The Hankel Transform of the Sum of Consecutive Generalized ...

3is given by−n)/2h n =L(n2 2 n+1√ L 2 + 4·{( √ L 2 + 4 + L)( √ L 2 + 4 + L + 2) n + ( √ L 2 + 4 − L)(L + 2 − √ }L 2 + 4) n .(6)From now till the end, let us denote byNow, we can writeξ = √ L 2 + 4, t 1 = L + 2 + ξ, t 2 = L + 2 − ξ. (7)h n = Ln(n−1)/22 n+1 ξ· ((ξ + L)t n 1 + (ξ − L)t n 2) . (8)Or, introducingϕ n = t n 1 + t n 2, ψ n = t n 1 − t n 2 (n ∈ N 0 ), (9)the final statement can be expressed byh n = Ln(n−1)/22 n+1 ξ· (Lψ n + ξϕ n ) . (10)Lemma 1.1. The values ϕ n and ψ n satisfy the relationsandϕ j · ϕ k = ϕ j+k + (4L) j ϕ k−j , ψ j · ψ k = ϕ j+k − (4L) j ϕ k−j (0 ≤ j ≤ k) (11)ϕ j · ψ k = ψ j+k + (4L) j ψ k−j , ψ j · ϕ k = ψ j+k − (4L) j ψ k−j (0 ≤ j ≤ k). (12)Corollary 1.1. The function h n = h n (L) is the next polynomialh n (L) = 2 −n L n(n−1)/2⎧⎨[(n−1)/2]∑·⎩i=0( n2i+1)L(L + 2) n−2i−1 (L 2 + 4) i +[n/2]∑i=0⎫( n) ⎬2i (L + 2) n−2i (L 2 + 4) i ⎭ .