Original entry on oeis.org
1, 5, 36, 281, 2245, 18061, 145601, 1174500, 9475901, 76455961, 616891945, 4977472781, 40161441636, 324048393905, 2614631600701, 21096536145301, 170220478472105, 1373448758774436, 11081871650713781, 89415697915538545, 721463601671126161, 5821234309893001301, 46969478172465070500, 378980086070257592201, 3057856106268358639861
Offset: 0
- Alois P. Heinz, Table of n, a(n) for n = 0..400
- N. Allegra, Exact solution of the 2d dimer model: Corner free energy, correlation functions and combinatorics, arXiv:1410.4131 [cond-mat.stat-mech], 2014. See Table 1.
- Index entries for linear recurrences with constant coefficients, signature (11,-25,11,-1).
-
ft:=(m,n)->
2^(m*n/2)*mul( mul(
(cos(Pi*i/(n+1))^2+cos(Pi*j/(m+1))^2), j=1..m/2), i=1..n/2);
gt:=(m,n)->round(evalf(ft(m,n),300));
tt:=[seq(gt(4,2*n),n=0..10)];
# second Maple program:
a:= n-> (<<0|1|0|0>, <0|0|1|0>, <0|0|0|1>, <-1|11|-25|11>>^n.
<<1, 5, 36, 281>>)[1, 1]:
seq(a(n), n=0..30); # Alois P. Heinz, Oct 28 2012
-
LinearRecurrence[{11, -25, 11, -1}, {1, 5, 36, 281}, 25] (* Jean-François Alcover, Jun 17 2018 *)
-
x='x+O('x^200); Vec((1-x)*(x^2-5*x+1)/(x^4-11*x^3+25*x^2-11*x+1)) \\ Altug Alkan, Mar 23 2016
A099390
Array T(m,n) read by antidiagonals: number of domino tilings (or dimer tilings) of the m X n grid (or m X n rectangle), for m>=1, n>=1.
Original entry on oeis.org
0, 1, 1, 0, 2, 0, 1, 3, 3, 1, 0, 5, 0, 5, 0, 1, 8, 11, 11, 8, 1, 0, 13, 0, 36, 0, 13, 0, 1, 21, 41, 95, 95, 41, 21, 1, 0, 34, 0, 281, 0, 281, 0, 34, 0, 1, 55, 153, 781, 1183, 1183, 781, 153, 55, 1, 0, 89, 0, 2245, 0, 6728, 0, 2245, 0, 89, 0, 1, 144, 571, 6336, 14824, 31529, 31529, 14824, 6336, 571, 144, 1
Offset: 1
0, 1, 0, 1, 0, 1, ...
1, 2, 3, 5, 8, 13, ...
0, 3, 0, 11, 0, 41, ...
1, 5, 11, 36, 95, 281, ...
0, 8, 0, 95, 0, 1183, ...
1, 13, 41, 281, 1183, 6728, ...
- S. R. Finch, Mathematical Constants, Cambridge, 2003, pp. 406-412.
- P. E. John, H. Sachs, and H. Zernitz, Problem 5. Domino covers in square chessboards, Zastosowania Matematyki (Applicationes Mathematicae) XIX 3-4 (1987), 635-641.
- R. P. Stanley, Enumerative Combinatorics, Vol. 1, Cambridge University Press, 2nd ed., pp. 547 and 570.
- Darko Veljan, Kombinatorika: s teorijom grafova (Croatian) (Combinatorics with Graph Theory) mentions the value 12988816 = 2^4*901^2 for the 8 X 8 case on page 4.
- Alois P. Heinz, Table of n, a(n) for n = 1..1035
- M. Aanjaneya and S. P. Pal, Faultfree tromino tilings of rectangles, arXiv:math/0610925 [math.CO], 2006.
- Mudit Aggarwal and Samrith Ram, Generating Functions for Straight Polyomino Tilings of Narrow Rectangles, J. Int. Seq., Vol. 26 (2023), Article 23.1.4.
- F. Ardila and R. P. Stanley, Tilings, arXiv:math/0501170 [math.CO], 2005.
