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

CHAPTER 5.92THE NUMBER OF SPANNING TREES OF CERTAIN FAMILIES OFPLANAR MAPSCorollary 5.6.3 (The n-Tent chains) The complexity of the n-Tent chains T n is givenby the following formula:τ(T n ) = √ 1 (21( 5 + √ 212• The n-Hexagonal chains (k = 1 and h = 6)) n+1 − ( 5 − √ 21)), n+1 n ≥ 1.2If we take h = 6 in the sequence of maps C nn-Hexagonal chains H n (see Figure 5.10).in Figure 5.6, we obtain the sequenceFigure 5.10: The n-Hexagonal chains H nCorollary 5.6.4 (The n-Hexagonal chains) The complexity of the n-Hexagonal chainsH n is given by the following formula:τ(H n ) = 1 ((34 √ + 2 √ 2) n+1 − (3 − 2 √ )2) n+1 , n ≥ 1.2• The n-Eight chains (k = 1 and h = 8)If we take h = 8 in the sequence of maps C n in Figure 5.6, we obtain the sequence n-Eightchains E n (see Figure 5.11).Figure 5.11: The n-Eight chains E nCorollary 5.6.5 (The n-Eight chains) The complexity of the n-Eight chains E n is givenby the following formula:τ(E n ) = 1 ((42 √ + √ 15) n+1 − (4 − √ )15) n+1 , n ≥ 1.15