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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

(list; graph; refs; listen; history; text; internal format)

OFFSET

5,6

COMMENTS

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

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

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

LINKS

Table of n, a(n) for n=5..49.

Christopher Hunt Gribble, C++ program

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

EXAMPLE

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:

._______________ _______________ _______________

| | | | |_______| | | |

| | | | | | | |_______|

| . | . | | . | | | . | |

| | | | | . | | | |

|_______|_______| |_______| | |_______| . |

| | | | |_______| | | |

| | | | | | | |_______|

| . | | | . | | | . | |

| | | | | | | | |

|_______|_______| |_______|_______| |_______|_______|

CROSSREFS