- M. Ciucu, Enumeration of perfect matchings in graphs with reflective symmetry, Journal of Combinatorial Theory, Series A, Volume 77, Issue 1, January 1997, Pages 67-97.
- Henry Cohn, 2-adic behavior of numbers of domino tilings, arXiv:math/0008222 [math.CO], 2000.
- Henry Cohn, 2-adic behavior of numbers of domino tilings, Electronic Journal of Combinatorics, 6 (1999), #R14.
- F. Faase, Counting Hamiltonian cycles in product graphs
- F. Faase, Results from the counting program
- Steven R. Finch, The Dimer Problem [From Steven Finch, Apr 20 2019]
- Steven R. Finch, Two Dimensional Monomer Dimer Constant [Broken link]
- M. E. Fisher, Statistical mechanics of dimers on a plane lattice, Physical Review, 124 (1961), 1664-1672.
- P. Flajolet and R. Sedgewick, Analytic Combinatorics, 2009; see page 363.
- Laura Florescu, Daniela Morar, David Perkinson, Nicholas Salter, and Tianyuan Xu, Sandpiles and Dominos, Electronic Journal of Combinatorics, Volume 22(1), 2015.
- W. Jockusch, Perfect matchings and perfect squares J. Combin. Theory Ser. A 67 (1994), no. 1, 100-115.
- Peter E. John and Horst Sachs, On a strange observation in the theory of the dimer problem, arXiv:math/9801094 [math.CO], 1998.
- Peter E. John and Horst Sachs, On a strange observation in the theory of the dimer problem, Discrete Math. 216 (2000), no. 1-3, 211-219.
- Yuhi Kamio, Junnosuke Koizumi, and Toshihiko Nakazawa, Quadratic residues and domino tilings, arXiv:2311.13597 [math.NT], 2023.
- David Klarner and Jordan Pollack, Domino tilings of rectangles with fixed width, Disc. Math. 32 (1980) 45-52, Table 1.
- Per Hakan Lundow, Enumeration of matchings in polygraphs, 1998.
- P. W. Kasteleyn, The statistics of dimers on a lattice, I. the number of dimer arrangements on a quadratic lattice, Physica 27 (1961), 1209-1225.
- Douglas M. McKenna, The Art of Space-Filling Domino Curves, Bridges Conference Proceedings, 2024, pp. 319-326.
- L. Pachter, Combinatorial approaches and conjectures for 2-divisibility problems concerning domino tilings of polyominoes, Electronic Journal of Combinatorics 4 (1997), #R29.
- J. Propp, Dimers and Dominoes
- J. Propp, Enumeration of Matchings: Problems and Progress, arXiv:math/9904150v2 [math.CO], 1999.
- Jaime Rangel-Mondragon, Polyominoes and Related Families, The Mathematica Journal, 9:3 (2005), 609-640.
- R. C. Read, A Note on Tiling Rectangles with Dominoes, The Fibonacci Quarterly, 18.1 (1980), 24-27.
- R. P. Stanley, A combinatorial miscellany
- H. N. V. Temperley and Michael E. Fisher, Dimer problem in statistical mechanics -- an exact result, Philos. Mag. (8) 6 (1961), 1061-1063.
- Herman Tulleken, Polyominoes 2.2: How they fit together, (2019).
- Eric Weisstein's World of Mathematics, Domino Tiling
- Eric Weisstein, Illustration for T(4,4) = 36, from Domino Tilings web page (see previous link) [Included with permission]
- Index entries for sequences related to dominoes
See also
A004003 for more literature on the dimer problem.
Rows 2-13, 16 (without zeros) are
A000045,
A001835,
A005178,
A003775,
A028468,
A028469,
A028470,
A028471,
A028472,
A028473,
A028474,
A241908,
A340532.
-
(Maple code for the even-numbered rows from N. J. A. Sloane, Mar 15 2015. This is not totally satisfactory since it uses floating point. However, it is useful for getting the initial values quickly.)
