn,more,hard
AUTHOR
Eduard I. Vatutin, Nov 04 2020
EXTENSIONS
Zero terms for even orders removed by Andrew Howroyd, May 26 2021
STATUS
approved