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A236939
Number T(n,k) of equivalence classes of ways of placing k 10 X 10 tiles in an n X n square under all symmetry operations of the square; irregular triangle T(n,k), n>=10, 0<=k<=floor(n/10)^2, read by rows.
9
1, 1, 1, 1, 1, 3, 1, 3, 1, 6, 1, 6, 1, 10, 1, 10, 1, 15, 1, 15, 1, 21, 36, 6, 1, 1, 21, 113, 80, 14, 1, 28, 261, 461, 174, 1, 28, 483, 1665, 1234, 1, 36, 819, 4725, 6124, 1, 36, 1266, 11193, 23259, 1, 45, 1878, 23646, 73204
OFFSET
10,6
LINKS
Christopher Hunt Gribble, C++ program
Christopher Hunt Gribble, Example graphics
FORMULA
It appears that:
T(n,0) = 1, n>= 10
T(n,1) = (floor((n-10)/2)+1)*(floor((n-10)/2+2))/2, n >= 10
T(c+2*10,2) = A131474(c+1)*(10-1) + A000217(c+1)*floor(10^2/4) + A014409(c+2), 0 <= c < 10, c even
T(c+2*10,2) = A131474(c+1)*(10-1) + A000217(c+1)*floor((10-1)(10-3)/4) + A014409(c+2), 0 <= c < 10, c odd
T(c+2*10,3) = (c+1)(c+2)/2(2*A002623(c-1)*floor((10-c-1)/2) + A131941(c+1)*floor((10-c)/2)) + S(c+1,3c+2,3), 0 <= c < 10 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
A236915(23,3), c = 7
A236936(26,3), c = 8
A236939(29,3), c = 9
EXAMPLE
The first 17 rows of T(n,k) are:
.\ k 0 1 2 3 4
n
10 1 1
11 1 1
12 1 3
13 1 3
14 1 6
15 1 6
16 1 10
17 1 10
18 1 15
19 1 15
20 1 21 36 6 1
21 1 21 113 80 14
22 1 28 261 461 174
23 1 28 483 1665 1234
24 1 36 819 4725 6124
25 1 36 1266 11193 23259
26 1 45 1878 23646 73204
.
T(20,3) = 6 because the number of equivalence classes of ways of placing 3 10 X 10 square tiles in a 20 X 20 square under all symmetry operations of the square is 6.
KEYWORD
tabf,nonn
AUTHOR
STATUS
approved