A340185 - OEIS

%I #43 Feb 28 2023 23:47:29

%S 1,1,15,2639,5100561,105518291153,23067254643457375,

%T 52901008815129395889375,1266973371422697144030728637409,

%U 315937379766837559600972497421046382689,818563964325891485548944567913851815851212484079

%N Number of spanning trees in the halved Aztec diamond HOD_n.

%C                                               *

%C                                               |

%C                       *                   *---*---*

%C                       |                   |   |   |

%C       *           *---*---*           *---*---*---*---*

%C       |           |   |   |           |   |   |   |   |

%C   *---*---*   *---*---*---*---*   *---*---*---*---*---*---*

%C     HOD_1           HOD_2                   HOD_3

%C -------------------------------------------------------------

%C                   *

%C                   |

%C               *---*---*

%C               |   |   |

%C           *---*---*---*---*

%C           |   |   |   |   |

%C       *---*---*---*---*---*---*

%C       |   |   |   |   |   |   |

%C   *---*---*---*---*---*---*---*---*

%C                 HOD_4

%H Seiichi Manyama, <a href="/A340185/b340185.txt">Table of n, a(n) for n = 0..40</a>

%H Mihai Ciucu, <a href="https://arxiv.org/abs/0710.4500">Symmetry classes of spanning trees of Aztec diamonds and perfect matchings of odd squares with a unit hole</a>, arXiv:0710.4500 [math.CO], 2007. See Corollary 3.7.

%F a(n) = Product_{1<=j<k<=2*n and j+k<=2*n} (4 - 4*cos(j*Pi/(2*n+1))*cos(k*Pi/(2*n+1))).

%F From _Seiichi Manyama_, Jan 02 2021: (Start)

%F a(n) = 4^((n-1)*n) * Product_{1<=j<k<=n} (1 - cos(j*Pi/(2*n+1))^2 * cos(k*Pi/(2*n+1))^2).

%F a(n) = A340052(n) * A065072(n) = (1/2^n) * sqrt(A127605(n) * A004003(n) / (2*n+1)). (End)

%F 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

%t 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 *)

%o (PARI) default(realprecision, 120);

%o {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)))))}

%o (PARI) default(realprecision, 120);

%o {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

%o (Python)

%o # Using graphillion

%o from graphillion import GraphSet

%o def make_HOD(n):

%o     s = 1

%o     grids = []

%o     for i in range(2 * n + 1, 1, -2):

%o         for j in range(i - 2):

%o             a, b, c = s + j, s + j + 1, s + i + j

%o             grids.extend([(a, b), (b, c)])

%o         grids.append((s + i - 2, s + i - 1))

%o         s += i

%o     return grids

%o def A340185(n):

%o     if n == 0: return 1

%o     universe = make_HOD(n)

%o     GraphSet.set_universe(universe)

%o     spanning_trees = GraphSet.trees(is_spanning=True)

%o     return spanning_trees.len()

%o print([A340185(n) for n in range(7)])

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

%K nonn

%O 0,3

%A _Seiichi Manyama_, Dec 31 2020