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

5.8. The number of spanning trees in some particular planar maps 97Proof: τ(L 1 ) = 30, τ(L 2 ) = 819, in the sequence of maps Light chains L n , we cutthe last Light (see Figure. 5.14) and we use Theorem 4.3.31, then we obtain: τ(L n ) =30 × τ(L n−1 ) − 9 × 9 × τ(L n−2 ), hence, we obtain the following system:⎧⎨⎩τ(L n ) = 30τ(L n−1 ) − 81τ(L n−2 )τ(L 1 ) = 30τ(L 2 ) = 819The characteristic quadratic equation is r 2 −30r+81 = 0; so, the solutions of this equationare: r 1 = 3 and r 2 = 27, hence: τ(L n ) = α(3) n + β(27) n , α, β ∈ R, n ≥ 1. Using theinitial condition τ(L 1 ) = 30, τ(L 2 ) = 819, we obtain: α = −1,β = 9, hence the result. 8 8□Numerical results The Table 5.4 illustrates some of the values of the number of spanningtrees in the n-Light chains τ(L n ) by using the formula given in Theorem 5.7.3:n τ(L n )1 302 8193 221404 5978615 161424806 4358476597 117678888908 3177330063219 857879118954010 231627362174199Table 5.4: Some values of τ(L n )5.8 The number of spanning trees in some particularplanar mapsIn this section, we derive simple formulae for the number of spanning trees in some specialplanar maps. In the following cases of particular planar maps, each case need anotherplanar map for help us in the proof, for this reason we shall obtain in each case twoformulae for calculation the number of spanning trees in it.