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

5Now,Also,∞∑T (2n, n; L) t n =n=0∞∑T (2n + 2, n; L) t n =n=0∞∑n=0∞∑n=0P (0,0)nP (2,0)n((L+1L−1L+1L−1∞∑T (2n, n − 1; L) t n = t ·n=0∞∑T (2n + 2, n + 1; L) t n = 1 t ·n=0) ((L− 1)t) n= G(0,0)(L+1L−1 , (L − 1)t ),) ((L− 1)t) n= G(2,0)(L+1L−1 , (L − 1)t ).() }{G (2,0) L+1, (L − 1)t − 1 ,L−1() }{G (0,0) L+1, (L − 1)t − 1 .L−1The generating function G(t; L) for the sequence {a n } n≥0 is given byG(t; L) =∞∑a n t nn=0(14)= t + 1 ( )( )G (0,0) L+1L−1t, (L − 1)t − (t + 1)G (2,0) L+1, (L − 1)t − 1 L−1t .The function()L+1ρ(t; L) = φ , (L − 1)t = √ 1 − 2(L + 1)t + (L − 1)L−1 2 t 2 (15)has domainorD ρ =(−∞, 1 − 2√ L + L) (√1 + 2 L + L)∪1 − 2L + L 2 1 − 2L + L , +∞ 2D ρ = (−∞, 1/4) (L = 1).(L ≠ 1),Theorem 2.1. The generating function G(t; L) for the sequence {a n } n≥0 isG(t; L) = t + 1 { }1ρ(t; L) t − 4− 1 (1 − (L − 1)t + ρ(t; L)) 2 t . (16)Example 2.1. For L = 1, we getG(t; 1) =and for L = 2, we find∞∑a n (1) t n = 1 tn=0( (1 −√ 1 − 4t)(1 + t)2t)− 1 . (17)G(t; 2) =∞∑n=0a n (2) t n = − 1 { }t + t + 1 1√t2 − 6t + 1 t − 4(1 − t + √ . (18)t 2 − 6t + 1) 2