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%I #13 Nov 30 2023 12:03:50
%S 1,1,0,1,1,0,1,2,1,0,1,3,6,1,0,1,4,24,140,1,0,1,5,70,4995,16456,1,0,1,
%T 6,165,65824,10763361,8390720,1,0,1,7,336,489125,1073758336,
%U 211822552035,17179934976,1,0,1,8,616,2521476,38147070625,281474993496064,37523658921114744,140737496748032,1,0
%N Array read by antidiagonals: T(n,k) is the number of k-colorings of an n X n grid, up to rotational symmetry.
%H Andrew Howroyd, <a href="/A343095/b343095.txt">Table of n, a(n) for n = 0..860</a>
%H Peter Kagey and William Keehn, <a href="https://arxiv.org/abs/2311.13072">Counting tilings of the n X m grid, cylinder, and torus</a>, arXiv:2311.13072 [math.CO], 2023. See p. 3.
%F T(n,k) = (k^(n^2) + 2*k^((n^2 + 3*(n mod 2))/4) + k^((n^2 + (n mod 2))/2))/4.
%e Array begins:
%e ====================================================================
%e n\k | 0 1 2 3 4 5
%e ----+---------------------------------------------------------------
%e 0 | 1 1 1 1 1 1 ...
%e 1 | 0 1 2 3 4 5 ...
%e 2 | 0 1 6 24 70 165 ...
%e 3 | 0 1 140 4995 65824 489125 ...
%e 4 | 0 1 16456 10763361 1073758336 38147070625 ...
%e 5 | 0 1 8390720 211822552035 281474993496064 74505806274453125 ...
%e ...
%t {{1}}~Join~Table[Function[n, (k^(n^2) + 2*k^((n^2 + 3 #)/4) + k^((n^2 + #)/2))/4 &[Mod[n, 2] ] ][m - k + 1], {m, 0, 8}, {k, m + 1, 0, -1}] // Flatten (* _Michael De Vlieger_, Nov 30 2023 *)
%o (PARI) T(n,k) = (k^(n^2) + 2*k^((n^2 + 3*(n%2))/4) + k^((n^2 + (n%2))/2))/4
%Y Rows 0..5 are A000012, A001477, A006528, A282613, A283027, A283031.
%Y Columns 0..10 are A000007, A000012, A047937, A047938, A047939, A047940, A047941, A047942, A047943, A047944, A047945.
%Y Main diagonal is A343096.
%Y Cf. A182406, A246106, A343097, A343874.
%K nonn,tabl
%O 0,8
%A _Andrew Howroyd_, Apr 14 2021
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