A236757 Number T(n,k) of equivalence classes of ways of placing k 4 X 4 tiles in an n X n square under all symmetry operations of the square; irregular triangle T(n,k), n>=4, 0<=k<=floor(n/4)^2, read by rows.

1, 1, 1, 1, 1, 3, 1, 3, 1, 6, 9, 3, 1, 1, 6, 29, 35, 14, 1, 10, 75, 209, 174, 1, 10, 147, 765, 1234, 1, 15, 270, 2340, 7639, 6169, 1893, 242, 17, 1, 1, 15, 438, 5806, 34342, 79821, 80722, 36569, 7106, 459

Offset: 4

Author

Christopher Hunt Gribble, Jan 30 2014

Keywords

• nonn
• tabf

Comments

The first 10 rows of T(n,k) are:

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

n

4 1 1

5 1 1

6 1 3

7 1 3

8 1 6 9 3 1

9 1 6 29 35 14

10 1 10 75 209 174

11 1 10 147 765 1234

12 1 15 270 2340 7639 6169 1893 242 17 1

13 1 15 438 5806 34342 79821 80722 36569 7106 459

Examples

			T(8,3) = 3 because the number of equivalence classes of ways of placing 3 4 X 4 square tiles in an 8 X 8 square under all symmetry operations of the square is 3. The portrayal of an example from each equivalence class is:
._____________          _____________          _____________
|      |      |        |      |______|        |      |      |
|   .  |   .  |        |   .  |      |        |   .  |______|
|      |      |        |      |   .  |        |      |      |
|______|______|        |______|      |        |______|   .  |
|      |      |        |      |______|        |      |      |
|   .  |      |        |   .  |      |        |   .  |______|
|      |      |        |      |      |        |      |      |
|______|______|        |______|______|        |______|______|
		

Links

• Christopher Hunt Gribble, C++ program

Crossrefs

Cf. A054252, A236679, A236560, A236800, A236829, A236865, A236915, A236936, A236939.

Formula

It appears that:

T(n,0) = 1, n>= 4

T(n,1) = (floor((n-4)/2)+1)*(floor((n-4)/2+2))/2, n >= 4

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

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

T(c+2*4,3) = (c+1)(c+2)/2(2*A002623(c-1)*floor((4-c-1)/2) + A131941(c+1)*floor((4-c)/2)) + S(c+1,3c+2,3), 0 <= c < 4 where

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

A054252(2,3), c = 0

A236679(5,3), c = 1

A236560(8,3), c = 2

A236757(11,3), c = 3