A236560 Number T(n,k) of equivalence classes of ways of placing k 3 X 3 tiles in an n X n square under all symmetry operations of the square; irregular triangle T(n,k), n>=3, 0<=k<=floor(n/3)^2, read by rows.

1, 1, 1, 1, 1, 3, 1, 3, 6, 2, 1, 1, 6, 21, 29, 14, 1, 6, 53, 161, 174, 1, 10, 111, 665, 1713, 1549, 608, 107, 11, 1, 1, 10, 201, 1961, 9973, 24267, 29437, 17438, 4756, 459

Offset: 3

Author

Christopher Hunt Gribble, Jan 30 2014

Keywords

• tabf
• nonn

Comments

The first 8 rows of T(n,k) are:

.\ k 0 1 2 3 4 5 6 7 8 9

n

3 1 1

4 1 1

5 1 3

6 1 3 6 2 1

7 1 6 21 29 14

8 1 6 53 161 174

9 1 10 111 665 1713 1549 608 107 11 1

10 1 10 201 1961 9973 24267 29437 17438 4756 459

Examples

			T(6,2) = 6 because the number of equivalence classes of ways of placing 2 3 X 3 square tiles in a 6 X 6 square under all symmetry operations of the square is 6. The portrayal of an example from each equivalence class is:
.___________      ___________      ___________
|     |     |    |     |_____|    |     |     |
|  .  |  .  |    |  .  |     |    |  .  |_____|
|_____|_____|    |_____|  .  |    |_____|     |
|           |    |     |_____|    |     |  .  |
|           |    |           |    |     |_____|
|___________|    |___________|    |_____|_____|
.
.___________      ___________      ___________
|     |     |    |_____ _____|    |_____      |
|  .  |     |    |     |     |    |     |_____|
|_____|_____|    |  .  |  .  |    |  .  |     |
|     |     |    |_____|_____|    |_____|  .  |
|     |  .  |    |           |    |     |_____|
|_____|_____|    |___________|    |_____|_____|
		

Links

• Christopher Hunt Gribble, C++ program

Crossrefs

Cf. A054252, A236679, A236757, A236800, A236829, A236865, A236915, A236936, A236939.

Formula

It appears that:

T(n,0) = 1, n>= 3

T(n,1) = (floor((n-3)/2)+1)*(floor((n-3)/2+2))/2, n >= 3

T(c+2*3,2) = A131474(c+1)*(3-1) + A000217(c+1)*floor(3^2/4) + A014409(c+2), 0 <= c < 3, c even

T(c+2*3,2) = A131474(c+1)*(3-1) + A000217(c+1)*floor((3-1)(3-3)/4) + A014409(c+2), 0 <= c < 3, c odd

T(c+2*3,3) = (c+1)(c+2)/2(2*A002623(c-1)*floor((3-c-1)/2) + A131941(c+1)*floor((3-c)/2)) + S(c+1,3c+2,3), 0 <= c < 3 where

S(c+1,3c+2,3) =

A054252(2,3), c = 0

A236679(5,3), c = 1

A236560(8,3), c = 2