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%I #21 Feb 17 2014 12:59:17
%S 1,1,1,1,1,3,1,3,1,6,1,6,1,10,1,10,1,15,1,15,30,5,1,1,21,96,74,14,1,
%T 21,221,413,174,1,28,417,1525,1234,1,28,705,4290,6124,1,36,1107,10269,
%U 23259,1,36,1638,21630,73204,1,45,2334,41790,199436
%N Number T(n,k) of equivalence classes of ways of placing k 9 X 9 tiles in an n X n square under all symmetry operations of the square; irregular triangle T(n,k), n>=9, 0<=k<=floor(n/9)^2, read by rows.
%H Christopher Hunt Gribble, <a href="/A236936/a236936.cpp.txt">C++ program</a>
%H Christopher Hunt Gribble, <a href="/A236936/a236936.txt">Example graphics</a>
%F It appears that:
%F T(n,0) = 1, n>= 9
%F T(n,1) = (floor((n-9)/2)+1)*(floor((n-9)/2+2))/2, n >= 9
%F T(c+2*9,2) = A131474(c+1)*(9-1) + A000217(c+1)*floor(9^2/4) + A014409(c+2), 0 <= c < 9, c even
%F T(c+2*9,2) = A131474(c+1)*(9-1) + A000217(c+1)*floor((9-1)(9-3)/4) + A014409(c+2), 0 <= c < 9, c odd
%F T(c+2*9,3) = (c+1)(c+2)/2(2*A002623(c-1)*floor((9-c-1)/2) + A131941(c+1)*floor((9-c)/2)) + S(c+1,3c+2,3), 0 <= c < 9 where
%F S(c+1,3c+2,3) =
%F A054252(2,3), c = 0
%F A236679(5,3), c = 1
%F A236560(8,3), c = 2
%F A236757(11,3), c = 3
%F A236800(14,3), c = 4
%F A236829(17,3), c = 5
%F A236865(20,3), c = 6
%F A236915(23,3), c = 7
%F A236936(26,3), c = 8
%e The first 17 rows of T(n,k) are:
%e .\ k 0 1 2 3 4
%e n
%e 9 1 1
%e 10 1 1
%e 11 1 3
%e 12 1 3
%e 13 1 6
%e 14 1 6
%e 15 1 10
%e 16 1 10
%e 17 1 15
%e 18 1 15 30 5 1
%e 19 1 21 96 74 14
%e 20 1 21 221 413 174
%e 21 1 28 417 1525 1234
%e 22 1 28 705 4290 6124
%e 23 1 36 1107 10269 23259
%e 24 1 36 1638 21630 73204
%e 25 1 45 2334 41790 199436
%e .
%e T(18,3) = 5 because the number of equivalence classes of ways of placing 3 9 X 9 square tiles in an 18 X 18 square under all symmetry operations of the square is 5.
%Y Cf. A054252, A236679, A236560, A236757, A236800, A236829, A236865, A236915, A236939.
%K tabf,nonn
%O 9,6
%A _Christopher Hunt Gribble_, Feb 01 2014