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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.
9
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
4,6
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
LINKS
Christopher Hunt Gribble, C++ program
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
EXAMPLE
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:
._____________ _____________ _____________
| | | | |______| | | |
| . | . | | . | | | . |______|
| | | | | . | | | |
|______|______| |______| | |______| . |
| | | | |______| | | |
| . | | | . | | | . |______|
| | | | | | | | |
|______|______| |______|______| |______|______|
KEYWORD
nonn,tabf
AUTHOR
STATUS
approved