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%I #27 Nov 28 2022 10:21:59
%S 1,1,1,1,1,1,1,3,3,1,1,7,11,7,1,1,17,49,49,17,1,1,41,229,373,229,41,1,
%T 1,99,1081,3105,3105,1081,99,1,1,239,5123,26515,44930,26515,5123,239,
%U 1,1,577,24323,227441,674292,674292,227441,24323,577,1
%N Square array T(n,k), n >= 2, k >= 2, read by antidiagonals, where T(n,k) is the number of self-avoiding closed paths on an n X k grid which pass through four corners ((0,0), (0,k-1), (n-1,k-1), (n-1,0)).
%H Seiichi Manyama, <a href="/A333513/b333513.txt">Antidiagonals n = 2..15, flattened</a>
%F T(n,k) = T(k,n).
%e Square array T(n,k) begins:
%e 1, 1, 1, 1, 1, 1, ...
%e 1, 1, 3, 7, 17, 41, ...
%e 1, 3, 11, 49, 229, 1081, ...
%e 1, 7, 49, 373, 3105, 26515, ...
%e 1, 17, 229, 3105, 44930, 674292, ...
%e 1, 41, 1081, 26515, 674292, 17720400, ...
%o (Python)
%o # Using graphillion
%o from graphillion import GraphSet
%o import graphillion.tutorial as tl
%o def A333513(n, k):
%o universe = tl.grid(n - 1, k - 1)
%o GraphSet.set_universe(universe)
%o cycles = GraphSet.cycles()
%o for i in [1, k, k * (n - 1) + 1, k * n]:
%o cycles = cycles.including(i)
%o return cycles.len()
%o print([A333513(j + 2, i - j + 2) for i in range(11 - 1) for j in range(i + 1)])
%Y Column k=2-7 give: A000012, A001333(n-2), A333514, A333515, A358712, A358713.
%Y Main diagonal gives A333466.
%Y Cf. A333758.
%K nonn,tabl
%O 2,8
%A _Seiichi Manyama_, Mar 25 2020
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