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 5, 7 and 11 all pandiagonal Latin squares are cyclic, so a(n) = A338562(n) for n < 6. For n=6 (order 13) this is not true (from Dabbaghian and Wu).
Pandiagonal Latin squares exist only for odd orders not divisible by 3. - Andrew Howroyd, May 26 2021
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.
FORMULA
a(n) = A338620(n) * (2*n+1)!.
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, Mar 08 2021
EXTENSIONS
Zero terms for even orders removed by Andrew Howroyd, May 26 2021
STATUS
approved