Re: Collatz Question


bill wrote:
mensanator@xxxxxxxxxxx wrote:
bill wrote:
martin cohen wrote:
bill wrote:

The Collatz sequence must;
1. Reach a value of 1
2 Increase indefinitely
3. Oscillate within a fixed range

4. Profit!

explain?

btw, what's the question?

Just blowing off steam! But a couple of questions were intimated..

Has the question of a loop been resolved?

Generally, no. But in my forthcoming paper

Blueprint for Failure: How to Construct a Counterexample to the Collatz
Conjecture

this issue will be addressed. It turns out that you cannot prove
that a counterexample cannot be constructed. While that doesn't
mean there IS a counterexample, it is important to know you can't
rule one out.

For my benefit, a counterexample is a sequence that does not go to 1?

Yes. Although there are none known in the positive domain of 3n+1,
you can see examples in similar systems, such as 3n+5 or 3n+1 in
the negative domain. And keep in mind that when dealing with 3n+C,
the termination point is C, not 1 (unless C=1, of course).

So if we start at 16 in 3n+5, we get

16, 8, 4, 2, 1, 8, 4, 2, 1, 8, ...

and we never reach the proper cycle C, 4C, 2C, C such as
when we start with 40

40, 20, 10, 5, 20, 10, 5, 20, 10, 5, ...

The fact that there are at least two distinct loop cycles is
why the conjecture fails for 3n+5. You'll find that it fails for
all C that are not powers of 3 and seems to work for all that
are powers of 3. Thus the conjecture is provably false for
3n+5, 3n+7, 3n+11, 3n+13, ...
and not known to be false for
3n+1, 3n+3, 3n+9, 3n+27, 3n+81, ...

In the negative domain, the conjecture is false for all C
because there are at least 3 distinct loop cycles:
-C, -5C and -17C
(more when C is not a power of 3).


For instance, loop cycles are constructed in one of two ways.
The first way is how loops in 3n+C systems end up failing.
But the mechanism of such failure (factors of C>3) doesn't apply
to 3n+1 since 1 doesn't have any factors. Thus, we can rule out
the Factor Congruence mechanism completely for 3n+1.

Now, it would appear that the second mechanism might not apply
either, which is necessary for 3n+1 to be true. BUT, the loop
cycle in the negative domain (-17) PROVES that you CANNOT
rule out the second mechanism becuase it is indepedent of the
domain!

A proof that there are no constructible counterexamples via the
second mechanism in the positive domain would also prove there
are none in the negative domain. But -17 IS a counterexample in
the negative domain, so no such prrof can exist.


Does the Collatz sequence obey the laws of nature?

It obeys the laws of mathematics

IE Is the Collatz
sequence
likely to generate fewer numbers divisible by 4, 8, 16, 32, ... , 2^k
than one would
expect based on simple probabilities?

No. All numbers are generated. The real issue is whether
the numbers are all on the same tree, i.e, whether there is a
pathway from 1 to every positive integer. For instance,
the system 3n+5 generates every integer, but the pathway
built upwards from 5 only contains multiples of 5. All other
integers are found on different pathways.

Another problem is that if you build a Collatz tree upwards from
the root, there are always going to be more even numbers than
odd. But that doesn't mean there are actually more even than odd
or that Collatz isn't generating them all, they'll catch up at
infinity.

In the long run, the 3C+1 sequence should generate about twice as many
even numbers as odd numbers. But I haven't found a long run.
What is the longest known sequence with root less than 10^90?

I don't have such statistics handy but it can easily be shown
that sequences can be arbitrarily long. Here's a rule of thumb:

assuming a random number of m binary bits, the quantity
of odd numbers in the sequence will be 2.41*m (and on
average, there will be twice as many even numbers as odd).

So 10**90 would need 299 bits, so the average 299 bit number
would have a sequence 299*2.41 or 720 odd numbers plus
1440 even numbers.

But that's just the mean, so it's nowhere near the longest run.
2**m-1 will have about twice the mean. And the maximum could
be as much as 10-12 times the mean.

Examples:

10**90 has [857, 352] even, odd numbers. But look at the pattern
in binary

11111011010110000111100001001001010010101100111000111010010111110000010010101011010010001010000001000000011001011100011100100011111001100100110011010111100000011000110101011001101110000111111111001101110111001000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

way too many 0s, so not surprising that it's below the mean.

Whereas, 2**299-1 has [3111, 1774] even, odd numbers,
a little higher than expected from the formula, but 4.819*m
is just a statistic, actual mileage will vary:

<http://members.aol.com/mensanator666/Page.htm>

I don't need the whole sequence , just the root!

Well, I can guarantee there are numbers smaller than

1018517988167243043134222844204689080525734196832968125318070224677190649881668353091698687

with longer sequences, but finding one might be hard.


Thanks. Bill J


Is this approach really a new idea?

.

Relevant Pages

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