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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.
9
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
5,6
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
LINKS
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:
._______________ _______________ _______________
| | | | |_______| | | |
| | | | | | | |_______|
| . | . | | . | | | . | |
| | | | | . | | | |
|_______|_______| |_______| | |_______| . |
| | | | |_______| | | |
| | | | | | | |_______|
| . | | | . | | | . | |
| | | | | | | | |
|_______|_______| |_______|_______| |_______|_______|
KEYWORD
tabf,nonn
AUTHOR
STATUS
approved