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

2with∑n−k( )( )k n − kT (n, k; L) =L j . (4)j jj=0Herea 0 = L + 1. (5)Example 1.1. Let L = 1. Vandermonde’s convolution identity implies that( n=k)∑ ( )( )k n − k.j jjHence we haveT (2n, n; 1) =wherefrom we get Catalan numbers( ) 2nc(n) = −nanda n = c(n) + c(n + 1) =( )( )2n 2n, T (2n, n − 1; 1) = ,n n − 1( ) 2n= 1 ) 2nn − 1 n + 1(n(2n)!(5n + 4)n!(n + 2)!(n = 0, 1, 2, . . .).In the paper [?] the authors have proved that the Hankel transform of a n equalssequence of Fibonacci numbers with odd indices1h n = F 2n+1 = √{( √ 5 + 1)(3 + √ 5) n + ( √ 5 − 1)(3 − √ }5) n .5 2n+1Example 1.2. For L = 2 we get like a n (2) the next numbersand the Hankel transform h n :One of us, P. Barry conjectured thath n (2) = 2 n2 −n2 −23, 8, 28, 112, 484, . . . ,3, 20, 272, 7424, 405504, . . . .{(2 + √ 2) n+1 + (2 − √ 2) n+1 }.In general, he made the conjecture, which we will prove through this paper.Theorem 1.1. (The main result) For the generalized Pascal triangle associatedto the sequence n ↦→ L n , the Hankel transform of the sequencec(n; L) + c(n + 1; L)