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 1-7 of 7 results.

A340166 a(n) = 4^(2*(n-1)^2) * Product_{1<=i,j<=n-1} (1 - sin(i*Pi/(2*n))^2 * sin(j*Pi/(2*n))^2).

Original entry on oeis.org

1, 12, 17745, 2958176256, 54090331699622625, 107181043200192494332800000, 22868509031094388112997259982567521313, 523389340935243821042846225254323436248483571433472
Offset: 1

Views

Author

Seiichi Manyama, Dec 30 2020

Keywords

Links

Crossrefs

Cf. A007725, A340139, A340167.

Programs

  • Mathematica
    Table[4^(2*(n-1)^2) * Product[Product[1 - Sin[i*Pi/(2*n)]^2 * Sin[j*Pi/(2*n)]^2, {i, 1, n-1}], {j, 1, n-1}], {n, 1, 10}] // Round (* Vaclav Kotesovec, Dec 31 2020 *)
  • PARI
    default(realprecision, 120);
{a(n) = round(4^(2*(n-1)^2)*prod(i=1, n-1, prod(j=1, n-1, 1-(sin(i*Pi/(2*n))*sin(j*Pi/(2*n)))^2)))}

Formula

a(n) = 4^(2*(n-1)^2) * Product_{1<=i,j<=n-1} (1 - cos(i*Pi/(2*n))^2 * cos(j*Pi/(2*n))^2).
a(n) = 4^(2*(n-1)^2) * Product_{1<=i,j<=n-1} (1 - sin(i*Pi/(2*n))^2 * cos(j*Pi/(2*n))^2).
a(n) ~ Gamma(1/4) * exp(8*G*n^2/Pi) / (Pi^(3/4) * sqrt(n) * 2^(6*n - 2)), where G is Catalan's constant A006752. - Vaclav Kotesovec, Jan 05 2021

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

Richard Stanley

Keywords

References

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

Links

Crossrefs

Cf. A007725, A007341, A065072, A340052.

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

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

Links

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

A185442 Triangle T(n,k), n>=1, 0<=k<=2n(n+1), read by rows: row n gives the coefficients of the chromatic polynomial of the Aztec diamond graph of order n, highest powers first.

Original entry on oeis.org

1, -4, 6, -3, 0, 1, -16, 120, -555, 1755, -3978, 6588, -7965, 6885, -4050, 1458, -243, 0, 1, -36, 630, -7127, 58476, -370128, 1876942, -7818056, 27208798, -80059990, 200769740, -431267475, 795531116, -1260437072, 1711682175, -1983112401, 1945239399, -1597006926, 1079055243, -585362106, 245489859, -74816136, 14762007, -1416933, 0
Offset: 1

Views

Author

Alois P. Heinz, Feb 03 2011

Keywords

Comments

The Aztec diamond graph of order n has 2*n*(n+1) vertices with integer coordinates (x,y) obeying |x-1/2| + |y-1/2| <= n and (2*n)^2 edges connecting vertices having Euclidean distance 1. It can be derived from the Aztec diamond using vertices to represent tiles and edges to connect vertices of neighboring tiles. The chromatic polynomial has 2*n*(n+1)+1 coefficients.

Examples

			2 example graphs:      o-o
.                      | |
.                    o-o-o-o
.                    | | | |
.          o-o       o-o-o-o
.          | |         | |
.          o-o         o-o
Order:      1           2
Vertices:   4          12
Edges:      4          16
The Aztec diamond graph of order 1 is the cycle graph C_4 with chromatic polynomial q^4 -4*q^3 +6*q^2 -3*q => [1, -4, 6, -3, 0].
Triangle T(n,k) begins:
  1,   -4,     6,      -3,        0;
  1,  -16,   120,    -555,     1755,      -3978,        6588,  ...
  1,  -36,   630,   -7127,    58476,    -370128,     1876942,  ...
  1,  -64,  2016,  -41639,   633851,   -7578762,    74074918,  ...
  1, -100,  4950, -161659,  3917248,  -75096624,  1186008180,  ...
  1, -144, 10296, -487283, 17170275, -480406458, 11115470152,  ...
  ...

Links

Crossrefs

Cf. A046092, A016742, A006125, A007725.

A340139 a(n) = 4^((n-2)*(n-1)) * Product_{1<=i

Original entry on oeis.org

1, 1, 13, 1904, 3016365, 50771120400, 8993476465721657, 16670531837245286776832, 322175275214070402711647486361, 64754609334534873770923002355900227840
Offset: 1

Views

Author

Seiichi Manyama, Dec 29 2020

Keywords

Links

Crossrefs

Cf. A007725.

Programs

  • Mathematica
    Table[4^((n-2)*(n-1)) * Product[Product[1 - Sin[i*Pi/(2*n)]^2 * Sin[j*Pi/(2*n)]^2, {i, 1, j-1}], {j, 2, n-1}], {n, 1, 12}] // Round (* Vaclav Kotesovec, Dec 31 2020 *)
  • PARI
    default(realprecision, 120);
{a(n) = round(4^((n-2)*(n-1))*prod(j=2, n-1, prod(i=1, j-1, 1-(sin(i*Pi/(2*n))*sin(j*Pi/(2*n)))^2)))}

Formula

a(n) = 4^((n-2)*(n-1)) * Product_{1<=i
a(n) ~ sqrt(Gamma(1/4)) * exp(4*G*n^2/Pi) / (Pi^(3/8) * n^(3/4) * 2^(3*n - 9/4) * (1 + sqrt(2))^n), where G is Catalan's constant A006752. - Vaclav Kotesovec, Jan 05 2021

A340176 Number of spanning trees in the halved Aztec diamond HMD_n.

