years ago i read in a math puzzle book, the name of which i forgot, that was translated from russian that if you choose any positive integer greater than 1 and perform the following procedure you'll allways reach 1.
procedure of operations:
if the integer you chose is even - keep dividing it by 2 until the quotient is 1 or until the quotient is an odd number greater than 1.
if the ouotient of division by 2 is an odd number greater than 1 - multiply it by 3 and than add 1 to the result and divide it by 2 repeating the above procedure.if you initially choose an odd integer multiply it by 3,add 1 to the result and continue as described in the previous paragraph.
example 1: integer chosen: 17
17*3 1=52, 52/2=26, 26/2=13, 13*3 1=40, 40/2=20, 20/2=10, 10/2=5,
5*3 1=16, 16/2=8, 8/2=4, 4/2=2, 2/2=1.
example 2: integer chosen:31
31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479,
1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154,
577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1.
according to the above mentioned book there was no proof of the generality of this rule but nobody has found a contrary example.also there was no known rule concerning the number of steps needed to reach the final result (1) as a function of the initial (chosen) integer.
i would be very grateful if you could kindly tell me the name of the book as well as if there was found a proof for the above rule or if anyone managed to find a contrary example.