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

130 CHAPTER 7. MAXIMAL PLANAR MAPSProof: We put f n = τ(F n ) and g n = τ(G n ). f 2 = 2, g 2 = 1, in the map F n , we cut thefirst cycle (see Figure. 7.11), and we use Theorem 4.3.31 (the same goes for the mapG n ), then we obtain: τ(G n ) = 3τ(F n−1 ) − τ(G n−1 ), τ(F n ) = 2τ(G n ) − τ(F n−1 ) therefore,we have the following system:{fn = 2g n − f n−1g n = 3f n−1 − g n−1 with f 2 = 2 and g 2 = 1,we replace by the value of g n in the first equation, we get:{fn = 5f n−1 − 2g n−1g n = 3f n−1 − g n−1 with f 2 = 2 and g 2 = 1(fn)= Mg n)= Mg n(fn(fn−1), where M =g n−1)g n−1(fn−1( 5 -23 -1= ... = M n−2 (f2g 2),then, we compute M n−2 :det (M − λI 2 ) = λ 2 −4λ + 1 = 0, λ 1 = 2 − √ 3 and λ 2 = 2 + √ 3, λ 1 ≠ λ 2 then there isP invertible such that M = P DP −1 where( )λ1 0D =,0 λ 2P is the transformation matrix formed by eigenvectors( 1 1P =M n−2 = P D n−2 P −1 whereD n−2 =3+ √ 32(P −1 = √ −133− √ 33− √ 322-1−3− √ 312),)( √ )(2 − 3)n−200 (2 + √ 3) n−2 ,from which we obtain M n−2 , then(fng n)= M n−2 (f2g 2),,),hence the result.□