Digits:=100;
p:=evalf(Pi);
z:=proc(h,d) global p; evalf(cos( h*p/(2*d+1) )); end;
T:=proc(m,n) global z; round(mul( mul( 4*z(h,m)^2+4*z(k,n)^2, k=1..n), h=1..m)); end;
[seq(T(1,n),n=0..10)]; # A001519
[seq(T(2,n),n=0..10)]; # A188899
[seq(T(3,n),n=0..10)]; # A256044
[seq(T(n,n),n=0..10)]; # A004003
-
T[?OddQ, ?OddQ] = 0;
T[m_, n_] := Product[2*(2+Cos[2j*Pi/(m+1)]+Cos[2k*Pi/(n+1)]), {k, 1, n/2}, {j, 1, m/2}];
Flatten[Table[Round[T[m-n+1, n]], {m, 1, 12}, {n, 1, m}]] (* Jean-François Alcover, Nov 25 2011, updated May 28 2022 *)
-
{T(n, k) = sqrtint(abs(polresultant(polchebyshev(n, 2, x/2), polchebyshev(k, 2, I*x/2))))} \\ Seiichi Manyama, Apr 13 2020
A187596
Array T(m,n) read by antidiagonals: number of domino tilings of the m X n grid (m>=0, n>=0).
Original entry on oeis.org
1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 2, 0, 1, 1, 1, 3, 3, 1, 1, 1, 0, 5, 0, 5, 0, 1, 1, 1, 8, 11, 11, 8, 1, 1, 1, 0, 13, 0, 36, 0, 13, 0, 1, 1, 1, 21, 41, 95, 95, 41, 21, 1, 1, 1, 0, 34, 0, 281, 0, 281, 0, 34, 0, 1, 1, 1, 55, 153, 781, 1183, 1183, 781, 153, 55, 1, 1, 1, 0, 89, 0, 2245, 0, 6728, 0, 2245, 0, 89, 0, 1, 1, 1, 144, 571, 6336
Offset: 0
Array begins:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
1, 0, 3, 0, 11, 0, 41, 0, 153, 0, 571, ...
1, 1, 5, 11, 36, 95, 281, 781, 2245, 6336, 18061, ...
1, 0, 8, 0, 95, 0, 1183, 0, 14824, 0, 185921, ...
1, 1, 13, 41, 281, 1183, 6728, 31529, 167089, 817991, 4213133, ...
1, 0, 21, 0, 781, 0, 31529, 0, 1292697, 0, 53175517, ...
- R. P. Stanley, Enumerative Combinatorics, Vol. 1, Cambridge University Press, 1997.
-
with(LinearAlgebra):
T:= proc(m,n) option remember; local i, j, t, M;
if m<=1 or n<=1 then 1 -irem(n*m, 2)
elif irem(n*m, 2)=1 then 0
elif mAlois P. Heinz, Apr 11 2011
-
t[m_, n_] := Product[2*(2+Cos[2*j*Pi/(m+1)]+Cos[2*k*Pi/(n+1)]), {k, 1, n/2}, {j, 1, m/2}]; t[?OddQ, ?OddQ] = 0; Table[t[m-n, n] // FullSimplify, {m, 0, 13}, {n, 0, m}] // Flatten (* Jean-François Alcover, Jan 07 2014, after A099390 *)
A239264
Number A(n,k) of domicule tilings of a k X n grid; square array A(n,k), n>=0, k>=0, read by antidiagonals.
Original entry on oeis.org
1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 3, 0, 1, 1, 1, 5, 5, 1, 1, 1, 0, 11, 0, 11, 0, 1, 1, 1, 21, 43, 43, 21, 1, 1, 1, 0, 43, 0, 280, 0, 43, 0, 1, 1, 1, 85, 451, 1563, 1563, 451, 85, 1, 1, 1, 0, 171, 0, 9415, 0, 9415, 0, 171, 0, 1, 1, 1, 341, 4945, 55553, 162409, 162409, 55553, 4945, 341, 1, 1
Offset: 0
A(3,2) = 5:
+-----+ +-----+ +-----+ +-----+ +-----+
|o o-o| |o o o| |o o o| |o o o| |o-o o|
|| | || X | || | || | X || | ||
|o o-o| |o o o| |o o o| |o o o| |o-o o|
+-----+ +-----+ +-----+ +-----+ +-----+
A(4,3) = 43:
+-------+ +-------+ +-------+ +-------+ +-------+
|o o o o| |o o o-o| |o o-o o| |o o-o o| |o o-o o|
|| X || | X | | \ / | || || | \ ||
|o o o o| |o o o o| |o o o o| |o o o o| |o o o o|
| | | X | || || | \ \ | || \ |
|o-o o-o| |o-o o o| |o o-o o| |o-o o o| |o o-o o|
+-------+ +-------+ +-------+ +-------+ +-------+ ...
