cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 14 results. Next

A004003 Number of domino tilings (or dimer coverings) of a 2n X 2n square.

Original entry on oeis.org

1, 2, 36, 6728, 12988816, 258584046368, 53060477521960000, 112202208776036178000000, 2444888770250892795802079170816, 548943583215388338077567813208427340288, 1269984011256235834242602753102293934298576249856
Offset: 0

Views

Author

Keywords

Comments

A099390 is the main entry for domino tilings (or dimer tilings) of a rectangle.
The numbers of domino tilings in A006253, A004003, A006125 give the number of perfect matchings in the relevant graphs. There are results of Jockusch and Ciucu that if a planar graph has a rotational symmetry then the number of perfect matchings is a square or twice a square - this applies to these 3 sequences. - Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 12 2001
Christine Bessenrodt points out that Pachter (1997) shows that a(n) is divisible by 2^n (cf. A065072).
a(n) is the number of different ways to cover a 2n X 2n lattice with 2n^2 dominoes. John and Sachs show that a(n) = 2^n*B(n)^2, where B(n) == n+1 (mod 32) when n is even and B(n) == (-1)^((n-1)/2)*n (mod 32) when n is odd. - Yong Kong (ykong(AT)curagen.com), May 07 2000

Examples

			The 36 solutions for the 4 X 4 board, from Neven Juric, May 14 2008:
A01 = {(1,2), (3,4), (5,6), (7,8), (9,10), (11,12), (13,14), (15,16)}
A02 = {(1,2), (3,4), (5,6), (7,11), (9,10), (8,12), (13,14), (15,16)}
A03 = {(1,2), (3,4), (5,9), (6,7), (10,11), (8,12), (13,14), (15,16)}
A04 = {(1,2), (3,4), (5,9), (6,10), (7,8), (11,12), (13,14), (15,16)}
A05 = {(1,2), (3,4), (5,9), (6,10), (7,11), (8,12), (13,14), (15,16)}
A06 = {(1,2), (3,4), (5,6), (7,8), (9,10), (13,14), (11,15), (12,16)}
A07 = {(1,2), (3,4), (5,9), (6,10), (7,8), (11,15), (13,14), (12,16)}
A08 = {(1,2), (3,4), (5,6), (7,8), (9,13), (10,14), (11,12), (15,16)}
A09 = {(1,2), (3,4), (5,6), (7,11), (8,12), (9,13), (10,14), (15,16)}
A10 = {(1,2), (3,4), (5,6), (7,8), (9,13), (10,11), (14,15), (12,16)}
A11 = {(1,2), (3,4), (5,6), (7,8), (9,13), (10,14), (11,15), (12,16)}
A12 = {(1,2), (5,6), (3,7), (4,8), (9,10), (11,12), (13,14), (15,16)}
A13 = {(1,2), (3,7), (4,8), (5,9), (6,10), (11,12), (13,14), (15,16)}
A14 = {(1,2), (5,6), (3,7), (4,8), (9,10), (13,14), (11,15), (12,16)}
A15 = {(1,2), (3,7), (4,8), (6,10), (5,9), (11,15), (12,16), (13,14)}
A16 = {(1,2), (3,7), (4,8), (5,6), (9,13), (10,14), (11,12), (15,16)}
A17 = {(1,2), (3,7), (4,8), (5,6), (9,13), (10,11), (14,15), (12,16)}
A18 = {(1,2), (5,6), (3,7), (4,8), (9,13), (10,14), (11,15), (12,16)}
A19 = {(1,5), (2,6), (3,4), (7,8), (9,10), (11,12), (13,14), (15,16)}
A20 = {(1,5), (2,6), (3,4), (7,11), (8,12), (9,10), (13,14), (15,16)}
A21 = {(1,5), (3,4), (2,6), (9,10), (7,8), (11,15), (13,14), (12,16)}
A22 = {(1,5), (2,6), (3,4), (7,8), (9,13), (10,14), (11,12), (15,16)}
A23 = {(1,5), (2,6), (3,4), (7,11), (8,12), (9,13), (10,14), (15,16)}
A24 = {(1,5), (2,6), (3,4), (7,8), (9,13), (10,11), (14,15), (12,16)}
A25 = {(1,5), (2,6), (3,4), (7,8), (9,13), (10,14), (11,15), (12,16)}
A26 = {(1,5), (2,3), (6,7), (4,8), (9,10), (11,12), (13,14), (15,16)}
A27 = {(1,5), (2,6), (3,7), (4,8), (9,10), (11,12), (13,14), (15,16)}
A28 = {(1,5), (2,3), (6,7), (4,8), (9,10), (11,15), (13,14), (12,16)}
A29 = {(1,5), (2,6), (3,7), (4,8), (9,10), (13,14), (11,15), (12,16)}
A30 = {(1,5), (2,3), (6,7), (4,8), (9,13), (10,14), (11,12), (15,16)}
A31 = {(1,5), (2,6), (3,7), (4,8), (9,13), (10,14), (11,12), (15,16)}
A32 = {(1,5), (2,6), (3,7), (4,8), (9,13), (10,14), (11,15), (12,16)}
A33 = {(1,5), (2,6), (3,7), (4,8), (9,13), (10,11), (14,15), (12,16)}
A34 = {(1,5), (2,3), (4,8), (6,10), (7,11), (9,13), (14,15), (12,16)}
A35 = {(1,5), (2,3), (6,7), (4,8), (9,13), (10,14), (11,15), (12,16)}
A36 = {(1,5), (2,3), (6,7), (4,8), (9,13), (10,11), (14,15), (12,16)}
		

