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

CHAPTER 5.104THE NUMBER OF SPANNING TREES OF CERTAIN FAMILIES OFPLANAR MAPSNumerical results The Table 5.7 illustrates some of the values of the number of spanningtrees in the n-Envelope chains E n and Q n by using the formula given in Theorem5.8.4:n τ(E n ) τ(Q n )1 114 552 9971 49503 864444 4296054 74918341 372339005 6492826374 32268951556 562702973111 2796599668507 48766840757904 242368243167058 4226394508982281 21004924580838009 366281888829371034 18203987900001425510 31743941981547513851 15776546789615064750Table 5.7: Some values of τ(E n ) and τ(Q n )5.8.4 Formulae for the number of spanning trees in the n-Diphenylene chainsWe derive two simple formulae for the number of spanning trees of a special family ofplanar maps which is called the n-Diphenylene chains.Theorem 5.8.5 (The n-Diphenylene chains) [87] The complexity of the n-Diphenylenechains D n and Q n (see Figure. 5.19) is given by the following formulae:τ(D n ) = 12 √ 30 (126 + 23√ 30)((11 + 2 √ 30) n−1 − (11 − 2 √ )30) n+2 ,andτ(Q n ) = 3 ((112 √ + 2 √ 30) n − (11 − 2 √ )30) n , n ≥ 1.30Proof: We put d n = τ(D n ) and q n = τ(Q n ). d 1 = 23, q 1 = 6, in the sequence of mapsD n , we cut the last Diphenylene (see Figure 5.19), and we use Theorem 4.3.31 (the samegoes for the sequence of maps Q n ), then we obtain: τ(Q n ) = 6τ(D n−1 )−τ(Q n−1 ), τ(D n ) =23τ(D n−1 ) − 4τ(Q n−1 ) therefore, we have the following system:{dn = 23d n−1 − 4q n−1q n = 6d n−1 − q n−1 with d 1 = 23 and q 1 = 6