A129682 -id:A129682 - cp's OEIS frontend

A219866 Number A(n,k) of tilings of a k X n rectangle using dominoes and straight (3 X 1) trominoes; square array A(n,k), n>=0, k>=0, read by antidiagonals.

1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 4, 4, 1, 1, 1, 2, 7, 14, 7, 2, 1, 1, 2, 15, 41, 41, 15, 2, 1, 1, 3, 30, 143, 184, 143, 30, 3, 1, 1, 4, 60, 472, 1069, 1069, 472, 60, 4, 1, 1, 5, 123, 1562, 5624, 9612, 5624, 1562, 123, 5, 1

Offset: 0

Alois P. Heinz, Nov 30 2012

			A(2,3) = A(3,2) = 4, because there are 4 tilings of a 3 X 2 rectangle using dominoes and straight (3 X 1) trominoes:
  .___.   .___.   .___.   .___.
  | | |   |___|   | | |   |___|
  | | |   |___|   |_|_|   | | |
  |_|_|   |___|   |___|   |_|_|
Square array A(n,k) begins:
  1,  1,  1,    1,     1,      1,        1,         1, ...
  1,  0,  1,    1,     1,      2,        2,         3, ...
  1,  1,  2,    4,     7,     15,       30,        60, ...
  1,  1,  4,   14,    41,    143,      472,      1562, ...
  1,  1,  7,   41,   184,   1069,     5624,     29907, ...
  1,  2, 15,  143,  1069,   9612,    82634,    707903, ...
  1,  2, 30,  472,  5624,  82634,  1143834,  15859323, ...
  1,  3, 60, 1562, 29907, 707903, 15859323, 354859954, ...

Liang Kai, Antidiagonals n = 0..35, flattened (0..22 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.

Columns (or rows) k=0-10 give: A000012, A000931(n+3), A129682, A219867, A219862, A219868, A219869, A219870, A219871, A219872, A219873.
Main diagonal gives: A219874.
Cf. A219987, A364457.

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=3, l))+ b(n, subsop(k=2, l))+
         `if`(k `if`(n>=k, b(n, [0$k]), b(k, [0$n])):
seq(seq(A(n, d-n), n=0..d), d=0..10);

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], k = Position[l, 0, 1][[1, 1]]; b[n, ReplacePart[l, k -> 3]] + b[n, ReplacePart[l, k -> 2]] + If[k < Length[l] && l[[k+1]] == 0, b[n, ReplacePart[l, {k -> 1, k+1 -> 1}]], 0] + If[k+1 < Length[l] && l[[k+1]] == 0 && l[[k+2]] == 0, b[n, ReplacePart[l, {k -> 1, k+1 -> 1, 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, 10}] // Flatten (* Jean-François Alcover, Dec 16 2013, translated from Maple *)

A233313 Number of tilings of a 2 X 3 X n box using bricks of shape 3 X 1 X 1 and 2 X 1 X 1.

Original entry on oeis.org

1, 4, 45, 717, 9787, 148414, 2282036, 34688229, 530613082, 8119995275, 124183342755, 1899899589557, 29066650643742, 444678773140018, 6803102237763707, 104079849391557116, 1592303310404361651, 24360457647669398381, 372687643806340329749, 5701702230014416236396

Offset: 0

Alois P. Heinz, Dec 07 2013

nonn

			a(1) = A219866(3,2) = A129682(3) = A219866(2,3) = A219867(2) = 4:
._____.  ._____.  ._____.  ._____.
|_____|  | | | |  |___| |  | |___|
|_____|  |_|_|_|  |___|_|  |_|___|.

