we obtain Hn ¼An K. Thus we have Hn ¼Gn K: In Theorem 3, letting k ¼ 2. Then we have l1 n l2 n l1 n1 l2 n2 ¼ g1 n g2 n g1 n1 g2 n2 12 0 1 : Therefore, l2 n ¼ 2g1 n g2 n. Since g1 n ¼g2 nþ1 for all n2 Z, we havel2 n ¼ 2g2 nþ1 g2 n, where l2 n and g2 n is the usual Lucas and Fibonacci numbers, respectively. Indeed, we generalize a relation Lucas and Fibonacci numbers, i.e., Ln ¼ 2Fnþ1 Fn (see p. 176, [4]).