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A107991
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Complexity (number of maximal spanning trees) in an unoriented simple graph with nodes {1,2,...,n} and edges {i,j} if i + j > n.
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3
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1, 1, 1, 3, 8, 40, 180, 1260, 8064, 72576, 604800, 6652800, 68428800, 889574400, 10897286400, 163459296000, 2324754432000, 39520825344000, 640237370572800, 12164510040883200, 221172909834240000, 4644631106519040000, 93666727314800640000
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OFFSET
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1,4
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COMMENTS
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Proof of the formula: check that the associated combinatorial laplacian has eigenvalues {0,..n-1}\ {floor((n+1)/2)} by exhibiting a basis of eigenvectors (which are very simple).
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REFERENCES
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N. Biggs, Algebraic Graph Theory, Cambridge University Press (1974).
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LINKS
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FORMULA
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a(n) = (n-1)!/floor((n+1)/2).
1/a(n+1) is the coefficient of the power series of 3*exp(x)/4 + 1/4*exp(-x) + x/2*exp(x) ; this function is the sum of f_n(x) where f_0(x)=cosh(x) and f_{n+1} is the primitive of f_n. - Pierre-Alain Sallard, Dec 15 2018
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EXAMPLE
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a(1)=a(2)=a(3)=1 because the corresponding graphs are trees.
a(4)=3 because the corresponding graph is a is a triangle with one of its vertices adjacent to a fourth vertex.
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MAPLE
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a:=n->(n-1)!/floor((n+1)/2);
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MATHEMATICA
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Function[x, 1/x] /@
CoefficientList[Series[3*Exp[x]/4 + 1/4*Exp[-x] + x/2*Exp[x], {x, 0, 10}], x] (* Pierre-Alain Sallard, Dec 15 2018 *)
Table[(n - 1)! / Floor[(n + 1) / 2], {n, 1, 30}] (* Vincenzo Librandi, Dec 15 2018 *)
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PROG
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(Magma) [Factorial(n-1)/Floor((n+1)/2): n in [1..25]]; // Vincenzo Librandi, Dec 15 2018
(GAP) List([1..20], n->Factorial(n-1)/Int((n+1)/2)); # Muniru A Asiru, Dec 15 2018
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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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