enumeration of the number of spanning trees in some ... - Toubkal

CHAPTER 5.94THE NUMBER OF SPANNING TREES OF CERTAIN FAMILIES OFPLANAR MAPSFigure 5.12: The n-Home chains H nτ(H n ) = 11τ(H n−1 ) − 3 × 3τ(H n−2 ), hence we obtain the system:⎧⎨ τ(H n ) = 11τ(H n−1 ) − 9τ(H n−2 )τ(H 1 ) = 11⎩τ(H 2 ) = 112The characteristic quadratic equation is r 2 − 11r + 9 = 0, so the solutions of this equationare: r 1 = 11−√ 85and r2 2 = 11+√ 85, hence:2τ(H n ) = α( 11 − √ 852) n + β( 11 + √ 85) n , α, β ∈ R, n ≥ 1.2Using the initial condition τ(H 1 ) = 11, τ(H 2 ) = 112, we obtain: α = 1 − 11√ 85,2 170β = 1 + 11√ 85, hence the result. □2 170Numerical results The Table 5.2 illustrates some of the values of the number of spanningtrees in the n-Home chains H n by using the formula given in Theorem 5.7.1:n τ(H n )1 112 1123 11334 114555 1158086 11707937 118364518 1196638249 120977400510 12230539639Table 5.2: Some values of τ(H n )