Alois P. Heinz, Table of n, a(n) for n = 0..500

Cf. A000931, A129682, A219866, A219867, A233505, A233506, A233507, A233509.

s:= subsop:
b:= proc(n, l) option remember; local k, t; t:= min(l[]);
      if n=0 then 1 elif t>0 then b(n-t, map(h->h-t, l))
    else for k while l[k]>0 do od; add(`if`(n>=j,
         b(n, s(k=j, l)), 0), j=2..3)+ `if`(k<=4 and l[k+2]=0,
         b(n, s(k=1, k+2=1, l))+ `if`(k<=2 and l[k+4]=0,
         b(n, s(k=1, k+2=1, k+4=1, l)), 0), 0)+ `if`(
         irem(k, 2)>0 and l[k+1]=0, b(n, s(k=1, k+1=1, l)), 0)
      fi
    end:
a:=n-> b(n, [0$6]):
seq(a(n), n=0..25);

b[n_, l_] := b[n, l] = Module[{k, t}, t = Min[l]; If [n == 0, 1, If[t > 0, b[n-t, l-t], k = 1; While[l[[k]] > 0 , k++]; Sum[If[n >= j, b[n, ReplacePart[l, k -> j]], 0], {j, 2, 3}] + If[k <= 4 && l[[k+2]] == 0, b[n, ReplacePart[l, {k -> 1, k+2 -> 1}]] + If[k <= 2 && l[[k+4]] == 0, b[n, ReplacePart[l, {k -> 1, k+2 -> 1, k+4 -> 1}]], 0], 0] + If[Mod[k, 2] > 0 && l[[k+1]] == 0, b[n, ReplacePart[l, {k -> 1, k+1 -> 1}]], 0] ] ] ]; a[n_] := b[n, Array[0&, 6]]; Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Dec 16 2013, translated from Maple *)

A233505 Number of tilings of a 2 X 2 X n box using bricks of shape 3 X 1 X 1 and 2 X 1 X 1.

Original entry on oeis.org

1, 2, 9, 45, 201, 1062, 5564, 28859, 152012, 799387, 4202165, 22117465, 116385352, 612443308, 3223118545, 16961953022, 89264218645, 469766599585, 2472212575433, 13010374123502, 68468999197712, 360328057238019, 1896278913278432, 9979444454056631

Offset: 0

Alois P. Heinz, Dec 11 2013

Alois P. Heinz, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5, 0, 19, -62, 28, -141, 20, -92, 156, 76, 184, 62, 66, 41, 27, -12, 9, 2, -2, -1).

Cf. A000931, A129682, A219866, A219867, A233313, A233506, A233507, A233509.

a:= n-> coeff(series((x^17 +3*x^16 -x^15 -5*x^14 +8*x^13 -8*x^12 -22*x^10 -20*x^9 -62*x^8 -20*x^7 -42*x^6 +18*x^5 +19*x^3 +x^2 +3*x -1)/(-x^20 -2*x^19 +2*x^18 +9*x^17 -12*x^16 +27*x^15 +41*x^14 +66*x^13 +62*x^12 +184*x^11 +76*x^10 +156*x^9 -92*x^8 +20*x^7 -141*x^6 +28*x^5 -62*x^4 +19*x^3 +5*x -1), x, n+1), x, n):
seq(a(n), n=0..40);

G.f.: (x^17 +3*x^16 -x^15 -5*x^14 +8*x^13 -8*x^12 -22*x^10 -20*x^9 -62*x^8 -20*x^7 -42*x^6 +18*x^5 +19*x^3 +x^2 +3*x -1) / (-x^20 -2*x^19 +2*x^18 +9*x^17 -12*x^16 +27*x^15 +41*x^14 +66*x^13 +62*x^12 +184*x^11 +76*x^10 +156*x^9 -92*x^8 +20*x^7 -141*x^6 +28*x^5 -62*x^4 +19*x^3 +5*x -1).

A233506 Number of tilings of a 3 X 3 X n box using bricks of shape 3 X 1 X 1 and 2 X 1 X 1.

