A331905 Number of spanning trees in the multigraph cube of an n-cycle.
1, 4, 12, 128, 605, 3072, 16807, 82944, 412164, 2035220, 9864899, 47579136, 227902597, 1084320412, 5134860060, 24207040512, 113664879137, 531895993344, 2481300851179, 11543181696640, 53565699079956, 248005494380204, 1145875775104967, 5284358088818688
Offset: 1
Keywords
Examples
The multigraph cube of a 4-cycle has four vertices, with two edges between each pair of distinct vertices - i.e., it is a doubled-edge cover of the complete graph on 4 vertices. The complete graph on 4 vertices has 4^2 = 16 spanning trees, and each of those spanning trees corresponds to 8 spanning trees of the multigraph tree because there are independent choices of 2 multigraph edges to be made for each of the three edges in the graph's spanning tree. So a(4) = 16 * 8 = 128.
Links
- Alois P. Heinz, Table of n, a(n) for n = 1..1546
- G. Baron et al., The number of spanning trees in the square of a cycle, Fib. Quart., 23 (1985), 258-264.
- Yoshiaki Doi et al., A note on spanning trees
- Min Li, Zhibing Chen, Xiaoqing Ruan, and Xuerong Yong, The formulas for the number of spanning trees in circulant graphs, Disc. Math. 338 (11) (2015) 1883-1906, Lemma 1.
Programs
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Maple
a:= n-> ((<<0|1|0|0>, <0|0|1|0>, <0|0|0|1>, <-1|4|1|4>>^iquo(n, 2, 'd'). <[<0, 1, 4, 16>, <1, 2, 11, 49>][d+1]>)[1, 1])^2*n*(2-irem(n, 2)): seq(a(n), n=1..30); # Alois P. Heinz, Feb 06 2020
Formula
The following formulas were provided by Tsuyoshi Miezaki on Feb 05 2020 (see Doi et al. link). Let z1=(-3+sqrt(-7))/4, z2=(-3-sqrt(-7))/4; T(n,z) = cos(n*arccos(z)). Then a(n) = (2*n/7)*(T(n,z1)-1)*(T(n,z2)-1). Furthermore a(n) = 2*n*A005822(n)^2 if n is even, or n*A005822(n)^2 if n is odd. - N. J. A. Sloane, Feb 06 2020
Extensions
More terms from Alois P. Heinz, Feb 06 2020
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