A107991 Complexity (number of maximal spanning trees) in an unoriented simple graph with nodes {1,2,...,n} and edges {i,j} if i + j > n.

1, 1, 1, 3, 8, 40, 180, 1260, 8064, 72576, 604800, 6652800, 68428800, 889574400, 10897286400, 163459296000, 2324754432000, 39520825344000, 640237370572800, 12164510040883200, 221172909834240000, 4644631106519040000, 93666727314800640000

Offset: 1

Author

Roland Bacher, Jun 13 2005

Keywords

• nonn

Comments

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).

Examples

			a(1)=a(2)=a(3)=1 because the corresponding graphs are trees.
a(4)=3 because the corresponding graph is a triangle with one of its vertices adjacent to a fourth vertex.
		

References

• N. Biggs, Algebraic Graph Theory, Cambridge University Press (1974).

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..450
• Niall Byrnes, Gary R. W. Greaves, and Matthew R. Foreman, Bootstrapping cascaded random matrix models: correlations in permutations of matrix products, arXiv:2405.02541 [math-ph], 2024. See p. 7.
• Pierre-Alain Sallard, Coefficients of repeated integrals of hyperbolic cosine.

Crossrefs

Cf. A001113, A073742, A073743.

Programs

GAP

List([1..20],n->Factorial(n-1)/Int((n+1)/2)); # Muniru A Asiru, Dec 15 2018

Magma

[Factorial(n-1)/Floor((n+1)/2): n in [1..25]]; // Vincenzo Librandi, Dec 15 2018

Maple

a:=n->(n-1)!/floor((n+1)/2);

Mathematica

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 *)

PARI

A107991(n)=(n-1)!/round(n/2) \\ M. F. Hasler, Apr 21 2015

SageMath

[factorial(n-1)/floor((n+1)/2) for n in range(1,24)] # Stefano Spezia, May 10 2024

Formula

a(n) = (n-1)!/floor((n+1)/2).

a(n+1) = n!/floor(n/2 + 1). - M. F. Hasler, Apr 21 2015

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

Sum_{n>=1} 1/a(n) = (e + sinh(1))/2 + cosh(1). - Amiram Eldar, Aug 15 2025