References

  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 569.
  • S. R. Finch, Mathematical Constants, Cambridge, 2003, pp. 406-412.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • 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.

Crossrefs

Main diagonal of array A099390 or A187596.

Programs

  • Maple
    f := n->2^(2*n^2)*product(product(cos(i*Pi/(2*n+1))^2+cos(j*Pi/(2*n+1))^2,j=1..n),i=1..n); for k from 0 to 12 do round(evalf(f(k),300)) od;
  • Mathematica
    a[n_] := Round[ N[ Product[ 2*Cos[(2i*Pi)/(2n+1)] + 2*Cos[(2j*Pi)/(2n+1)] + 4,  {i, 1, n}, {j, 1, n}], 300] ]; Table[a[n], {n, 0, 12}] (* Jean-François Alcover, Jan 04 2012, after Maple *)
    Table[Sqrt[Resultant[ChebyshevU[2*n, x/2], ChebyshevU[2*n, I*x/2], x]], {n, 0, 12}] (* Vaclav Kotesovec, Dec 30 2020 *)
  • PARI
    {a(n) = sqrtint(polresultant(polchebyshev(2*n, 2, x/2), polchebyshev(2*n, 2, I*x/2)))} \\ Seiichi Manyama, Apr 13 2020
    
  • Python
    from math import isqrt
    from sympy.abc import x
    from sympy import I, resultant, chebyshevu
    def A004003(n): return isqrt(resultant(chebyshevu(n<<1,x/2),chebyshevu(n<<1,I*x/2))) if n else 1 # Chai Wah Wu, Nov 07 2023

Formula

a(n) = A099390(2n,2n).
a(n) = Product_{j=1..n} Product_{k=1..n} (4*cos(j*Pi/(2*n+1))^2 + 4*cos(k*Pi/(2*n+1))^2). - N. J. A. Sloane, Mar 16 2015
a(n) = 2^n * A065072(n)^2. - Alois P. Heinz, Nov 22 2018
a(n)^2 = Resultant(U(2*n,x/2), U(2*n,i*x/2)), where U(n,x) is a Chebyshev polynomial of the second kind and i = sqrt(-1). - Seiichi Manyama, Apr 13 2020
a(n) ~ 2 * (sqrt(2)-1)^(2*n+1) * exp(G*(2*n+1)^2/Pi), where G is Catalan's constant A006752. - Vaclav Kotesovec, Dec 30 2020

Extensions

Corrected and extended by David Radcliffe

A007726 Number of spanning trees of quarter Aztec diamonds of order n.

Original entry on oeis.org

1, 1, 4, 56, 2640, 411840, 210613312, 351102230528, 1901049105201408, 33349238079515381760, 1892086487183556298556416, 346728396311328694807284940800, 205021218459835103075295973360128000, 390870571052378289975757743555515137130496
Offset: 1

Views

Author

Keywords

References

  • Mihai Ciucu (ciucu(AT)math.gatech.edu), in preparation, 2001.

