cp's OEIS Frontend

A220054 Number A(n,k) of tilings of a k X n rectangle using right trominoes and 1 X 1 tiles; square array A(n,k), n>=0, k>=0, read by antidiagonals.

%I A220054 #29 Apr 25 2025 21:00:48
%S A220054 1,1,1,1,1,1,1,1,1,1,1,1,5,1,1,1,1,11,11,1,1,1,1,33,39,33,1,1,1,1,87,
%T A220054 195,195,87,1,1,1,1,241,849,2023,849,241,1,1,1,1,655,3895,16839,16839,
%U A220054 3895,655,1,1,1,1,1793,17511,151817,249651,151817,17511,1793,1,1
%N A220054 Number A(n,k) of tilings of a k X n rectangle using right trominoes and 1 X 1 tiles; square array A(n,k), n>=0, k>=0, read by antidiagonals.
%H A220054 Liang Kai, <a href="/A220054/b220054.txt">Antidiagonals n = 0..35, flattened</a> (antidiagonals n = 0..29 from Alois P. Heinz)
%H A220054 Kai Liang, <a href="https://arxiv.org/abs/2503.17698">Solving tiling enumeration problems by tensor network contractions</a>, arXiv:2503.17698 [math.CO], 2025. See p. 25, Table 4.
%H A220054 Wikipedia, <a href="https://en.wikipedia.org/wiki/Tromino">Tromino</a>
%e A220054 A(2,2) = 5, because there are 5 tilings of a 2 X 2 rectangle using right trominoes and 1 X 1 tiles:
%e A220054   ._._.   ._._.   .___.   .___.   ._._.
%e A220054   |_|_|   | |_|   | ._|   |_. |   |_| |
%e A220054   |_|_|   |___|   |_|_|   |_|_|   |___|
%e A220054 Square array A(n,k) begins:
%e A220054   1,  1,   1,     1,       1,        1,          1,            1, ...
%e A220054   1,  1,   1,     1,       1,        1,          1,            1, ...
%e A220054   1,  1,   5,    11,      33,       87,        241,          655, ...
%e A220054   1,  1,  11,    39,     195,      849,       3895,        17511, ...
%e A220054   1,  1,  33,   195,    2023,    16839,     151817,      1328849, ...
%e A220054   1,  1,  87,   849,   16839,   249651,    4134881,     65564239, ...
%e A220054   1,  1, 241,  3895,  151817,  4134881,  128938297,   3814023955, ...
%e A220054   1,  1, 655, 17511, 1328849, 65564239, 3814023955, 207866584389, ...
%p A220054 b:= proc(n, l) option remember; local k, t;
%p A220054       if max(l[])>n then 0 elif n=0 or l=[] then 1
%p A220054     elif min(l[])>0 then t:=min(l[]); b(n-t, map(h->h-t, l))
%p A220054     else for k do if l[k]=0 then break fi od; b(n, subsop(k=1, l))+
%p A220054          `if`(k>1 and l[k-1]=1, b(n, subsop(k=2, k-1=2, l)), 0)+
%p A220054          `if`(k<nops(l) and l[k+1]=1, b(n, subsop(k=2, k+1=2, l)), 0)+
%p A220054          `if`(k<nops(l) and l[k+1]=0, b(n, subsop(k=1, k+1=2, l))+
%p A220054             b(n, subsop(k=2, k+1=1, l))+ b(n, subsop(k=2, k+1=2, l)), 0)+
%p A220054          `if`(k+1<nops(l) and l[k+1]=0 and l[k+2]=0,
%p A220054             b(n, subsop(k=2, k+1=2, k+2=2, l)), 0)
%p A220054       fi
%p A220054     end:
%p A220054 A:= (n, k)-> `if`(n>=k, b(n, [0$k]), b(k, [0$n])):
%p A220054 seq(seq(A(n, d-n), n=0..d), d=0..14);
%t A220054 b[n_, l_] := b[n, l] = Module[{k, t}, Which[ Max[l] > n , 0, n == 0 || l == {} , 1 , Min[l] > 0 , t := Min[l]; b[n - t, l - t] , True, For[k = 1, True, k++, If[ l[[k]] == 0 , Break[] ] ]; b[n, ReplacePart[l, k -> 1]] + 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 -> 2}]] + b[n, ReplacePart[l, {k -> 2, k + 1 -> 1}]] + b[n, ReplacePart[l, {k -> 2, k + 1 -> 2}]], 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}]], 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 09 2013, translated from Maple *)
%Y A220054 Columns (or rows) k=0+1, 2-10 give: A000012, A127864, A127867, A127870, A220055, A220056, A220057, A220058, A220059, A220060.
%Y A220054 Main diagonal gives: A220061.
%K A220054 nonn,tabl
%O A220054 0,13
%A A220054 _Alois P. Heinz_, Dec 03 2012