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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. 12
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 (list; table; graph; refs; listen; history; text; internal format)
OFFSET

0,13

LINKS

Alois P. Heinz, Antidiagonals n = 0..31, flattened

Wikipedia, Domino

Wikipedia, Tromino

EXAMPLE

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, ...

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<nops(l) and l[k+1]=1, b(n, subsop(k=2, k+1=2, l)), 0)+

`if`(k<nops(l) and l[k+1]=0, b(n, subsop(k=1, k+1=1, l))+

b(n, subsop(k=1, k+1=2, l))+b(n, subsop(k=2, k+1=1, l)), 0)+

`if`(k+1<nops(l) and l[k+1]=0 and l[k+2]=0,

b(n, subsop(k=2, k+1=2, k+2=2, l))+

b(n, subsop(k=2, k+1=2, k+2=1, l)), 0)

fi

end:

A:= (n, 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 *)

PROG

(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

CROSSREFS

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

Main diagonal gives: A219994.

Sequence in context: A262124 A199954 A333580 * A077614 A336396 A280379

Adjacent sequences: A219984 A219985 A219986 * A219988 A219989 A219990

KEYWORD

nonn,tabl

AUTHOR

Alois P. Heinz, Dec 02 2012

STATUS

approved

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Last modified December 9 23:05 EST 2022. Contains 358710 sequences. (Running on oeis4.)