Original entry on oeis.org

1, 1, 4, 208, 121856, 772189440, 51989627289600, 36837279603595907072, 273129993621426778551615488, 21114078836429317912110529666154496, 16975032309392309949804839529585109326888960
Offset: 0

Views

Author

Seiichi Manyama, Dec 31 2020

Keywords

Comments

*---*
| |
*---* *---*---*---*
| | | | | |
*---* *---*---*---* *---*---*---*---*---*
HMD_1 HMD_2 HMD_3
-------------------------------------------------
*---*
| |
*---*---*---*
| | | |
*---*---*---*---*---*
| | | | | |
*---*---*---*---*---*---*---*
HMD_4

Examples

			a(2) = 4;
      *   *           *---*           *---*           *---*
      |   |               |           |               |   |
  *---*---*---*   *---*---*---*   *---*---*---*   *---*   *---*

Links

Crossrefs

Cf. A007341, A007725, A007726, A334088, A334089, A340139, A340166, A340185 (halved Aztec diamond HOD_n).

Programs

  • PARI
    default(realprecision, 120);
{a(n) = round(prod(j=1, 2*n-1, prod(k=j+1, 2*n-1-j, 4-4*cos(j*Pi/(2*n))*cos(k*Pi/(2*n)))))}
  • PARI
    {a007341(n) = polresultant(polchebyshev(n-1, 2, x/2), polchebyshev(n-1, 2, (4-x)/2))};
{a334088(n) = sqrtint(polresultant(polchebyshev(2*n, 1, x/2), polchebyshev(2*n, 1, I*x/2)))};
{a(n) = if(n==0, 1, sqrtint(a007341(n)*a334088(n)/n))}
  • PARI
    default(realprecision, 120);
{a(n) = if(n==0, 1, round(4^((n-1)^2)*prod(j=1, n-1, prod(k=j+1, n-1, 1-(cos(j*Pi/(2*n))*cos(k*Pi/(2*n)))^2))))} \\ Seiichi Manyama, Jan 02 2021
  • Python
    # Using graphillion
from graphillion import GraphSet
def make_HMD(n):
    s = 1
    grids = []
    for i in range(2 * n, 0, -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 A340176(n):
    if n == 0: return 1
    universe = make_HMD(n)
    GraphSet.set_universe(universe)
    spanning_trees = GraphSet.trees(is_spanning=True)
    return spanning_trees.len()
print([A340176(n) for n in range(7)])

Formula

a(n) = Product_{1<=j
a(n) = 2^(n-1) * A007726(n) * A334089(n) = sqrt(A007341(n) * A334088(n) / n) for n > 0.
a(n) = 4^(n-1) * A340139(n) = 4^((n-1)^2) * Product_{1<=j 0. - Seiichi Manyama, Jan 02 2021
a(n) ~ sqrt(Gamma(1/4)) * exp(4*G*n^2/Pi) / (Pi^(3/8) * n^(3/4) * 2^(n - 1/4) * (1 + sqrt(2))^n), where G is Catalan's constant A006752. - Vaclav Kotesovec, Jan 05 2021

A340352 Number of spanning trees of odd Aztec diamond OD_n.

Original entry on oeis.org

1, 192, 4542720, 12116689944576, 3544863978266468352000, 112387469554685044937510092800000, 383669915612621265759587438135691539652804608, 140496256399491641572818822014023027580848616806252629983232
Offset: 1

Views

Author

Seiichi Manyama, Jan 05 2021

Keywords

Comments

R. P. Stanley conjectured that the even Aztec diamond has exactly four times as many spanning trees as the odd Aztec diamond. This conjecture was first proved by D. E. Knuth.
*
|
* *---*---*
| | | |
* *---*---* *---*---*---*---*
| | | | | | | | |
*---*---* *---*---*---*---* *---*---*---*---*---*---*
| | | | | | | | |
* *---*---* *---*---*---*---*
| | | |
* *---*---*
|
*
OD_1 OD_2 OD_3

Links

Crossrefs

Cf. A007725 (even Aztec diamond), A340166, A340185 (halved Aztec diamond HOD_n).

Programs

  • PARI
    default(realprecision, 120);
{a(n) = round(4^(2*(n-1)*n)*prod(j=1, n-1, prod(k=1, n-1, 1-(sin(j*Pi/(2*n))*sin(k*Pi/(2*n)))^2)))}

Formula

a(n) = 4^(2*(n-1)) * A340166(n) = 4^(2*(n-1)*n) * Product_{1<=j,k<=n-1} (1 - sin(j*Pi/(2*n))^2 * sin(k*Pi/(2*n))^2).
a(n) ~ Gamma(1/4) * exp(8*G*n^2/Pi) / (Pi^(3/4) * sqrt(n) * 2^(2*n + 2)), where G is Catalan's constant A006752. - Vaclav Kotesovec, Jan 06 2021
Showing 1-7 of 7 results.

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