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

1, 1, 15, 2639, 5100561, 105518291153, 23067254643457375, 52901008815129395889375, 1266973371422697144030728637409, 315937379766837559600972497421046382689, 818563964325891485548944567913851815851212484079

Offset: 0

Author

Seiichi Manyama, Dec 31 2020

Keywords

• nonn

Comments

*

|

* *---*---*

| | | |

* *---*---* *---*---*---*---*

| | | | | | | | |

*---*---* *---*---*---*---* *---*---*---*---*---*---*

HOD_1 HOD_2 HOD_3

-------------------------------------------------------------

*

|

*---*---*

| | |

*---*---*---*---*

| | | | |

*---*---*---*---*---*---*

| | | | | | |

*---*---*---*---*---*---*---*---*

HOD_4

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..40
• Mihai Ciucu, Symmetry classes of spanning trees of Aztec diamonds and perfect matchings of odd squares with a unit hole, arXiv:0710.4500 [math.CO], 2007. See Corollary 3.7.

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

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.