Crossrefs

Programs

  • Mathematica
    Table[Product[Product[4 - 2*Cos[j*Pi/n] - 2*Cos[k*Pi/n], {j, 1, k-1}], {k, 2, n-1}], {n, 1, 15}] // Round (* Vaclav Kotesovec, Dec 30 2020 *)
    Table[Sqrt[Resultant[ChebyshevU[n-1, x/2], ChebyshevU[n-1, (4-x)/2], x] / (n * 2^(n-1))], {n, 1, 15}] (* Vaclav Kotesovec, Dec 30 2020 *)
  • PARI
    default(realprecision, 120);
    {a(n) = round(prod(j=2, n-1, prod(i=1, j-1, 4*sin(i*Pi/(2*n))^2+4*sin(j*Pi/(2*n))^2)))} \\ Seiichi Manyama, Dec 29 2020

Formula

a(n) = Product_{0Sean A. Irvine, Jan 20 2018
From Vaclav Kotesovec, Dec 30 2020: (Start)
a(n) ~ sqrt(Gamma(1/4)) * 2^(5/8) * exp(2*G*n^2/Pi) / (Pi^(3/8) * n^(3/4) * 2^(n/2) * (1 + sqrt(2))^n), where G is Catalan's constant A006752.
a(n) = sqrt(A007341(n) / (n * 2^(n-1))). (End)

Extensions

More terms from Sean A. Irvine, Jan 20 2018

A256043 Order of all-2s configuration on the 2n X 2n sandpile grid graph.

Original entry on oeis.org

1, 3, 29, 901, 89893, 5758715, 22687425
Offset: 1

Views

Author

N. J. A. Sloane, Mar 14 2015

Keywords

Crossrefs

Bisection of main diagonal of A256045.

Extensions

a(7) added by Scott R. Shannon, Apr 23 2019

A360499 Number of ways to tile an n X n square using rectangles with distinct dimensions.

Original entry on oeis.org

1, 1, 21, 269, 4489, 82981, 2995185, 118897973
Offset: 1

Views

Author

Scott R. Shannon, Feb 09 2023

Keywords

Comments

All possible tilings are counted, including those identical by symmetry. Note that distinct dimensions means that, for example, a 1 x 3 rectangle can only be used once, regardless of if it lies horizontally or vertically.

Examples

			a(1) = 1 as the only way to tile a 1 x 1 square is with a square with dimensions 1 x 1.
a(2) = 1 as the only way to tile a 2 x 2 square is with a square with dimensions 2 x 2.
a(3) = 21. The possible tilings, excluding those equivalent by symmetry, are:
.
  +---+---+---+   +---+---+---+   +---+---+---+   +---+---+---+
  |           |   |   |       |   |       |   |   |           |
  +           +   +---+---+---+   +---+---+   +   +---+---+---+
  |           |   |           |   |       |   |   |           |
  +           +   +           +   +       +   +   +           +
  |           |   |           |   |       |   |   |           |
  +---+---+---+   +---+---+---+   +---+---+---+   +---+---+---+
.
The first tiling can occur in 1 way, the second in 8 different ways, the third in 8 different ways and the fourth in 4 different ways, giving 21 ways in total.
		

Crossrefs

Cf. A360498 (oblongs), A182275 (not necessarily distinct dimensions), A004003, A099390, A065072, A233320, A230031.

A340052 a(n) = Product_{1<=i

Original entry on oeis.org

1, 1, 5, 91, 5661, 1173821, 801125065, 1786768287095, 12964564030176889, 305121026002697122873, 23243604301636717923421133, 5722586073277932639539150258131, 4548248834078776410469611991220703125
Offset: 0

Views

Author

Seiichi Manyama, Dec 29 2020

Keywords

Crossrefs

Programs

  • Mathematica
    Table[2^(n*(n-1)) * Product[Product[Sin[i*Pi/(2*n + 1)]^2 + Sin[j*Pi/(2*n + 1)]^2, {i, 1, j-1}], {j, 2, n}], {n, 0, 15}] // Round (* Vaclav Kotesovec, Dec 30 2020 *)
  • PARI
    default(realprecision, 120);
    {a(n) = round(prod(j=2, n, prod(i=1, j-1, 4*sin(i*Pi/(2*n+1))^2+4*sin(j*Pi/(2*n+1))^2)))}

Formula

a(n)^2 = A127605(n)/(2^n * (2*n+1)).
a(n) ~ sqrt(Gamma(1/4)) * exp(G*(2*n+1)^2/(2*Pi)) / (2^(n/2 + 5/4) * Pi^(3/8) * n^(3/4)), where G is Catalan's constant A006752. - Vaclav Kotesovec, Dec 30 2020

A340185 Number of spanning trees in the halved Aztec diamond HOD_n.

