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

5.7. Other uses 93Numerical results The Table 5.1 illustrates some of the values of the number of spanningtrees in the n-Fan chains F n , the n-Grid chains G n , the n-Tent chains T n , then-Hexagonal chains H n and in the n-Eight chains E n , by using the formulae given in theprevious Corollaries:n 1 2 3 4 5 6 7 8 9 10τ(F n ) 3 8 21 55 144 377 987 2584 6765 17711τ(G n ) 4 15 56 209 780 2911 10864 40545 151316 564719τ(T n ) 5 24 115 551 2640 12649 60605 290376 1391275 6665999τ(H n ) 6 35 204 1189 6930 40391 235416 1372105 7997214 46611179τ(E n ) 8 63 496 3905 30744 242047 1905632 15003009 118118440 929944511Table 5.1: Some values of τ(F n ), τ(G n ), τ(T n ), τ(H n ) and τ(E n )5.7 Other usesIn this section, we firstly consider some other cases of maps which are not cycles. Wethen show that the number of spanning trees in some special planar maps which calledthe n-Home chains, the n-Barrel chains and the n-Light chains always satisfy recurrencerelations and describe how to derive these relations, i.e., we find three simple formulae forcalculating the number of spanning trees in the n-Home chains, the n-Barrel chains andthe n-Light chains.5.7.1 Formula for the Number of Spanning Trees in the n-HomechainsIn this section, we derive a simple formula for the number of spanning trees of a specialfamily of maps called the n-Home chains.Now, by applying Theorem 4.3.31, we obtain the following results:Theorem 5.7.1 (The n-Home chains) The complexity of the n-Home chains H n (seeFigure. 5.12) is given by the following formula:τ(H n ) = ( 1 2 − 11√ 85170 ) (( 11 − √ 852) n − 1 9 (11 + √ 85)), n+2 n ≥ 1.2Proof: τ(H 1 ) = 11, τ(H 2 ) = 112, in the sequence of maps Home chains H n , wecut the last Home (see Figure. 5.12) and we use Theorem 4.3.31, then we obtain: