Only the best!

[mlewis]

Try this one:

m must be > 0

n=m
do
if (n mod 2) <> 0 then n=3*n+1
else n=n/2
loop until n=1


Now prove that this will always end for any number > 0..

Good luck!

[VBOB]

All you're doing is turning odd numbers into even numbers (3*oddnumber) + 1, and then dividing that number by 2 until you return to another odd number. Eventaully you will create an even number that is a power of 2 and hence end the loop.

I know its not exactly a proof, have you a mathematical proof for it.

[mlewis]

There is no mathematical proof. This is known as the Collatz problem; nobody has ever proved it. Logically, your argument sounds correct, but there have been many similar ones that have never been proved.

The formula's been tested to somewhere near 5x10^500 or something like that, so chances are good that it will always end, but who knows?

[DavidHooper]

that sounds pretty stupid - what a crippled problem! mr collatz obviously can't understand the basics of odd/even multiplication:

<dots are units>
...
(x3) ... ... ...
(+1) ... ... ... .

[mlewis]

yes but

352
352x3 = 1056
+1 = 1057
1057 is odd so:
1057x3 = 3168
+1 = 3169
is odd:
3169x3 = 9507
+1 = 9508
is even, divide by 2
4754
divide by 2 again
2377
is odd:
2377x3 = 7132

......

(this actually takes several thousand reps to end up in 1)

see? you can't prove that it will always get to 1

[Guv]

M Lewis: Your post seems to have typos or and/or arithmetic errors. On the general idea, I am on your side. It looks valid but difficult to prove. A bit of research indicates that here are those who have hopes of proving that one is an attractor.

I wrote a simple program for my HP hand calculator to check a few numbers mentioned in articles I found. See below. For simple algorithims, the HP 48GX is more convenient than VB on a PC.

David Hooper: Check what happens if you start with 27. The first few values are the following. It is claimed that it takes 111 loops before terminating.

27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91 . . .

The following terminates rapidly.

29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1

Does the above suggest that maybe it requires a formal proof rather than intuition? At first glance, it seems obvious that one can be proven to be an attractor. After looking at series like the above, one becomes less certain.

I did some research and it seems that termination of the series at one is as yet unproven. It has been experimentally tested extensively. One source claims that all numbers up to 29,300 * 10^12 have been checked. Another claims that it has been checked for numbers up to 5.6 * 10^13. A third source claims verification only up to 7 * 10^11. I wonder if the people investigating this problem talk to each other and compare results.

When you study the roller coaster nature of the some series, you begin to wonder if there is some other odd number which is an attractor.

[mlewis]

Yes, it needs a formal proof, and no, no such proof has been written yet.

Numbers are very wierd.

[NotLKH]

I just wonder about this. Perhaps a little insite could be gleaned
assuming an M = (2^k -1)/3, Then another M = (2^k2 - m0)/3,
and eventually leading to all numbers are of the series (2^kn -
Mn)/3, where Mn is similar.

[MHENK]

Does this work for you'se guyz?

also, where did this thread get's its name? i failed to make a connection?

[mlewis]

The thread was named to direct it to the best math minds in the forum, as the best math minds in the world have been unable to solve this.

[da_silvy]

Originally posted by MHENK
Does this work for you'se guyz?

also, where did this thread get's its name? i failed to make a connection?

Yeah, very nice

What's it for though?

[NotLKH]

Found an Interesting Seive effect.

[mlewis]

The same effects have been seen by many others; as well as some very fascinating ones that involve lots of numbers. There is a theory that the above algorithm will always end in the digits

4 2 1

which makes sense; but again a formal proof has yet to be established.


Btw NotLKH, the links in your signature are broken.

[NotLKH]

Originally posted by mlewis
....There is a theory that the above algorithm will always end in the digits

4 2 1

which makes sense; but again a formal proof has yet to be established.

Btw NotLKH, the links in your signature are broken.

It does seem to be more than a theory, since to get to 1, there is
no number*3 + 1 = 1, excluding 0, so something must divide by 2
to become 1, and that is the number 2, so now we have 2 -> 1,
and there is no odd number*3 + 1 = 2, so something Must be
divided by 2 to get 2, and that would be 4, so now we have 4 -> 2 -> 1: Now Outside of 1, I would postulate that the final series
must always end as 8->4->2->1, at least.

BTW, Thanks for the Links heads up. They were good, but the threads don't exist any more.

-Lou