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

CHAPTER 5.88THE NUMBER OF SPANNING TREES OF CERTAIN FAMILIES OFPLANAR MAPSTheorem 5.4.1 For the sequence maps C n in Figure 5.4, we have:⎧⎨ τ(C n ) = h n τ(C n−1 ) − τ(C n−2 )kn−1, 2 with n ≥ 3(1) τ(C 1 ) = h 1 ,⎩τ(C 2 ) = h 1 h 2 − k12where k n is the number of common edges between the i − th and the (i + 1) − th cyclesof C n and h i is the number of edges of i − th cycle (h i ≥ 3, for i = 1, ..., n).Proof: τ(C 1 ) = h 1 , τ(C 2 ) = h 1 h 2 − k1, 2 in the sequence of maps C n , we cut the last cycleof length h n (see Figure 5.4) and we use Theorem 4.3.34, then we obtain:τ(C n ) = τ(C n−1 )h n − τ(C n−2 )k 2 n−1 = h n τ(C n−1 ) − τ(C n−2 )k 2 n−1,hence we obtain the system (1).□5.5 Particular casesIn this section, we apply the previous Theorem 5.4.1 to obtain some particular cases ofk i and h i .5.5.1 The case of k i = k and h i = h (h ≥ 2k + 1)Theorem 5.5.1 In the previous Theorem 5.4.1, if we take k i = k for i = 1, ..., n − 1 andh i = h (h ≥ 2k + 1) for i = 1, ..., n, n ≥ 1, we obtain the sequence of maps C n in Figure5.5, thenFigure 5.5: Case of n cycles whose lengths are the same with h i = h and k i = kτ(C n ) =(1√h2 − 4k 2( h + √ h 2 − 4k 22) n+1 − ( h − √ h 2 − 4k 2)), n+1 n ≥ 1.2