Original entry on oeis.org

1, 14, 717, 62253, 4732061, 382882762, 31449389548, 2571574546111, 210607584419520, 17254476918858789, 1413637025226131703, 115812392270890399373, 9488271882367228634756, 777357166136453697810804, 63686950935296529029018801, 5217741644362129948411085318

Offset: 0

Alois P. Heinz, Dec 11 2013

nonn

Alois P. Heinz, Table of n, a(n) for n = 0..100

Cf. A000931, A129682, A219866, A219867, A233313, A233505, A233507, A233509.

b:= proc(n, l) option remember; local k, t; t:= min(l[]);
      if n=0 then 1 elif t>0 then b(n-t, map(h->h-t, l))
    else for k while l[k]>0 do od;
         add(`if`(n>=j, b(n, s(k=j, l)), 0), j=2..3)+
         `if`(k<=6 and l[k+3]=0, b(n, s(k=1, k+3=1, l)), 0)+
         `if`(k<=3 and l[k+3]=0 and l[k+6]=0,
            b(n, s(k=1, k+3=1, k+6=1, l)), 0)+
         `if`(irem(k, 3)>0 and l[k+1]=0,
            b(n, s(k=1, k+1=1, l)), 0)+
         `if`(irem(k, 3)=1 and l[k+1]=0 and l[k+2]=0,
            b(n, subsop(k=1, k+1=1, k+2=1, l)), 0)
      fi
    end:
a:=n-> b(n, [0$9]): s:=subsop:
seq(a(n), n=0..10);

b[n_, l_] := b[n, l] = Module[{k, t}, t := Min[l]; If [n == 0, 1, If[t > 0, b[n-t, l-t], k = 1; While[l[[k]] > 0, k++ ]; Sum[If[n >= j, b[n, ReplacePart[l, k -> j]], 0], {j, 2, 3}] + If[k <= 6 && l[[k+3]] == 0, b[n, ReplacePart[l, {k -> 1, k+3 -> 1}]], 0] + If[k <= 3 && l[[k+3]] == 0 && l[[k+6]] == 0, b[n, ReplacePart[l, {k -> 1, k+3 -> 1, k+6 -> 1}]], 0] + If[Mod[k, 3] > 0 && l[[k+1]] == 0, b[n, ReplacePart[l, {k -> 1, k+1 -> 1}]], 0] + If[Mod[k, 3] == 1 && l[[k+1]] == 0 && l[[k+2]] == 0, b[n, ReplacePart[l, {k -> 1, k+1 -> 1, k+2 -> 1}]], 0] ]] ]; a[n_] := b[n, Array[0&, 9]]; Table[a[n], {n, 0, 10}] // Flatten (* Jean-François Alcover, Dec 16 2013, translated from Maple *)

A233507 Number of tilings of a 2 X 4 X n box using bricks of shape 3 X 1 X 1 and 2 X 1 X 1.

Original entry on oeis.org

1, 7, 201, 9787, 379688, 16512483, 726964790, 31549810845, 1378740599284, 60239603421159, 2630166605483293, 114886450998314920, 5017916294582867990, 219163121582772423673, 9572435654283943792842, 418094220600909382190818, 18261053013117932038592765

Offset: 0

Alois P. Heinz, Dec 11 2013

nonn

			a(1) = A219866(4,2) = A129682(4) = A219866(2,4) = A219862(2) = 7:
._______. ._______. ._______. ._______.
|_____| | | |_____| | | | | | |___| | |
|_____|_| |_|_____| |_|_|_|_| |___|_|_|
._______. ._______. ._______.
| |___| | | | |___| |___|___|
|_|___|_| |_|_|___| |___|___|.

Alois P. Heinz, Table of n, a(n) for n = 0..200

Cf. A000931, A129682, A219866, A219867, A233313, A233505, A233506, A233509.

b:= proc(n, l) option remember; local k, t; t:= min(l[]);
      if n=0 then 1
    elif t>0 then b(n-t, map(h->h-t, l))
    else for k while l[k]>0 do od;
         add(`if`(n>=j, b(n, s(k=j, l)), 0), j=2..3)+
         `if`(k<=6 and l[k+2]=0, b(n, s(k=1, k+2=1, l)), 0)+
         `if`(k<=4 and l[k+2]=0 and l[k+2*2]=0, b(n, s(k=1,
         k+2=1, k+2*2=1, l)), 0)+ `if`(irem(k, 2)>0 and
         l[k+1]=0, b(n, s(k=1, k+1=1, l)), 0)
      fi
    end:
a:=n-> b(n, [0$8]): s:= subsop:
seq(a(n), n=0..10);

