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A340185
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Number of spanning trees in the halved Aztec diamond HOD_n.
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6
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1, 1, 15, 2639, 5100561, 105518291153, 23067254643457375, 52901008815129395889375, 1266973371422697144030728637409, 315937379766837559600972497421046382689, 818563964325891485548944567913851815851212484079
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listen;
history;
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internal format)
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OFFSET
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0,3
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COMMENTS
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*
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* *---*---*
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* *---*---* *---*---*---*---*
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*---*---* *---*---*---*---* *---*---*---*---*---*---*
HOD_1 HOD_2 HOD_3
-------------------------------------------------------------
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*---*---*
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*---*---*---*---*
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*---*---*---*---*---*---*
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*---*---*---*---*---*---*---*---*
HOD_4
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LINKS
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FORMULA
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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))).
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).
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
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MATHEMATICA
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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 *)
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PROG
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(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
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)])
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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