Square array A(n,k) begins:
1, 1, 1, 1, 1, 1, 1, ...
1, 0, 1, 0, 1, 0, 1, ...
1, 1, 3, 5, 11, 21, 43, ...
1, 0, 5, 0, 43, 0, 451, ...
1, 1, 11, 43, 280, 1563, 9415, ...
1, 0, 21, 0, 1563, 0, 162409, ...
1, 1, 43, 451, 9415, 162409, 3037561, ...
Columns (or rows) k=0-10 give:
A000012,
A059841,
A001045(n+1),
A239265,
A239266,
A239267,
A239268,
A239269,
A239270,
A239271,
A239272.
Bisection of main diagonal gives:
A239273.
-
b:= proc(n, l) option remember; local d, f, k;
d:= nops(l)/2; f:=false;
if n=0 then 1
elif l[1..d]=[f$d] then b(n-1, [l[d+1..2*d][], true$d])
else for k to d while not l[k] do od;
`if`(k1 and l[k+d+1],
b(n, subsop(k=f, k+d+1=f, l)), 0)+
`if`(k>1 and n>1 and l[k+d-1],
b(n, subsop(k=f, k+d-1=f, l)), 0)+
`if`(n>1 and l[k+d], b(n, subsop(k=f, k+d=f, l)), 0)+
`if`(k `if`(irem(n*k, 2)>0, 0,
`if`(k>n, A(k, n), b(n, [true$(k*2)]))):
seq(seq(A(n, d-n), n=0..d), d=0..14);
-
b[n_, l_List] := b[n, l] = Module[{d = Length[l]/2, f = False, k}, Which [n == 0, 1, l[[1 ;; d]] == Array[f&, d], b[n-1, Join[l[[d+1 ;; 2*d]], Array[True&, d]]], True, For[k=1, !l[[k]], k++]; If[k1 && l[[k+d+1]], b[n, ReplacePart[l, {k -> f, k+d+1 -> f}]], 0] + If[k>1 && n>1 && l[[k+d-1]], b[n, ReplacePart[l, {k -> f, k+d-1 -> f}]], 0] + If[n>1 && l[[k+d]], b[n, ReplacePart[l, {k -> f, k+d -> f}]], 0] + If[k f, k+1 -> f}]], 0]]]; A[n_, k_] := If[Mod[n*k, 2]>0, 0, If[k>n, A[k, n], b[n, Array[True&, k*2]]]]; Table[Table[A[n, d-n], {n, 0, d}], {d, 0, 14}] // Flatten (* Jean-François Alcover, Feb 02 2015, after Alois P. Heinz *)
A187616
Triangle T(m,n) read by rows: number of domino tilings of the m X n grid (0 <= m <= n).
Original entry on oeis.org
1, 1, 0, 1, 1, 2, 1, 0, 3, 0, 1, 1, 5, 11, 36, 1, 0, 8, 0, 95, 0, 1, 1, 13, 41, 281, 1183, 6728, 1, 0, 21, 0, 781, 0, 31529, 0, 1, 1, 34, 153, 2245, 14824, 167089, 1292697, 12988816, 1, 0, 55, 0, 6336, 0, 817991, 0, 108435745, 0, 1, 1, 89, 571, 18061, 185921, 4213133, 53175517, 1031151241, 14479521761, 258584046368
Offset: 0
Triangle begins:
1
1 0
1 1 2
1 0 3 0
1 1 5 11 36
1 0 8 0 95 0
1 1 13 41 281 1183 6728
1 0 21 0 781 0 31529 0
1 1 34 153 2245 14824 167089 1292697 12988816
...
