Re: Proof of Collatz Conjecture?

Benjamin E. Cawaling Jr. wrote:
You wrote:
But this doesn't even begin to show that there
aren't any cycles of length bigger than 3, or
that there aren't any numbers that never wind
up in a cycle.

I believe that what I had established, by the arbitrariness
of n and a (or b) in my argument, is that every standard Collatz
sequence Cn has a length-3 cycle because there is always one
solution with b = 1

But you haven't proved a cycle can't exist with b>1.

to the expression
Cn = <n, m, l, ..., (i, h, ..., d, c, b)>
for any natural number n, i in { n, m, l, ..., d, c, b } and,
hence, every Cn is of the form
Cn = <n, m, l, ..., (4, 2, 1)>.
I stopped at length-3 cycle because a length-3 cycle could not
be included in a larger cycle with length greater than 3

Ok, (4,2,1) can't be included in a larger cycle. But you haven't
proved that a larger cycle must include (4,2,1).

-- although
two input terms to the standard Collatz function could give the same
output term, there is always only one succeeding term to any term in
any sequence or the cycle. Thus, all of your above concerns are answered

No.

-- all of them due to the arbitrariness of n and b in my argument.

If it is necessary to show that there are no cycles with length greater
than 3, then one could simply proceed as follows:
8. Suppose there exists a
Cn = <n, m, l, ..., (e, d, c, b)>
- that is, a length-4 standard Collatz cycle - for some
starting natural number n with
b = min(e,d,c,b).
Then, we simply rehash our earlier argument:
e = f(b) = 3b + 1
d = f(e) = (3b+1)/2 = 4b yields b = -1/5
c = f(d) = 3[(3b+1)/2] + 1 = 2b yields b = -1

Do you think it's a coincidence that the 4-cycle yields -1 just like
the 2-cycle does? You're making a big mistake here (the fallacy of
value-centric thinking). What is important here is that -1 is an
INTEGER (albeit negative). A cycle exists whenever you get an
integer. The sign simply tells you the cycle is in the negative domain.


-- therefore, there is no valid solution so that the cycle
(e,d,c,b) does not exist.

It exists in the negative domain.


Again, suppose there exists a
Cn = <n, m, l, ..., (f*,e,d,c,b)>
-- that is, a length-5 standard Collatz cycle -- for some
starting natural number n with
b = min(f*,e,d,c,b).
Then, we simply rehash our earlier argument:
f* = f(b) = 3b + 1
e = f(f*) = (3b+1)/2
d = f(e) = 3[(3b+1)/2] + 1 = (9b+5)/2 = 4b
yields b = -5

Again, we have an integer.

c = f(d) = 3[(9b+5)/2] + 1 = (27b+17)/2 = 2b
yields b = -17/23

Whoa! You made a big mistake here. Here's the correct analysis:

There are five ways to have a length 5 sequence (I've added
0=even 1=odd to the variable names) and named the paths
A, B, C, D, E. For example, path A is f0,e1,d0,c0,b1.


A B C D E
f0 f0 f1 f0 f1
| | / | /
e1 e0 e0
\ | |
d0 d1
\ /
c0
|
b1

If the sequence is a loop cycle, the predecessor of f is b
and since b is odd, f must be even, so we can rule out paths
C and E.

For path A, I get

3b+1 = (8b-2)/3
b = -5

For path B

3b+1 = 16b
b = 1/13

For path D

3b+1 = (8b-4)/3
b = -7


So not only is your -17/23 completely wrong, but you missed
the third path which yields b = -7.

Note that it, too is an integer. But it is part of the -5 cycle
meaning b is not minimum, so we generally don't count that as
a seperate loop cycle even though the above is mathematically
sound. And the 4-cycle? That's just two iterations of the 2-cycle.
We don't count multi-iterations seperately either.

Would you like to bet that were you to analyze the 6-cycle, you
would find BOTH a +1 result (twice around the 3-cycle) and a
-1 (three times around the 2-cycle)?

-- therefore, there is no valid solution so that the
cycle (f*,e,d,c,b) does not exist.

It exists in the negative domain.


I think that there is a pattern here (I did not explore this
further

And that was a mistake. You should have explored all the way out to
length 18 cycles where you'll find a lot of integer results, a bunch of
new negative ones and the result of 9 times around the 2-cycle (-1)
and 6 times around the 3-cycle (+1).

because I didn't think it is necessary since I believe
that the arbitrariness of n in my earlier argument already
covers these concerns) that we could easily generalize by
mathematical induction to prove that there are no cycles with
length greater than 3 (hence, there is also no divergent sequence).

Conclusion based on false premise.

.

Relevant Pages

  • Re: Proof of Collatz Conjecture?
    ... sequence Cn has a length-3 cycle because there is always one ... any sequence or the cycle. ... print 'PART II: Count legal sequences using partition generator:' ...
    (sci.math)
  • LSE64 - An example
    ... " Din" sequence{ ... " QRST" pulse ... # zero one high levels yields nothing ...
    (comp.lang.forth)
  • Re: An Argument for the Subroot Node
    ... each edge in the sequence is the initial vertex of the next edge ... If a graph has two distinct nodes with more than one path between ... it will contain a cycle. ...
    (sci.math)
  • Re: file editing
    ... > sequence, house, street, cycle, walk, read, filler, period, time); ...
    (comp.unix.questions)
  • Re: file editing
    ... > sequence, house, street, cycle, walk, read, filler, period, time); ...
    (comp.unix.shell)