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

5.5. Particular cases 89Proof: τ(C 1 ) = h, τ(C 2 ) = h 2 − k 2 , in the sequence of maps C n , we cut the last cycle oflength h (see Fig. 5.5) and we use Theorem 4.3.34, then we obtain: τ(C n ) = τ(C n−1 )h −τ(C n−2 )k 2 = hτ(C n−1 ) − τ(C n−2 )k 2 , hence we obtain the system:⎧⎨⎩τ(C n ) = hτ(C n−1 ) − τ(C n−2 )k 2 , with n ≥ 3τ(C 1 ) = h,τ(C 2 ) = h 2 − k 2The characteristic equation is r 2 − hr + k 2 = 0, ∆ = h 2 − 4k 2 , if h ≥ 2k ⇒ ∆ ≥ 0,therefore the solutions of this equation are: r 1 = h−√ h 2 −4k 2and r2 2 = h+√ h 2 −4k 2, hence:2τ(C n ) = α( h−√ h 2 −4k 2) n + β( h+√ h 2 −4k 2) n , α, β ∈ R, n ≥ 1 Using the initial conditions2 2τ(C 1 ) = h and τ(C 2 ) = h 2 − k 2 , we obtain:α = −h2 +h √ √ h 2 −4k 2 +2k 2h 2 −4k 2 (h− √ and β = h2 +h √ √ h 2 −4k 2 −2k 2h 2 −4k 2 ) h 2 −4k 2 (h+ √ , hence the result.h 2 −4k 2 )□5.5.2 The case of k i = 1 and h i = hTheorem 5.5.2 In the previous Theorem 5.5.1, if we take k i = 1 for i = 1, ..., n − 1 andh i = h (h ≥ 3) for i = 1, ..., n, n ≥ 1, we obtain the sequence of maps C n in Figure 5.6,Figure 5.6: Case of n cycles whose lengths are the same with h i = h and k i = 1thenτ(C n ) =(1√h2 − 4( h + √ h 2 − 42) n+1 − ( h − √ h 2 − 4)), n+1 n ≥ 1.2Proof: By replacing k i = 1 and h i = h for i = 1, ..., n in the system (1) of Theorem5.4.1, we get :⎧⎨ τ(C n ) = hτ(C n−1 ) − τ(C n−2 )τ(C 1 ) = h⎩τ(C 2 ) = h 2 − 1