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

6.4. An explicit formula for the number of spanning trees in S n,k 111and we apply Theorem 4.3.26, then we obtain:τ(S n,k ) = (k + 2) n−1 ∗ 2k + τ(S n−1,k ) ∗ (k + 2)τ(S n,k ) = 2k(k + 2) n−1 + (k + 2)τ(S n−1,k )(k + 2)τ(S n−1,k ) = 2k(k + 2) n−1 + (k + 2) 2 τ(S n−2,k )(k + 2) 2 τ(S n−2,k ) = 2k(k + 2) n−1 + (k + 2) 3 τ(S n−3,k )(k + 2) 3 τ(S n−3,k ) = 2k(k + 2) n−1 + (k + 2) 4 τ(S n−4,k ). . . . . . . . .(k + 2) n−4 τ(S 4,k ) = 2k(k + 2) n−1 + (k + 2) n−3 τ(S 3,k )(k + 2) n−3 τ(S 3,k ) = 2k(k + 2) n−1 + (k + 2) n−2 τ(S 2,k )(k + 2) n−2 τ(S 2,k ) = 2k(k + 2) n−1 + (k + 2) n−1 2kBy the sum of all the previous equations, we obtain:τ(S n,k ) = 2k(k + 2) n−1 (n − 1) + (k + 2) n −1 2kWe take 2k(k+2) n−1 as a factor, hence the result.□Particular cases:1) In the previous Theorem 6.4.1, if we take k = 1 (see the star flower planarmap S n,k shown in figure 6.3), then we obtain the star flower planar map S n,1 , which has2n vertices (n vertices of degree 4 and n of degree 2), 3n edges and n + 2 faces (n facesof degree 3, one face of degree n and the other face of degree 2n)(see Figure. 6.6):Figure 6.6: The star flower planar map S n,1Example 6.4.2 Here is an example of the star flower planar map S 2,1 with n = 2 (2triangles) and the star flower planar map S 3,1 with n = 3 (3 triangles) (see Figure. 6.7).