9, 9, 17, 17, 4, 12, 20, 20, 7, 7, 15, 15, 10, 23, 10, 111, 18, 18, 18, 106, 5, 26, 13, 13, 21, 21, 21, 34, 8, 109, 8, 29, 16, 16, 16, 104, 11, 24, 24, ... The starting value having the largest total stopping time while being less than 10 is 9, which has 19 steps, less than 100 is 97, which has 118 steps, less than 1000 is 871, which has 178 steps, less than 104 is 6171, which has 261 steps, less than 105 is , which has 350 steps, less than 106 is , which has 524 steps, less than 107 is , which has 685 steps, less than 108 is , which has 949 steps, less than 109 is , which has 986 steps, less than 1010 is , which has 1132 steps, less than 1011 is , which has 1228 steps, less than 1012 is , which has 1348 steps. These numbers are the lowest ones with the indicated step count, but not necessarily the only ones below the given limit. As an example, has 1132 steps, as does . The starting values having the smallest total stopping time with respect to their number of digits (in base 2) are the powers of two since is halved times to reach 1, and is never increased. Visualizations Supporting arguments Although the conjecture has not been proven, most mathematicians who have looked into the problem think the conjecture is true because experimental evidence and heuristic arguments support it. Experimental evidence , the conjecture has been checked by computer for all starting values up to 268 ≈ . All initial values tested so far eventually end in the repeating cycle of period 3. This computer evidence is still not rigorous proof that the conjecture is true for all starting values, as counterexamples may be found when considering very large (or possibly immense) positive integers, as in the case of the disproven Pólya conjecture. However, such verifications may have other implications. For example, one can derive additional constraints on the period and structural form of a non-trivial cycle. A probabilistic heuristic If one considers only the odd numbers in the sequence generated by the Collatz process, then each odd number is on average of the previous one. (More precisely, the geometric mean of the ratios of outcomes is .) This yields a heuristic argument that