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A236865
Number T(n,k) of equivalence classes of ways of placing k 7 X 7 tiles in an n X n square under all symmetry operations of the square; irregular triangle T(n,k), n>=7, 0<=k<=floor(n/7)^2, read by rows.
9
1, 1, 1, 1, 1, 3, 1, 3, 1, 6, 1, 6, 1, 10, 1, 10, 20, 4, 1, 1, 15, 65, 59, 14, 1, 15, 153, 329, 174, 1, 21, 295, 1225, 1234, 1, 21, 507, 3465, 6124, 1, 28, 810, 8358, 23259, 1, 28, 1214, 17710, 73204
OFFSET
7,6
COMMENTS
The first 13 rows of T(n,k) are:
.\ k 0 1 2 3 4 5 6 7 8 9
n
7 1 1
8 1 1
9 1 3
10 1 3
11 1 6
12 1 6
13 1 10
14 1 10 20 4 1
15 1 15 65 59 14
16 1 15 153 329 174
17 1 21 295 1225 1234
18 1 21 507 3465 6124
19 1 28 810 8358 23259
20 1 28 1214 17710 73204
LINKS
Christopher Hunt Gribble, C++ program
FORMULA
It appears that:
T(n,0) = 1, n>= 7
T(n,1) = (floor((n-7)/2)+1)*(floor((n-7)/2+2))/2, n >= 7
T(c+2*7,2) = A131474(c+1)*(7-1) + A000217(c+1)*floor(7^2/4) + A014409(c+2), 0 <= c < 7, c even
T(c+2*7,2) = A131474(c+1)*(7-1) + A000217(c+1)*floor((7-1)(7-3)/4) + A014409(c+2), 0 <= c < 7, c odd
T(c+2*7,3) = (c+1)(c+2)/2(2*A002623(c-1)*floor((7-c-1)/2) + A131941(c+1)*floor((7-c)/2)) + S(c+1,3c+2,3), 0 <= c < 7 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
A236865(20,3), c = 6
EXAMPLE
T(14,3) = 4 because the number of equivalent classes of ways of placing 3 7 X 7 square tiles in an 14 X 14 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