A236800 Number T(n,k) of equivalence classes of ways of placing k 5 X 5 tiles in an n X n square under all symmetry operations of the square; irregular triangle T(n,k), n>=5, 0<=k<=floor(n/5)^2, read by rows.

1, 1, 1, 1, 1, 3, 1, 3, 1, 6, 1, 6, 12, 3, 1, 1, 10, 40, 44, 14, 1, 10, 97, 245, 174, 1, 15, 193, 925, 1234, 1, 15, 339, 2640, 6124, 1, 21, 555, 6617, 27074, 19336, 4785, 461, 23, 1

Offset: 5

Author

Christopher Hunt Gribble, Jan 31 2014

Keywords

• tabf
• nonn

Comments

The first 11 rows of T(n,k) are:

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

n

5 1 1

6 1 1

7 1 3

8 1 3

9 1 6

10 1 6 12 3 1

11 1 10 40 44 14

12 1 10 97 245 174

13 1 15 193 925 1234

14 1 15 339 2640 6124

15 1 21 555 6617 27074 19336 4785 461 23 1

Examples

			T(10,3) = 3 because the number of equivalence classes of ways of placing 3 5 X 5 square tiles in an 10 X 10 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, A236757, A236829, A236865, A236915, A236936, A236939.

Formula

It appears that:

T(n,0) = 1, n>= 5

T(n,1) = (floor((n-5)/2)+1)*(floor((n-5)/2+2))/2, n >= 5

T(c+2*5,2) = A131474(c+1)*(5-1) + A000217(c+1)*floor(5^2/4) + A014409(c+2), 0 <= c < 5, c even

T(c+2*5,2) = A131474(c+1)*(5-1) + A000217(c+1)*floor((5-1)(5-3)/4) + A014409(c+2), 0 <= c < 5, c odd

T(c+2*5,3) = (c+1)(c+2)/2(2*A002623(c-1)*floor((5-c-1)/2) + A131941(c+1)*floor((5-c)/2)) + S(c+1,3c+2,3), 0 <= c < 5 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

A236800(14,3), c = 4