%I #20 Dec 18 2018 16:52:25
%S 1,1,1,1,1,1,1,1,1,1,1,1,2,1,1,1,1,3,3,1,1,1,1,5,8,5,1,1,1,1,8,18,18,
%T 8,1,1,1,1,13,44,68,44,13,1,1,1,1,21,107,233,233,107,21,1,1,1,1,34,
%U 257,838,1262,838,257,34,1,1,1,1,55,621,2989,6523,6523,2989,621,55,1,1
%N Number A(n,k) of tilings of a k X n rectangle using V (2m+1)-ominoes (m >= 0) in standard orientation; square array A(n,k), n>=0, k>=0, read by antidiagonals.
%C The shapes of the tiles are:
%C ._.
%C ._. | |
%C ._. | | | |
%C ._. | |_. | |_._. | |_._._.
%C |_| |___| |_____| |_______| ... .
%C .
%C The sequence of column k (or row k) satisfies a linear recurrence with constant coefficients of order 2^(k-1) for k>0.
%H Alois P. Heinz, <a href="/A322494/b322494.txt">Antidiagonals n = 0..23, flattened</a>
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Polyomino">Polyomino</a>
%e A(3,3) = 8:
%e ._____. ._____. ._____. ._____. ._____. ._____. ._____. ._____.
%e |_|_|_| | |_|_| |_|_|_| |_| |_| |_|_|_| |_| |_| | |_|_| | | |_|
%e |_|_|_| |___|_| | |_|_| |_|___| |_| |_| | |___| | |_|_| | |___|
%e |_|_|_| |_|_|_| |___|_| |_|_|_| |_|___| |___|_| |_____| |_____|.
%e .
%e Square array A(n,k) begins:
%e 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
%e 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
%e 1, 1, 2, 3, 5, 8, 13, 21, 34, ...
%e 1, 1, 3, 8, 18, 44, 107, 257, 621, ...
%e 1, 1, 5, 18, 68, 233, 838, 2989, 10687, ...
%e 1, 1, 8, 44, 233, 1262, 6523, 34468, 181615, ...
%e 1, 1, 13, 107, 838, 6523, 51420, 396500, 3086898, ...
%e 1, 1, 21, 257, 2989, 34468, 396500, 4577274, 52338705, ...
%e 1, 1, 34, 621, 10687, 181615, 3086898, 52338705, 888837716, ...
%p b:= proc(n, l) option remember; local k, m, r;
%p if n=0 or l=[] then 1
%p elif min(l)>0 then (t-> b(n-t, map(h->h-t, l)))(min(l))
%p elif l[-1]=n then b(n, subsop(-1=[][], l))
%p else for k while l[k]>0 do od; r:= 0;
%p for m from 0 while k+m<=nops(l) and l[k+m]=0 and n>m do
%p r:= r+b(n, [l[1..k-1][], 1$m, m+1, l[k+m+1..nops(l)][]])
%p od; r
%p fi
%p end:
%p A:= (n, k)-> b(max(n, k), [0$min(n, k)]):
%p seq(seq(A(n, d-n), n=0..d), d=0..14);
%t b[n_, l_] := b[n, l] = Module[{k, m, r}, Which[n == 0 || l == {}, 1, Min[l] > 0, Function[t, b[n-t, l-t]][Min[l]], l[[-1]] == n, b[n, ReplacePart[ l, -1 -> Nothing]], True, For[k=1, l[[k]] > 0, k++]; r = 0; For[m=0, k+m <= Length[l] && l[[k+m]] == 0 && n>m, m++, r = r + b[n, Join[l[[1 ;; k-1]], Array[1&, m], {m+1}, l[[k+m+1 ;; Length[l]]]]]]; r]];
%t A[n_, k_] := b[Max[n, k], Array[0&, Min[n, k]]];
%t Table[Table[A[n, d-n], {n, 0, d}], {d, 0, 14}] // Flatten (* _Jean-François Alcover_, Dec 18 2018, after _Alois P. Heinz_ *)
%Y Columns (or rows) k=0+1,2-10 give: A000012, A000045(n+1), A322496, A322497, A322498, A322499, A322500, A322501, A322502, A322503.
%Y Main diagonal gives A322495.
%Y Cf. A226444.
%K nonn,tabl
%O 0,13
%A _Alois P. Heinz_, Dec 12 2018