Original entry on oeis.org

1, 1, 15, 2639, 5100561, 105518291153, 23067254643457375, 52901008815129395889375, 1266973371422697144030728637409, 315937379766837559600972497421046382689, 818563964325891485548944567913851815851212484079
Offset: 0

Views

Author

Seiichi Manyama, Dec 31 2020

Keywords

Comments

*
|
* *---*---*
| | | |
* *---*---* *---*---*---*---*
| | | | | | | | |
*---*---* *---*---*---*---* *---*---*---*---*---*---*
HOD_1 HOD_2 HOD_3
-------------------------------------------------------------
*
|
*---*---*
| | |
*---*---*---*---*
| | | | |
*---*---*---*---*---*---*
| | | | | | |
*---*---*---*---*---*---*---*---*
HOD_4

Crossrefs

Cf. A004003, A007725, A007726, A065072, A127605, A340052, A340176 (halved Aztec diamond HMD_n).

Programs

  • Mathematica
    Table[4^((n-1)*n) * Product[Product[(1 - Cos[j*Pi/(2*n + 1)]^2*Cos[k*Pi/(2*n + 1)]^2), {k, j+1, n}], {j, 1, n}], {n, 0, 12}] // Round (* Vaclav Kotesovec, Jan 03 2021 *)
  • PARI
    default(realprecision, 120);
    {a(n) = round(prod(j=1, 2*n, prod(k=j+1, 2*n-j, 4-4*cos(j*Pi/(2*n+1))*cos(k*Pi/(2*n+1)))))}
    
  • PARI
    default(realprecision, 120);
    {a(n) = round(4^((n-1)*n)*prod(j=1, n, prod(k=j+1, n, 1-(cos(j*Pi/(2*n+1))*cos(k*Pi/(2*n+1)))^2)))} \\ Seiichi Manyama, Jan 02 2021
    
  • Python
    # Using graphillion
    from graphillion import GraphSet
    def make_HOD(n):
        s = 1
        grids = []
        for i in range(2 * n + 1, 1, -2):
            for j in range(i - 2):
                a, b, c = s + j, s + j + 1, s + i + j
                grids.extend([(a, b), (b, c)])
            grids.append((s + i - 2, s + i - 1))
            s += i
        return grids
    def A340185(n):
        if n == 0: return 1
        universe = make_HOD(n)
        GraphSet.set_universe(universe)
        spanning_trees = GraphSet.trees(is_spanning=True)
        return spanning_trees.len()
    print([A340185(n) for n in range(7)])

Formula

a(n) = Product_{1<=j
From Seiichi Manyama, Jan 02 2021: (Start)
a(n) = 4^((n-1)*n) * Product_{1<=j
a(n) = A340052(n) * A065072(n) = (1/2^n) * sqrt(A127605(n) * A004003(n) / (2*n+1)). (End)
a(n) ~ sqrt(Gamma(1/4)) * exp(G*(2*n+1)^2/Pi) / (Pi^(3/8) * n^(3/4) * 2^(n + 3/4) * (1 + sqrt(2))^(n + 1/2)), where G is Catalan's constant A006752. - Vaclav Kotesovec, Jan 03 2021

A360256 Number of ways to tile an n X n square using rectangles with distinct height X width dimensions.

Original entry on oeis.org

1, 1, 33, 513, 14409, 693025, 50447161
Offset: 1

Author

Scott R. Shannon, Feb 17 2023

Keywords

Comments

All possible tilings are counted, including those identical by symmetry. Note that distinct height X width dimensions means that, for example, a 1 X 3 rectangle can be used twice, once in a horizontal (1 X 3) and once in a vertical (3 X 1) direction.

Examples

			a(1) = 1 as the only way to tile a 1 X 1 square is with a square with dimensions 1 X 1.
a(2) = 1 as the only way to tile a 2 X 2 square is with a square with dimensions 2 X 2.
a(3) = 33. The possible tilings, excluding those equivalent by symmetry, are:
.
  +---+---+---+   +---+---+---+   +---+---+---+   +---+---+---+   +---+---+---+
  |   |       |   |   |       |   |       |   |   |           |   |   |       |
  +---+---+---+   +---+---+---+   +---+---+   +   +---+---+---+   +---+---+---+
  |   |       |   |           |   |       |   |   |           |   |       |   |
  +   +       +   +           +   +       +   +   +           +   +       +   +
  |   |       |   |           |   |       |   |   |           |   |       |   |
  +---+---+---+   +---+---+---+   +---+---+---+   +---+---+---+   +---+---+---+
.
The first tiling can occur in 4 different ways, the second in 8 different ways, the third in 8 different ways, the fourth in 4 different ways and the fifth in 8 different ways. There is also the single 3 X 3 rectangle. This gives 33 ways in total.
		

