A340176 Number of spanning trees in the halved Aztec diamond HMD_n.
1, 1, 4, 208, 121856, 772189440, 51989627289600, 36837279603595907072, 273129993621426778551615488, 21114078836429317912110529666154496, 16975032309392309949804839529585109326888960
Offset: 0
Keywords
Examples
a(2) = 4;
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Links
- Seiichi Manyama, Table of n, a(n) for n = 0..45
- 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 3.6.
- Wikipedia, Chebyshev polynomials
- Wikipedia, Resultant
Crossrefs
Programs
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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) = 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
Comments