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Table T(n,k), n>=1, k>=1, read by antidiagonals: T(n,k) = number of equivalence classes of ways of placing one 1 X 1 tile in an n X k rectangle under all symmetry operations of the rectangle.
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%I #72 Nov 30 2016 23:14:02

%S 1,1,1,2,1,2,2,2,2,2,3,2,4,2,3,3,3,4,4,3,3,4,3,6,4,6,3,4,4,4,6,6,6,6,

%T 4,4,5,4,8,6,9,6,8,4,5,5,5,8,8,9,9,8,8,5,5,6,5,10,8,12,9,12,8,10,5,6,

%U 6,6,10,10,12,12,12,12,10,10,6,6,7,6,12,10,15

%N Table T(n,k), n>=1, k>=1, read by antidiagonals: T(n,k) = number of equivalence classes of ways of placing one 1 X 1 tile in an n X k rectangle under all symmetry operations of the rectangle.

%C It appears that the number of equivalence classes of ways of placing one m X m tile in an n X k rectangle under all symmetry operations of the rectangle is T(n-m+1,k-m+1) for m >= 2, n >= m and k >= m, and zero otherwise. - _Christopher Hunt Gribble_, Oct 17 2014

%C The sum over each antidiagonal of A243866

%C = Sum_{j=1..n}(2*j + 1 - (-1)^j)*(2*(n - j + 1) + 1 - (-1)^(n - j + 1))/16

%C = (n + 2)*(2*n^2 + 8*n + 3 - 3*(-1)^n)/48

%C - see A006918. - _Christopher Hunt Gribble_, Apr 01 2015

%H Christopher Hunt Gribble, <a href="/A243866/b243866.txt">Table of n, a(n) for n = 1..9870</a>

%F Empirically,

%F T(n,k) = floor((n+1)/2)*floor((k+1)/2)

%F = (2*n+1-(-1)^n)*(2*k+1-(-1)^k)/4;

%F T(n,1) = A034851(n,1);

%F T(n,2) = A226048(n,1);

%F T(n,3) = A226290(n,1);

%F T(n,4) = A225812(n,1);

%F T(n,5) = A228022(n,1);

%F T(n,6) = A228165(n,1);

%F T(n,7) = A228166(n,1). - _Christopher Hunt Gribble_, May 01 2015

%e T(n,k) for 1<=n<=11 and 1<=k<=11 is:

%e k 1 2 3 4 5 6 7 8 9 10 11 ...

%e .n

%e .1 1 1 2 2 3 3 4 4 5 5 6

%e .2 1 1 2 2 3 3 4 4 5 5 6

%e .3 2 2 4 4 6 6 8 8 10 10 12

%e .4 2 2 4 4 6 6 8 8 10 10 12

%e .5 3 3 6 6 9 9 12 12 15 15 18

%e .6 3 3 6 6 9 9 12 12 15 15 18

%e .7 4 4 8 8 12 12 16 16 20 20 24

%e .8 4 4 8 8 12 12 16 16 20 20 24

%e .9 5 5 10 10 15 15 20 20 25 25 30

%e 10 5 5 10 10 15 15 20 20 25 25 30

%e 11 6 6 12 12 18 18 24 24 30 30 36

%e ...

%p b := proc (n,k);

%p floor((1/2)*n+1/2)*floor((1/2)*k+1/2)

%p end proc;

%p seq(seq(b(n, k-n+1), n = 1 .. k), k = 1 .. 140);

%Y Cf. A034851, A226048, A226290, A225812, A228022, A228165, A228166, A006918, A244306, A244307, A248011, A248016, A248059, A248060, A248017, A248027.

%K tabl,nonn

%O 1,4

%A _Christopher Hunt Gribble_, Jun 19 2014

%E Terms corrected by _Christopher Hunt Gribble_, Mar 27 2015