b[n_, l_] := b[n, l] = Module[{k, t}, t = Min[l]; Which[n == 0, 1, t > 0, b[n-t, l-t], True, For[k = 1, l[[k]] > 0, k++]; Sum[If[n >= j, b[n, ReplacePart[l, k -> j]], 0], {j, 2, 3}] + If[k <= 6 && l[[k + 2]] == 0, b[n, ReplacePart[l, {k -> 1, k+2 -> 1}]], 0] + If[k <= 4 && l[[k+2]] == 0 && l[[k+2*2]] == 0, b[n, ReplacePart[l, {k -> 1, k+2 -> 1, k+2*2 -> 1}]], 0] + If[Mod[k, 2] > 0 && l[[k+1]] == 0, b[n, ReplacePart[l, {k -> 1, k+1 -> 1}]], 0]]]; a[n_] := b[n, Array[0&, 8]]; Table[Print[an = a[n]]; an, {n, 0, 16}] (* Jean-François Alcover, Dec 30 2013, translated from Maple *)

A233509 Number of tilings of a 2 X 5 X n box using bricks of shape 3 X 1 X 1 and 2 X 1 X 1.

Original entry on oeis.org

1, 15, 1062, 148414, 16512483, 2043497465, 257251613508, 31941208907916, 3990164870713039, 498504394558488109, 62237975023439983192, 7773270324407375580946, 970802515607358269506951, 121240108673115249961266051, 15141593230837339625055971170

Offset: 0

Alois P. Heinz, Dec 11 2013

nonn

			a(1) = A219866(5,2) = A129682(5) = A219866(2,5) = A219868(2) = 15:
.___.  .___.  .___.  .___.  .___.  .___.  .___.  .___.
| | |  |___|  | | |  |___|  | | |  |___|  | | |  |___|
| | |  |___|  |_|_|  | | |  | | |  |___|  |_|_|  | | |
|_|_|  |___|  |___|  |_|_|  |_|_|  |___|  |___|  |_|_|
| | |  | | |  | | |  | | |  |___|  |___|  |___|  |___|
|_|_|  |_|_|  |_|_|  |_|_|  |___|  |___|  |___|  |___|
.___.  .___.  .___.  .___.  .___.  .___.  .___.
| | |  | | |  |___|  |___|  | | |  | | |  |___|
|_|_|  |_|_|  |___|  |___|  |_| |  | |_|  | | |
| | |  | | |  | | |  | | |  | |_|  |_| |  | | |
| | |  |_|_|  | | |  |_|_|  | | |  | | |  |_|_|
|_|_|  |___|  |_|_|  |___|  |_|_|  |_|_|  |___|.

Alois P. Heinz, Table of n, a(n) for n = 0..40

Cf. A000931, A129682, A219866, A219867, A233313, A233505, A233506, A233507.

b:= proc(n, l) option remember; local k, t; t:= min(l[]);
      if n=0 then 1 elif t>0 then b(n-t, map(h->h-t, l))
    else for k while l[k]>0 do od;
         add(`if`(n>=j, b(n, s(k=j, l)), 0), j=2..3)+
         `if`(k<=5 and l[k+5]=0, b(n, s(k=1, k+5=1, l)), 0)+
         `if`(irem(k, 5)>0 and l[k+1]=0, b(n, s(k=1, k+1=1, l)), 0)+
         `if`(irem(k, 5) in [$1..3] and l[k+1]=0 and l[k+2]=0,
         b(n, s(k=1, k+1=1, k+2=1, l)), 0)
      fi
    end:
a:=n-> b(n, [0$10]): s:=subsop:
seq(a(n), n=0..4);

b[n_, l_] := b[n, l] = Module[{k, t}, t = Min[l]; Which[n == 0, 1, t > 0, b[n-t, l-t], True, For[k = 1, l[[k]] > 0, k++]; Sum[If[n >= j, b[n, ReplacePart[l, k -> j]], 0], {j, 2, 3}] + If[k <= 5 && l[[k+5]] == 0, b[n, ReplacePart[l, {k -> 1, k+5 -> 1}]], 0] + If[Mod[k, 5] > 0 && l[[k+1]] == 0, b[n, ReplacePart[l, {k -> 1, k+1 -> 1}]], 0] + If[1 <= Mod[k, 5] <= 3 && l[[k+1]] == 0 && l[[k+2]] == 0, b[n, ReplacePart[l, {k -> 1, k+1 -> 1, k+2 -> 1}]], 0]]];a[n_] := b[n, Array[0&, 10]]; Table[Print[an = a[n]]; an, {n, 0, 14}] (* Jean-François Alcover, Dec 30 2013, translated from Maple *)

