A219987 Number A(n,k) of tilings of a k X n rectangle using dominoes and right trominoes; square array A(n,k), n>=0, k>=0, read by antidiagonals.

1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 2, 0, 1, 1, 1, 5, 5, 1, 1, 1, 0, 11, 8, 11, 0, 1, 1, 1, 24, 55, 55, 24, 1, 1, 1, 0, 53, 140, 380, 140, 53, 0, 1, 1, 1, 117, 633, 2319, 2319, 633, 117, 1, 1, 1, 0, 258, 1984, 15171, 21272, 15171, 1984, 258, 0, 1

Offset: 0

Author

Alois P. Heinz, Dec 02 2012

Keywords

• nonn
• tabl

Examples

			A(3,3) = 8, because there are 8 tilings of a 3 X 3 rectangle using dominoes and right trominoes:
  .___._.   .___._.   .___._.   .___._.
  |___| |   |___| |   |___| |   |_. | |
  | ._|_|   | | |_|   | |___|   | |_|_|
  |_|___|   |_|___|   |_|___|   |_|___|
  ._.___.   ._.___.   ._.___.   ._.___.
  | |___|   | | ._|   | |___|   | |___|
  |___| |   |_|_| |   |_|_. |   |_| | |
  |___|_|   |___|_|   |___|_|   |___|_|
Square array A(n,k) begins:
  1,  1,   1,    1,     1,       1,         1,          1, ...
  1,  0,   1,    0,     1,       0,         1,          0, ...
  1,  1,   2,    5,    11,      24,        53,        117, ...
  1,  0,   5,    8,    55,     140,       633,       1984, ...
  1,  1,  11,   55,   380,    2319,     15171,      96139, ...
  1,  0,  24,  140,  2319,   21272,    262191,    2746048, ...
  1,  1,  53,  633, 15171,  262191,   5350806,  100578811, ...
  1,  0, 117, 1984, 96139, 2746048, 100578811, 3238675344, ...
		

Links

• Liang Kai, Antidiagonals n = 0..35, flattened (Antidiagonals n = 0..31 from Alois P. Heinz)
• Kai Liang, Solving tiling enumeration problems by tensor network contractions, arXiv:2503.17698 [math.CO], 2025. See p. 25, Table 4.
• Wikipedia, Domino
• Wikipedia, Tromino

Crossrefs

Columns (or rows) k=0-10 give: A000012, A059841, A052980, A165716, A165791, A219988, A219989, A219990, A219991, A219992, A219993.

Main diagonal gives: A219994.

Cf. A219866, A364457.

Programs

Maple

b:= proc(n, l) option remember; local k, t;
      if max(l[])>n then 0 elif n=0 or l=[] then 1
    elif min(l[])>0 then t:=min(l[]); b(n-t, map(h->h-t, l))
    else for k do if l[k]=0 then break fi od;
         b(n, subsop(k=2, l))+
         `if`(k>1 and l[k-1]=1, b(n, subsop(k=2, k-1=2, l)), 0)+
         `if`(k `if`(n>=k, b(n, [0$k]), b(k, [0$n])):
seq(seq(A(n, d-n), n=0..d), d=0..14);

Mathematica

b[n_, l_] := b[n, l] = Module[{k, t}, If[Max[l] > n, 0, If[n == 0 || l == {}, 1, If[Min[l] > 0, t = Min[l]; b[n-t, l-t], For[k = 1, True, k++, If[l[[k]] == 0, Break[]]]; b[n, ReplacePart[l, k -> 2]] + If[k > 1 && l[[k-1]] == 1, b[n, ReplacePart[l, {k -> 2, k-1 -> 2}]], 0] + If[k < Length[l] && l[[k+1]] == 1, b[n, ReplacePart[l, {k -> 2, k+1 -> 2}]], 0] + If[k < Length[l] && l[[k+1]] == 0, b[n, ReplacePart[l, {k -> 1, k+1 -> 1}]] + b[n, ReplacePart[l, {k -> 1, k+1 -> 2}]] + b[n, ReplacePart[l, {k -> 2, k+1 -> 1}]], 0] + If[k+1 < Length[l] && l[[k+1]] == 0 && l[[k+2]] == 0, b[n, ReplacePart[l, {k -> 2, k+1 -> 2, k+2 -> 2}]] + b[n, ReplacePart[l, {k -> 2, k+1 -> 2, k+2 -> 1}]], 0]]]]]; a[n_, k_] := If[n >= k, b[n, Array[0&, k]], b[k, Array[0&, n]]]; Table[Table[a[n, d-n], {n, 0, d}], {d, 0, 14}] // Flatten (* Jean-François Alcover, Dec 05 2013, translated from Alois P. Heinz's Maple program *)

Sage

from sage.combinat.tiling import TilingSolver, Polyomino
def A(n,k):
    p = Polyomino([(0,0), (0,1)])
    q = Polyomino([(0,0), (0,1), (1,0)])
    T = TilingSolver([p,q], box=[n,k], reusable=True, reflection=True)
    return T.number_of_solutions()
# Ralf Stephan, May 21 2014