Collatz conjecture
Jim Spriggs
or the other, but is it An Important Mathematical Problem?
A N Niel
Spriggs <jim.s...@ANTISPAMbtinternet.com.invalid> wrote:
> Is anybody interested? Well, I suppose it would be nice to know one way
> or the other, but is it An Important Mathematical Problem?
There was a column in the Monthly on unsolved problems a few years back.
Collatz was under the heading: "Do not work on this problem".
Larry Lard
Isn't that what everyone told Wiles, though? :)
--
Larry Lard
Replies to group please
robert j. kolker
A N Niel wrote:
>
> There was a column in the Monthly on unsolved problems a few years back.
> Collatz was under the heading: "Do not work on this problem".
Was a reason given?
Bob Kolker
robert j. kolker
Larry Lard wrote:
>
> Isn't that what everyone told Wiles, though? :)
I doubt it.
Bob Kolker
>
robert j. kolker
Jim Spriggs wrote:
> Is anybody interested? Well, I suppose it would be nice to know one way
> or the other, but is it An Important Mathematical Problem?
What makes a mathematical problem Important is its connection to many
other branches of mathematics. For example FLT and the Riemann
Hypothesis. I do not know if the Collatz has this characterstic.
There is no doubt that it is a Hard Problem. Paul Erdos one said of the
Collatz that mathematics is not ready for problems of this sort.
Bob Kolker
Dave Seaman
> Larry Lard wrote:
>>
>> Isn't that what everyone told Wiles, though? :)
> I doubt it.
Nobody knew what Wiles was working on.
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&bookid=228>
robert j. kolker
Dave Seaman wrote:
>
> Nobody knew what Wiles was working on.
Wiles was pretty cagey about his attack on FLT. Nothing was breathed
about it until he was nearly ready to present his work.
Bob Kolker
>
>
Larry Lard
Dave Seaman wrote:
> On Wed, 09 Feb 2005 08:45:46 -0500, robert j. kolker wrote:
>
>
> > Larry Lard wrote:
> >>
> >> Isn't that what everyone told Wiles, though? :)
>
> > I doubt it.
>
> Nobody knew what Wiles was working on.
OK I checked, cos this wasn't an original thought (I have few), and I
dug up the transcript of the BBC 'Horizon' programme. You're right that
once proving FLT became a matter of proving Taniyama-Shimura, from
*that* point Wiles was working in secrecy, but bearing in mind that
this had been an interest of his from childhood:
>>
NARRATOR: ... After centuries of failing to find a proof,
mathematicians began to abandon Fermat in favour of more serious maths.
In the 70s Fermat was no longer in fashion. At the same time Andrew
Wiles was just beginning his career as a mathematician. He went to
Cambridge as a research student under the supervision of Professor John
Coates.
JOHN COATES: I've been very fortunate to have Andrew as a student, and
even as a research student he, he was a wonderful person to work with.
He had very deep ideas then and it was always clear he was a
mathematician who would do great things.
NARRATOR: But not with Fermat. Everyone thought Fermat's last theorem
was impossible, **so Professor Coates encouraged Andrew to forget his
childhood dream and work on more mainstream maths. **
ANDREW WILES: The problem with working on Fermat is that you could
spend years getting nothing so when I went to Cambridge my advisor,
John Coates, was working on Iwasawa theory and elliptic curves and I
started working with him.
>>
(my emphasis; text taken from
<http://web.archive.org/web/20020601180301/www.bbc.co.uk/science/horizon/fermattran.shtml>)
So yeah, not 'everyone' and not explicitly 'do not work on this
problem', more 'work on what I'm working on, apprentice!', but the
points stands I think.
po...@sat.inesc-id.pt
like to do some research on my favourite topics. Collatz is one case,
I've read some papers some months ago about it but nothing really
interesting.
I think we need some more mathematical developments to start working
with it...
The main issue, I think, is that the problem is so simple to state that
it is VERY, VERY difficult to know where to start.
I don't know about its interest in other branches of mathematics but it
is sure an excelent problem. I hope to live till the day someone proves
it!
Oh, my interest made me program an analysis problems for collatz
sequences and numbers in Scheme (it was also a nice example of the
plotting powers of DrScheme):
http://mega.ist.utl.pt/~pocm/collatz.scm
Have fun!
Paulo Matos
Larry Hammick
news:1107959729.4...@f14g2000cwb.googlegroups.com...
> Well, I'm interested in Collatz, although I'm not a Mathematician I do
> like to do some research on my favourite topics. Collatz is one case,
> I've read some papers some months ago about it but nothing really
> interesting.
> I think we need some more mathematical developments to start working
> with it...
> The main issue, I think, is that the problem is so simple to state that
> it is VERY, VERY difficult to know where to start.
exists, it could go overlooked for years, simply because
no one has ever seen its like before. Collatz wouldn't be
the first such problem to suddenly and unexpectedly bite
the dust. Compare e.g. the result the that two knots in a
string cannot "cancel each other out". And look how many
years passed before anybody noticed a connection
between the elliptic curve stuff and FLT (Kenneth Ribet,
1986).
LH
ste...@math.bgsu.edu
One good place to start is to try to solve the finite cycles
conjecture.
There is a very nice online paper by DeWeger in which he proves
that there is no finite cycle for the Collatz problem with less than
69 circuits. (Actually, I can extend it to 71).
Note that if we change the multiplier to 5(instead of 3), this is no
longer true: We have the cycle 13,33,83 in that case.
To me this is the hardest part of the Collatz conjecture.
