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.

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

Author

Christopher Hunt Gribble, Feb 01 2014

Keywords

• tabf
• nonn

Examples

			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.
		

Links

• Christopher Hunt Gribble, C++ program
• Christopher Hunt Gribble, Example graphics

Crossrefs

Cf. A054252, A236679, A236560, A236757, A236800, A236829, A236865, A236915, A236936

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