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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 two 1 X 1 tiles in an n X k rectangle under all symmetry operations of the rectangle.
6

%I #26 Nov 30 2016 22:18:17

%S 0,1,1,2,3,2,4,6,6,4,6,10,13,10,6,9,15,22,22,15,9,12,21,34,36,34,21,

%T 12,16,28,48,56,56,48,28,16,20,36,65,78,88,78,65,36,20,25,45,84,106,

%U 123,123,106,84,45,25,30,55,106,136,168,171,168,136,106,55,30

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

%H Christopher Hunt Gribble, <a href="/A244306/b244306.txt">Table of n, a(n) for n = 1..9870</a>

%F Empirically,

%F T(n,k) = (4*k^2*n^2 + 2*k^2 + 8*k*n + 2*n^2 - 4*k - 4*n - 1 - (2*k^2 - 4*k - 1)*(-1)^n - (2*n^2 - 4*n - 1)*(-1)^k - (-1)^k*(-1)^n)/32.

%F T(1,k) = A002620(k) = floor(k^2/4).

%F T(2,k) = A000217(k) = k*(k+1)/2.

%F = T(1,k) + T(1,k+1) = floor(k^2/4) + floor((k+1)^2/4).

%F T(3,k) = 2*A000217(k) + A024206(k-2)

%F = k*(k+1) + floor((k-1)^2/4) - 1.

%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 0 1 2 4 6 9 12 16 20 25 30

%e .2 1 3 6 10 15 21 28 36 45 55 66

%e .3 2 6 13 22 34 48 65 84 106 130 157

%e .4 4 10 22 36 56 78 106 136 172 210 254

%e .5 6 15 34 56 88 123 168 216 274 335 406

%e .6 9 21 48 78 123 171 234 300 381 465 564

%e .7 12 28 65 106 168 234 321 412 524 640 777

%e .8 16 36 84 136 216 300 412 528 672 820 996

%e .9 20 45 106 172 274 381 524 672 856 1045 1270

%e 10 25 55 130 210 335 465 640 820 1045 1275 1550

%e 11 30 66 157 254 406 564 777 996 1270 1550 1885

%Y Cf. A034851, A226048, A226290, A225812, A228022, A228165, A228166, A243866, A006918, A244307, A248011, A248016, A248059, A248060, A248017, A248027.

%K tabl,nonn

%O 1,4

%A _Christopher Hunt Gribble_, Jun 25 2014

%E Terms corrected and extended by _Christopher Hunt Gribble_, Apr 02 2015