
Max. power of 2 and number of odd integers in a loop in the Collatz problem
Posted:
Jun 3, 2009 5:43 AM


I have found, that if there exists a loop, then if n = number of odd integers in the loop k = maximum power of 2 that divides a number in the loop then, the following conclusions hold... for positive integers k <= n+1 for negative integers k < n log3//log2  n + 1 Consider the negative loop starting with 17 50 25 74 37 110 55 164 82 41 122 61 182 91 272 136 68 34 and then back to 17 Here... n = 7 k = 4 4 < 7 * log3/log2 7 + 1 = 5.09...

