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A383570
Number of transversals in pine Latin squares of order 4n.
2
8, 384, 76032, 62881792
OFFSET
1,1
COMMENTS
A pine Latin square is a not necessarily canonical composite Latin square of order N=2*K formed from specially arranged cyclic Latin squares of order K.
By construction, pine Latin square is determined one-to-one by the cyclic square used, so number of pine Latin squares of order N is equal to number of cyclic Latin squares of order N/2.
All pine Latin squares are horizontally symmetric column-inverse Latin squares.
All pine Latin squares for selected order N are isomorphic one to another as Latin squares, so they have same properties (number of transversals, intercalates, etc.).
Pine Latin squares have interesting properties, for example, maximum known number of intercalates (see A383368 and A092237) for some orders N (at least N in {2, 4, 6, 10, 18}).
Pine Latin squares do not exist for odd orders because they must be horizontally symmetric.
Hypothesis: number of transversals in pine Latin squares of all orders N=4k+2 is zero (verified for orders N<=18).
EXAMPLE
For order N=8 pine Latin square
0 1 2 3 4 5 6 7
1 2 3 0 7 4 5 6
2 3 0 1 6 7 4 5
3 0 1 2 5 6 7 4
4 5 6 7 0 1 2 3
5 6 7 4 3 0 1 2
6 7 4 5 2 3 0 1
7 4 5 6 1 2 3 0
has 384 transversals.
.
For order N=10 pine Latin square
0 1 2 3 4 5 6 7 8 9
1 2 3 4 0 9 5 6 7 8
2 3 4 0 1 8 9 5 6 7
3 4 0 1 2 7 8 9 5 6
4 0 1 2 3 6 7 8 9 5
5 6 7 8 9 0 1 2 3 4
6 7 8 9 5 4 0 1 2 3
7 8 9 5 6 3 4 0 1 2
8 9 5 6 7 2 3 4 0 1
9 5 6 7 8 1 2 3 4 0
has no transversals.
.
For order N=12 pine Latin square
0 1 2 3 4 5 6 7 8 9 10 11
1 2 3 4 5 0 11 6 7 8 9 10
2 3 4 5 0 1 10 11 6 7 8 9
3 4 5 0 1 2 9 10 11 6 7 8
4 5 0 1 2 3 8 9 10 11 6 7
5 0 1 2 3 4 7 8 9 10 11 6
6 7 8 9 10 11 0 1 2 3 4 5
7 8 9 10 11 6 5 0 1 2 3 4
8 9 10 11 6 7 4 5 0 1 2 3
9 10 11 6 7 8 3 4 5 0 1 2
10 11 6 7 8 9 2 3 4 5 0 1
11 6 7 8 9 10 1 2 3 4 5 0
has 76032 transversals.
CROSSREFS
KEYWORD
nonn,more,hard
AUTHOR
Eduard I. Vatutin, Apr 30 2025
STATUS
approved