1. remove all even numbers from the system (see Richard E Crandall 1978, "On the "3x + 1" Problem" in Mathematics of Computation, Vol 32, Number 144, October 1978, pp 1281-1292), and

2. then use the halving version to show that more than half of n (odd are sufficient) go to less than n/2 in finitely many steps.

Then splitting D+ (odd positive integers) into intervals by powers of 2, the majority in an interval go to or below the next smaller interval.
A minority of n take longer, but there are not enough of them to form another tree.

Somewhere on your site I saw a remark asking why numbers like 98, 99, 100, 101 and 102 (a consecutive quintuple) all take the same time to reach 4, 2, 1, 4, ... ? The answer is that they all display the same 3^m/2^k approximant to reach < 4 for the first time in k + m steps, because of "coalescences".

I will send a picture to the Editors by email if it would help.

## Hailstone problem of Lothar Collatz

To prove it you need to

1. remove all even numbers from the system (see Richard E Crandall 1978, "On the "3x + 1" Problem" in Mathematics of Computation, Vol 32, Number 144, October 1978, pp 1281-1292), and

2. then use the halving version to show that more than half of n (odd are sufficient) go to less than n/2 in finitely many steps.

Then splitting D+ (odd positive integers) into intervals by powers of 2, the majority in an interval go to or below the next smaller interval.

A minority of n take longer, but there are not enough of them to form another tree.

Somewhere on your site I saw a remark asking why numbers like 98, 99, 100, 101 and 102 (a consecutive quintuple) all take the same time to reach 4, 2, 1, 4, ... ? The answer is that they all display the same 3^m/2^k approximant to reach < 4 for the first time in k + m steps, because of "coalescences".

I will send a picture to the Editors by email if it would help.

Respect

Chris Hewish