Crossrefs

A360498 Number of ways to tile an n X n square using oblongs with distinct dimensions.

Original entry on oeis.org

0, 0, 4, 12, 256, 3620, 87216, 2444084, 87181220
Offset: 1

Author

Scott R. Shannon, Feb 09 2023

Keywords

Comments

All possible tilings are counted, including those identical by symmetry. Note that distinct dimensions means that, for example, a 1 x 3 oblong can only be used once, regardless of if it lies horizontally or vertically.

Examples

			a(1) = 0 as no distinct oblongs can tile a square with dimensions 1 x 1.
a(2) = 0 as no distinct oblongs can tile a square with dimensions 2 x 2.
a(3) = 4. There is one tiling, excluding those equivalent by symmetry:
.
  +---+---+---+
  |           |
  +---+---+---+
  |           |
  +           +
  |           |
  +---+---+---+
.
This tiling can occur in 4 different ways, giving 4 ways in total.
a(4) = 12. The possible tilings, excluding those equivalent by symmetry, are:
.
  +---+---+---+---+   +---+---+---+---+
  |   |           |   |               |
  +   +           +   +---+---+---+---+
  |   |           |   |               |
  +---+---+---+---+   +               +
  |               |   |               |
  +               +   +               +
  |               |   |               |
  +---+---+---+---+   +---+---+---+---+
.
The first tiling can occur in 8 different way and the second in 4 different ways, giving 12 ways in total.
		

Crossrefs

Cf. A360499 (rectangles), A004003, A099390, A065072, A233320, A230031.

A334089 a(n) = sqrt(A334088(n)/2^(n-1)).

Original entry on oeis.org

1, 2, 13, 272, 18281, 3944920, 2732887529, 6077512159232, 43384923739812577, 994156445200670735008, 73125714588602035608260981, 17265651822746410593596262486016, 13085551252412040683513520733767180041, 31834381760532514451976501491991780699626368
Offset: 1

Author

Seiichi Manyama, Apr 14 2020

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Resultant[ChebyshevT[2*n, x/2], ChebyshevT[2*n, I*x/2], x]^(1/4) / 2^((n-1)/2), {n, 1, 15}] (* Vaclav Kotesovec, Apr 14 2020 *)
  • PARI
    {a(n) = sqrtint(sqrtint(polresultant(polchebyshev(2*n, 1, x/2), polchebyshev(2*n, 1, I*x/2)))/2^(n-1))}

Formula

a(n) ~ exp(2*G*n^2/Pi) / 2^(3*n/2 - 5/8), where G is Catalan's constant A006752. - Vaclav Kotesovec, Apr 14 2020

A360725 Number of ways to tile an n X n square using oblongs with distinct height x width dimensions.

Original entry on oeis.org

0, 0, 4, 36, 1056, 31052, 1473944, 87469884
Offset: 1

Author

Scott R. Shannon, Feb 18 2023

Keywords

Comments

All possible tilings are counted, including those identical by symmetry. Note that distinct height x width dimensions means that, for example, a 1 x 3 oblong can be used twice, once in a horizonal (1 x 3) and once in a vertical (3 x 1) direction.

Examples

			a(1) = 0 as no distinct oblongs can tile a square with dimensions 1 x 1.
a(2) = 0 as no distinct oblongs can tile a square with dimensions 2 x 2.
a(3) = 4. There is one tiling, excluding those equivalent by symmetry:
.
  +---+---+---+
  |           |
  +---+---+---+
  |           |
  +           +
  |           |
  +---+---+---+
.
This tiling can occur in 4 different ways, giving 4 ways in total.
a(4) = 36. The possible tilings, excluding those equivalent by symmetry, are:
.
  +---+---+---+---+   +---+---+---+---+   +---+---+---+---+   +---+---+---+---+
  |   |           |   |               |   |   |           |   |   |           |
  +   +           +   +---+---+---+---+   +   +---+---+---+   +   +---+---+---+
  |   |           |   |               |   |   |           |   |   |   |       |
  +---+---+---+---+   +               +   +   +           +   +   +   +       +
  |               |   |               |   |   |           |   |   |   |       |
  +               +   +               +   +---+---+---+---+   +---+---+       +
  |               |   |               |   |               |   |       |       |
  +---+---+---+---+   +---+---+---+---+   +---+---+---+---+   +---+---+---+---+
.
The first tiling can occur in 8 different ways, the second in 4 different ways, the third in 16 different ways and the fourth in 8 different ways. This gives 36 ways in total.
		

Crossrefs

Showing 1-10 of 14 results. Next