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A236829
Number T(n,k) of equivalence classes of ways of placing k 6 X 6 tiles in an n X n square under all symmetry operations of the square; irregular triangle T(n,k), n>=6, 0<=k<=floor(n/6)^2, read by rows.
9
1, 1, 1, 1, 1, 3, 1, 3, 1, 6, 1, 6, 1, 10, 16, 4, 1, 1, 10, 51, 50, 14, 1, 15, 125, 293, 174, 1, 15, 239, 1065, 1234, 1, 21, 423, 3075, 6124, 1, 21, 672, 7371, 23259, 1, 28, 1030, 16093, 81480, 51615, 10596, 808, 31, 1
OFFSET
6,6
COMMENTS
The first 13 rows of T(n,k) are:
.\ k 0 1 2 3 4 5 6 7 8 9
n
6 1 1
7 1 1
8 1 3
9 1 3
10 1 6
11 1 6
12 1 10 16 4 1
13 1 10 51 50 14
14 1 15 125 293 174
15 1 15 239 1065 1234
16 1 21 423 3075 6124
17 1 21 672 7371 23259
18 1 28 1030 16093 81480 51615 10596 808 31 1
LINKS
Christopher Hunt Gribble, C++ program
FORMULA
It appears that:
T(n,0) = 1, n>= 6
T(n,1) = (floor((n-6)/2)+1)*(floor((n-6)/2+2))/2, n >= 6
T(c+2*6,2) = A131474(c+1)*(6-1) + A000217(c+1)*floor(6^2/4) + A014409(c+2), 0 <= c < 6, c even
T(c+2*6,2) = A131474(c+1)*(6-1) + A000217(c+1)*floor((6-1)(6-3)/4) + A014409(c+2), 0 <= c < 6, c odd
T(c+2*6,3) = (c+1)(c+2)/2(2*A002623(c-1)*floor((6-c-1)/2) + A131941(c+1)*floor((6-c)/2)) + S(c+1,3c+2,3), 0 <= c < 6 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
A236829(17,3), c = 5
EXAMPLE
T(12,3) = 4 because the number of equivalence classes of ways of placing 3 6 X 6 square tiles in a 12 X 12 square under all symmetry operations of the square is 4. The portrayal of an example from each equivalence class is:
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KEYWORD
tabf,nonn
AUTHOR
STATUS
approved