%I #35 Apr 29 2019 04:45:45
%S 1,1,1,1,1,6,1,1,46,1926,1,12,648,92544,15767640,1,30,6312,3943710,
%T 2933201376,2061379857600,1,90,92400,192994200,577186150464,
%U 1605824110657800,5363188066566330000,1,318,1051140,10266445476,118129589107200,1340797019145183600
%N Triangle read by rows: T(n,m) is the number of inequivalent n X m matrices under action of the Klein group, with one-fourth each of 1s, 2s, 3s and 4s (ordered occurrences rounded up/down if n*m != 0 mod 4).
%C Computed using Polya's enumeration theorem for coloring.
%H María Merino, <a href="/A287020/b287020.txt">Rows n=0..42 of triangle, flattened</a>
%H M. Merino and I. Unanue, <a href="https://doi.org/10.1387/ekaia.17851">Counting squared grid patterns with Pólya Theory</a>, EKAIA, 34 (2018), 289-316 (in Basque).
%F G.f.: g(x1,x2,x3,x4)=(y1^(m*n) + 3*y2^(m*n/2))/4 for even n and m;
%F (y1^(m*n) + y1^n*y2^((m*n-m)/2) + 2*y2^(m*n/2))/4 for odd n and even m;
%F (y1^(m*n) + y1^m*y2^((m*n-n)/2) + 2*y2^(m*n/2))/4 for even n and odd m;
%F (y1^(m*n) + y1^n*y2^((m*n-n)/2) + y1^m*y2^((m*n-m)/2) + y1*y2^((m*n-1)/2))/4 for odd n and m; where coefficient correspond to y1=x1+x2+x3+x4, y2=x1^2+x2^2+x3^2+x4^2, and occurrences of numbers are ceiling(m*n/4) for the first k numbers and floor(m*n/4) for the last (4-k) numbers, if m*n = k mod 4.
%e For n = 4 and m = 2 the T(4,2) = 648 solutions are colorings of 4 X 2 matrices in 4 colors inequivalent under the action of the Klein group with exactly 2 occurrences of each color (coefficient of x1^2 x2^2 x3^2 x4^2).
%e Triangle begins:
%e ========================================================
%e n\m | 0 1 2 3 4 5
%e ----|---------------------------------------------------
%e 0 | 1
%e 1 | 1 1
%e 2 | 1 1 6
%e 3 | 1 1 46 1926
%e 4 | 1 12 648 92544 15767640
%e 5 | 1 30 6312 3943710 2933201376 2061379857600
%Y Cf. A283435, A286892, A287021, A287022, A287377, A287378, A287383, A287384.
%K nonn,tabl
%O 0,6
%A _María Merino_, Imanol Unanue, May 18 2017