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

%I A340185 #43 Feb 28 2023 23:47:29
%S A340185 1,1,15,2639,5100561,105518291153,23067254643457375,
%T A340185 52901008815129395889375,1266973371422697144030728637409,
%U A340185 315937379766837559600972497421046382689,818563964325891485548944567913851815851212484079
%N A340185 Number of spanning trees in the halved Aztec diamond HOD_n.
%C A340185                                               *
%C A340185                                               |
%C A340185                       *                   *---*---*
%C A340185                       |                   |   |   |
%C A340185       *           *---*---*           *---*---*---*---*
%C A340185       |           |   |   |           |   |   |   |   |
%C A340185   *---*---*   *---*---*---*---*   *---*---*---*---*---*---*
%C A340185     HOD_1           HOD_2                   HOD_3
%C A340185 -------------------------------------------------------------
%C A340185                   *
%C A340185                   |
%C A340185               *---*---*
%C A340185               |   |   |
%C A340185           *---*---*---*---*
%C A340185           |   |   |   |   |
%C A340185       *---*---*---*---*---*---*
%C A340185       |   |   |   |   |   |   |
%C A340185   *---*---*---*---*---*---*---*---*
%C A340185                 HOD_4
%H A340185 Seiichi Manyama, <a href="/A340185/b340185.txt">Table of n, a(n) for n = 0..40</a>
%H A340185 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 A340185 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 A340185 From _Seiichi Manyama_, Jan 02 2021: (Start)
%F A340185 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 A340185 a(n) = A340052(n) * A065072(n) = (1/2^n) * sqrt(A127605(n) * A004003(n) / (2*n+1)). (End)
%F A340185 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 A340185 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 A340185 (PARI) default(realprecision, 120);
%o A340185 {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 A340185 (PARI) default(realprecision, 120);
%o A340185 {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 A340185 (Python)
%o A340185 # Using graphillion
%o A340185 from graphillion import GraphSet
%o A340185 def make_HOD(n):
%o A340185     s = 1
%o A340185     grids = []
%o A340185     for i in range(2 * n + 1, 1, -2):
%o A340185         for j in range(i - 2):
%o A340185             a, b, c = s + j, s + j + 1, s + i + j
%o A340185             grids.extend([(a, b), (b, c)])
%o A340185         grids.append((s + i - 2, s + i - 1))
%o A340185         s += i
%o A340185     return grids
%o A340185 def A340185(n):
%o A340185     if n == 0: return 1
%o A340185     universe = make_HOD(n)
%o A340185     GraphSet.set_universe(universe)
%o A340185     spanning_trees = GraphSet.trees(is_spanning=True)
%o A340185     return spanning_trees.len()
%o A340185 print([A340185(n) for n in range(7)])
%Y A340185 Cf. A004003, A007725, A007726, A065072, A127605, A340052, A340176 (halved Aztec diamond HMD_n).
%K A340185 nonn
%O A340185 0,3
%A A340185 _Seiichi Manyama_, Dec 31 2020