A348212 Number of transversals in a cyclic diagonal Latin square of order 2n+1.

1, 0, 15, 133, 0, 37851, 1030367, 0, 1606008513, 87656896891, 0, 452794797220965, 41609568918940625

Offset: 1

Author

Eduard I. Vatutin, Oct 07 2021

Keywords

• nonn
• more
• hard

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

Examples

			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 . .
		

Links

• Eduard I. Vatutin, About the number of transversals in cyclic Latin and cyclic diagonal Latin squares (in Russian).
• Index entries for sequences related to Latin squares and rectangles.

Crossrefs

Cf. A006717, A011655, A287644, A287645, A338562, A342997, A342998.

Formula

a(n) = A006717(n) * A011655(n+1).