-
with(LinearAlgebra):
T:= proc(m,n) option remember; local i, j, t, M;
if m<=1 or n<=1 then 1 -irem(n*m, 2)
elif irem(n*m, 2)=1 then 0
else M:= Matrix(n*m, shape =skewsymmetric);
for i to n do
for j to m do
t:= (i-1)*m+j;
if jAlois P. Heinz, Apr 11 2011
-
T[m_, n_] := T[m, n] = Module[{i, j, t, M}, Which[m <= 1 || n <= 1, 1 - Mod[n*m, 2], Mod[n*m, 2] == 1, 0, True, M[i_, j_] /; j < i := -M[j, i]; M[, ] = 0; For[i = 1, i <= n, i++, For[j = 1, j <= m, j++, t = (i-1)*m+j; If[j < m, M[t, t+1] = 1]; If[i < n, M[t, t+m] = 1 - 2*Mod[j, 2]]]]; Sqrt[Det[Table[M[i, j], {i, 1, n*m}, {j, 1, n*m}]]]]]; Table[Table[T[m, n], {n, 0, m}], {m, 0, 10}] // Flatten (* Jean-François Alcover, Jan 07 2014, translated from Maple *)
A189006
Array A(m,n) read by antidiagonals: number of domino tilings of the m X n grid with upper left corner removed iff m*n is odd, (m>=0, n>=0).
Original entry on oeis.org
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 3, 3, 1, 1, 1, 1, 5, 4, 5, 1, 1, 1, 1, 8, 11, 11, 8, 1, 1, 1, 1, 13, 15, 36, 15, 13, 1, 1, 1, 1, 21, 41, 95, 95, 41, 21, 1, 1, 1, 1, 34, 56, 281, 192, 281, 56, 34, 1, 1, 1, 1, 55, 153, 781, 1183, 1183, 781, 153, 55, 1, 1, 1, 1, 89, 209, 2245, 2415, 6728, 2415, 2245, 209, 89, 1, 1
Offset: 0
A(3,3) = 4, because there are 4 domino tilings of the 3 X 3 grid with upper left corner removed:
. .___. . .___. . .___. . .___.
._|___| ._|___| ._| | | ._|___|
| |___| | | | | | |_|_| |___| |
|_|___| |_|_|_| |_|___| |___|_|
Array begins:
1, 1, 1, 1, 1, 1, 1, ...
1, 1, 1, 1, 1, 1, 1, ...
1, 1, 2, 3, 5, 8, 13, ...
1, 1, 3, 4, 11, 15, 41, ...
1, 1, 5, 11, 36, 95, 281, ...
1, 1, 8, 15, 95, 192, 1183, ...
1, 1, 13, 41, 281, 1183, 6728, ...
Rows m=0+1, 2-12 give:
A000012,
A000045(n+1),
A002530(n+1),
A005178(n+1),
A189003,
A028468,
A189004,
A028470,
A189005,
A028472,
A210724,
A028474.
-
with(LinearAlgebra):
A:= proc(m, n) option remember; local i, j, s, t, M;
if m=0 or n=0 then 1
elif m1 or j>1 or s=0 then
if j
-
A[1, 1] = 1; A[m_, n_] := A[m, n] = Module[{i, j, s, t, M}, Which[m == 0 || n == 0, 1, m < n, A[n, m], True, s = Mod[n*m, 2];M[i_, j_] /; j < i := -M[j, i]; M[, ] = 0; For[i = 1, i <= n, i++, For[j = 1, j <= m, j++, t = (i-1)*m+j-s; If[i > 1 || j > 1 || s == 0, If[j < m, M[t, t+1] = 1]; If[i < n, M[t, t+m] = 1-2*Mod[j, 2]]]]]; Sqrt[Det[Array[M, {n*m-s, n*m-s}]]]]]; Table[Table[A[m, d-m], {m, 0, d}], {d, 0, 15}] // Flatten (* Jean-François Alcover, Dec 26 2013, translated from Maple *)
A187618
Triangle T(m,n) read by rows: number of domino tilings of the 2m X 2n grid (0 <= m <= n).