A278815 Number of tilings of a 2 X n grid with monomers, dimers, and trimers.

Original entry on oeis.org

1, 2, 7, 29, 109, 416, 1596, 6105, 23362, 89415, 342193, 1309593, 5011920, 19180976, 73406985, 280933906, 1075154535, 4114694797, 15747237101, 60265824784, 230641706484, 882682631025, 3378090801226, 12928199853783, 49477163668857, 189352713633433

Offset: 0

Kathryn Haymaker, Nov 28 2016

The first three terms are the same as A030186 because there are only monomers and dimers in boards with n<3.

Alois P. Heinz, Table of n, a(n) for n = 0..1000
Kathryn Haymaker and Sara Robertson, Counting Colorful Tilings of Rectangular Arrays, Journal of Integer Sequences, Vol. 20 (2017), Article 17.5.8, Corollary 2.
Index entries for linear recurrences with constant coefficients, signature (3,2,5,-2,0,-1).

Cf. A030186, A052961, A129682, A219866.

a:=[1,2,7,29,109,416];; for n in [7..30] do a[n]:=3*a[n-1]+2*a[n-2] +5*a[n-3]-2*a[n-4]-a[n-6]; od; a; # G. C. Greubel, Oct 28 2019

R:=PowerSeriesRing(Integers(), 30); Coefficients(R!( (1-x-x^2-x^3)/(1-3*x-2*x^2-5*x^3+2*x^4+x^6) )); // G. C. Greubel, Oct 28 2019

seq(coeff(series((1-x-x^2-x^3)/(1-3*x-2*x^2-5*x^3+2*x^4+x^6), x, n+1), x, n), n = 0..30); # G. C. Greubel, Oct 28 2019

LinearRecurrence[{3,2,5,-2,0,-1}, {1,2,7,29,109,416}, 30] (* G. C. Greubel, Oct 28 2019 *)

my(x='x+O('x^30)); Vec((1-x-x^2-x^3)/(1-3*x-2*x^2-5*x^3+ 2*x^4 +x^6)) \\ G. C. Greubel, Oct 28 2019

def A278815_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( (1-x-x^2-x^3)/(1-3*x-2*x^2-5*x^3+2*x^4+x^6) ).list()
A278815_list(30) # G. C. Greubel, Oct 28 2019

a(n) = 3*a(n-1) + 2*a(n-2) + 5*a(n-3) - 2*a(n-4) - a(n-6).

G.f.: (1 - x - x^2 - x^3)/(1 - 3*x - 2*x^2 - 5*x^3 + 2*x^4 + x^6).

cp's OEIS Frontend

A219866 Number A(n,k) of tilings of a k X n rectangle using dominoes and straight (3 X 1) trominoes; square array A(n,k), n>=0, k>=0, read by antidiagonals.

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A233313 Number of tilings of a 2 X 3 X n box using bricks of shape 3 X 1 X 1 and 2 X 1 X 1.

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A233505 Number of tilings of a 2 X 2 X n box using bricks of shape 3 X 1 X 1 and 2 X 1 X 1.

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A233506 Number of tilings of a 3 X 3 X n box using bricks of shape 3 X 1 X 1 and 2 X 1 X 1.

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A233507 Number of tilings of a 2 X 4 X n box using bricks of shape 3 X 1 X 1 and 2 X 1 X 1.

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A233509 Number of tilings of a 2 X 5 X n box using bricks of shape 3 X 1 X 1 and 2 X 1 X 1.

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A278815 Number of tilings of a 2 X n grid with monomers, dimers, and trimers.

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