More hailstones...
In the last issue of PASS Maths we presented the hailstone sequence: an unsolved mathematical mystery. Many of our readers have tried out our hailstone sequence generator and others have asked for more information about the problem or simply why it is of interest.
The hailstone sequence is not the problem's only name. You may also see it referred to as the Syracuse problem, the Collatz problem or simply as the 3n+1 conjecture. A conjecture is simply a mathematical statement which has not yet been proved. To our knowledge the principal conjecture "the sequence terminates at the value 1 for all starting values greater than 0" is still unproved, despite rumours to the contrary!
To try out even longer sequences (using larger starting values than those offered by our own hailstone generator) you could try combining your Mathematical skills with your German and read Alfred Wassermann's page "Experiment mit der 3n+1 Folge" (literally "Experiment with the 3n+1 sequence").
More ambitious readers might like to read the paper "The 3x+1 problem and its generalizations" by Jeff Lagarias for a mathematical overview of some of the lines of attack on solving the problem and for an insight into other related conjectures.
- Mathematical mysteries: the hailstone sequence - Issue No 1, January 1997.
- Experiment mit der 3n+1 Folge by Alfred Wassermann.
- The 3x+1 problem and its generalizations by Jeff Lagarias.
Comments
Anonymous
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