It is stipulated that the "H operation" of positive integer n is: ① when n is odd, H = 3N + 13; ② when n is even, H = NX1 / 2x1 / 2 Where h is odd For example, the result of number 3 after one "H operation" is 22, the result after two "H operations" is 11, and the result after three "H operations" is 46 (1) Find 257, after 257 times of "H operation" to get the result (2) If the result of "H operation" is always constant a, find the value of A I want to solve the problem

It is stipulated that the "H operation" of positive integer n is: ① when n is odd, H = 3N + 13; ② when n is even, H = NX1 / 2x1 / 2 Where h is odd For example, the result of number 3 after one "H operation" is 22, the result after two "H operations" is 11, and the result after three "H operations" is 46 (1) Find 257, after 257 times of "H operation" to get the result (2) If the result of "H operation" is always constant a, find the value of A I want to solve the problem

(1)H1 = 784H2 = 49H3 = 160H4 = 5H5 = 28H6 = 7H7 = 34H8 = 17H9 = 64H10 = 1H11 = 16H12 = 1H13 = 16…… From the 10th operation, it is a cycle of 1, 16, 1, 16. (257 - 10) △ 2 is more than 1, so h257 = 16 (2) obviously, 3A + 13 = a × 2 ^