Original entry on oeis.org
1, 1, 2, 1, 5, 36, 1, 13, 281, 6728, 1, 34, 2245, 167089, 12988816, 1, 89, 18061, 4213133, 1031151241, 258584046368, 1, 233, 145601, 106912793, 82741005829, 65743732590821, 53060477521960000, 1, 610, 1174500, 2720246633, 6675498237130, 16848161392724969, 43242613716069407953, 112202208776036178000000
Offset: 0
Triangle begins:
1
1 2
1 5 36
1 13 281 6728
1 34 2245 167089 12988816
1 89 ...
A348566
Triangle read by rows: T(m, n) is the number of symmetric recurrent sandpiles on an m X n grid (m >= 0, 0 <= n <= m).
Original entry on oeis.org
1, 1, 4, 1, 3, 2, 1, 14, 7, 128, 1, 11, 5, 71, 36, 1, 52, 18, 1358, 539, 43264, 1, 41, 13, 769, 281, 17753, 6728, 1, 194, 47, 14852, 4271, 1452866, 434657, 151519232, 1, 153, 34, 8449, 2245, 603126, 167089, 46069729, 12988816, 1, 724, 123, 163534, 34276, 49704772, 10894561, 16236962114, 3625549353, 5475450241024
Offset: 0
The triangle begins:
1
1 4
1 3 2
1 14 7 128
1 11 5 71 36
1 52 18 1358 539 43264
1 41 13 769 281 17753 6728
...
See Fig. 9 of the paper by Florescu et al. for the T(4, 4) = 36 symmetric recurrent sandpiles on a 4x4 grid.
A340475
Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where T(n,k) = Product_{a=1..n} Product_{b=1..k} (4*sin(a*Pi/(2*n+1))^2 + 4*sin(b*Pi/(2*k+1))^2).
Original entry on oeis.org
1, 1, 1, 1, 6, 1, 1, 29, 29, 1, 1, 139, 500, 139, 1, 1, 666, 8329, 8329, 666, 1, 1, 3191, 138301, 463736, 138301, 3191, 1, 1, 15289, 2295701, 25543057, 25543057, 2295701, 15289, 1, 1, 73254, 38105729, 1404312491, 4614756624, 1404312491, 38105729, 73254, 1
Offset: 0
Square array begins:
1, 1, 1, 1, 1, ...
1, 6, 29, 139, 666, ...
1, 29, 500, 8329, 138301, ...
1, 139, 8329, 463736, 25543057, ...
1, 666, 138301, 25543057, 4614756624, ...
-
default(realprecision, 120);
{T(n, k) = round(prod(a=1, n, prod(b=1, k, 4*sin(a*Pi/(2*n+1))^2+4*sin(b*Pi/(2*k+1))^2)))}
A340476
Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where T(n,k) = Product_{a=1..n} Product_{b=1..k} (4*sin(a*Pi/(2*n+1))^2 + 4*cos(b*Pi/(2*k+1))^2).
Original entry on oeis.org
1, 1, 1, 1, 4, 1, 1, 19, 11, 1, 1, 91, 176, 29, 1, 1, 436, 2911, 1471, 76, 1, 1, 2089, 48301, 79808, 11989, 199, 1, 1, 10009, 801701, 4375897, 2091817, 97021, 521, 1, 1, 47956, 13307111, 240378643, 372713728, 53924597, 783511, 1364, 1
Offset: 0
Square array begins:
1, 1, 1, 1, 1, ...
1, 4, 19, 91, 436, ...
1, 11, 176, 2911, 48301, ...
1, 29, 1471, 79808, 4375897, ...
1, 76, 11989, 2091817, 372713728, ...
-
default(realprecision, 120);
{T(n, k) = round(prod(a=1, n, prod(b=1, k, 4*sin(a*Pi/(2*n+1))^2+4*cos(b*Pi/(2*k+1))^2)))}
-
{T(n, k) = sqrtint(4^k*polresultant(polchebyshev(2*n+1, 1, I*x/2), polchebyshev(2*k, 2, x/2)))}
Showing 1-10 of 10 results.
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