OFFSET
1,3
COMMENTS
All cyclic diagonal Latin squares of order n have same number of transversals. A similar statement for diagonal transversals is not true (see A342998 and A342997).
All broken diagonals and antidiagonals of cyclic Latin squares are transversals, so a(n) >= 2*n for all n > 1 for which cyclic diagonal Latin squares exist. - Eduard I. Vatutin, Mar 22 2022
All cyclic diagonal Latin squares are diagonal Latin squares, so A287645(2n+1) <= a(n) <= A287644(2n+1) for all orders in which cyclic diagonal Latin squares exist. - Eduard I. Vatutin, Mar 23 2022
LINKS
EXAMPLE
A cyclic diagonal 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
has a(3)=15 transversals:
0 . . . . 0 . . . . . 1 . . . . . . . 4
. 3 . . . . . . . 1 2 . . . . . 3 . . .
. . 1 . . . . . 2 . . . . . 3 . . . 2 .
. . . 4 . . . 3 . . . . . 4 . 1 . . . .
. . . . 2 . 4 . . . . . 0 . . ... . . 0 . .
CROSSREFS
KEYWORD
nonn,more,hard
AUTHOR
Eduard I. Vatutin, Oct 07 2021
STATUS
approved