“…α − 1 = β 3 1 , 2α − 1 = β3 2 and β 1 , β 2 are integers and the only solution of this equation is (β 1 , β 2 ) = (−1, −1) and this solution does not provide desired α.If n ≡ 4,14, 16, 26 (30), then h n = C iff(26) α(−α 2 + 3α − 1) = C,if n ≡ 9, 21 (30), then h n = C iff(27) (2α − 1) 2 (−α 2 + 3α − 1) = C, if n ≡ 12, 18 (30), then h n = C iff (28) (α − 1)(−α 2 + 3α − 1) 2 = C,by (3.8) or equivalently α 2 − 3α+ 1 = C. This last equation leads to equation (5.2). The solutions of this equation do not provide desired α.…”