Regards,
Ray Steiner
mensa...@aol.com
(require (lib "plot.ss" "plot")
(lib "class.ss")
(lib "plot-extend.ss" "plot")) ;; Para criar os plots
Are these libraries part of DrScheme? Program won't run without them.
But I never use DrScheme even though I have it installed, so it
might just be me.
Christian Bau
Jim Spriggs <jim.s...@ANTISPAMbtinternet.com.invalid> wrote:
> Is anybody interested? Well, I suppose it would be nice to know one way
> or the other, but is it An Important Mathematical Problem?
Any techniques that can be used to solve it would be interesting.
Gerry Myerson
Jim Spriggs <jim.s...@ANTISPAMbtinternet.com.invalid> wrote:
> Is anybody interested? Well, I suppose it would be nice to know one way
> or the other, but is it An Important Mathematical Problem?
Was Fermat's Last an important problem? Not in the sense of
there being lots of other results that depended on it, no;
but in the sense of lots of mathematics being developed in
the pursuit of it, yes.
It may be too early to tell whether 3n + 1 is important.
I think all we can say is, it's not hugely important yet.
--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
robert j. kolker
G>
> It may be too early to tell whether 3n + 1 is important.
> I think all we can say is, it's not hugely important yet.
>
Connections to other branches of mathematics from Fermat's Last came
pretty early on. Ditto for the Riemann Hypothesis. The Collatz is still
a pretty isolated problem. I seriously doubt that it will give way to
elementary methods isolated to number theory. Paul Erdos was of the
opinion that mathematicians are not yet ready for problems like the Collatz.
Bob Kolker
Mike Kent
> A N Niel wrote:
>>There was a column in the Monthly on unsolved problems a few years back.
>
>>Collatz was under the heading: "Do not work on this problem".
>
>
> Isn't that what everyone told Wiles, though? :)
Probably not everyone. At the AMS Summer Institute in Corvallis
(in 1977 I think) Serre offered a wager that FLT would be proved
within 10 years.
Thomas Nordhaus
>In article <420A0D4A...@ANTISPAMbtinternet.com.invalid>,
> Jim Spriggs <jim.s...@ANTISPAMbtinternet.com.invalid> wrote:
>
>> Is anybody interested? Well, I suppose it would be nice to know one way
>> or the other, but is it An Important Mathematical Problem?
>
>Was Fermat's Last an important problem? Not in the sense of
>there being lots of other results that depended on it, no;
>but in the sense of lots of mathematics being developed in
>the pursuit of it, yes.
Ah. Sort of like the moon-landing.
Thomas
pmatos
able to use it with language module with no problems at all. Sorry for
the portuguese comments in the code!
Bill Dubuque
>Gerry Myerson <ge...@maths.mq.edi.ai.i2u4email> wrote:
>>
>> Was Fermat's Last an important problem? Not in the sense of
>> there being lots of other results that depended on it, no;
>> but in the sense of lots of mathematics being developed in
>> the pursuit of it, yes.
In popular math history expositions one often finds claims that work
on FLT strongly motivated the development of algebraic number theory.
This is not true. While many leading mathematicians of the past had
a passing interest in FLT, most deemed it an isolated problem with
little if any connection to mainstream mathematics. The development
of algebraic number theory was motivated much more strongly by innate
open problems, such as the generalization of quadratic reciprocity
to higher reciprocity laws. This was a fertile ground that led to
beautiful developments with many interesting connections to other
branches of mathematics.
> Connections to other branches of mathematics from
> Fermat's Last came pretty early on.
Why do you think that? Perhaps your opinion is based upon
said erroneous popular historical expositions?
--Bill Dubuque
mensa...@aol.com
Thanks, I was on 2.06. Installing the latest version fixed the
library problem. But how do I run it? What PLT language do I use?
Textual? Graphical? Pretty Big? One of the other 18 choices?
And once I select a language, how do I get it to do anything?
One program I've got requires that I click the [RUN] button
to create the graphic window and then at the prompt type
(spiral 1000) to make it draw the graphic.
I can't figure out what I'm supposed to do with your program.
pmatos
- Open DrScheme (209)
- Choose language (module ...)
- Load file
- Click Run
- Now, on the interactions window, after the prompt you can call
functions like the following: (now you can choose to hide definitions
window)
----> For a report of the sequence generated by a given value:
(value-report k) where k is the value. For example, (value-report 1000)
will generate a report on the sequence started by 1000.
----> For a report of a sequence of sequences generated by a list of
values:
(seq-report init end step) where init, end and step are integers. For
example (seq-report 1 1000 1) will generate a report of all the
sequences started by 1, 2, ... 1000.
There are many other functions but these are the most important and
since the plot labels were in portuguese and function names were in
portuguese I changed the text to English in the most important places.
http://mega.ist.utl.pt/~pocm/collatz-en.scm
Hope I can be of help! :)
Have fun,
Paulo Matos
mensa...@aol.com
> Oh, ok, so for non-schemers... here's a fast tutorial:
> - Open DrScheme (209)
> - Choose language (module ...)
> - Load file
> - Click Run
> - Now, on the interactions window, after the prompt you can call
> functions like the following: (now you can choose to hide definitions
> window)
Thanks, I did manage to get it to work by typing
(require collatz)
at which point I could run the reports.
> ----> For a report of the sequence generated by a given value:
> (value-report k) where k is the value. For example, (value-report
1000)
> will generate a report on the sequence started by 1000.
>
> ----> For a report of a sequence of sequences generated by a list of
> values:
> (seq-report init end step) where init, end and step are integers. For
> example (seq-report 1 1000 1) will generate a report of all the
> sequences started by 1, 2, ... 1000.
>
> There are many other functions but these are the most important and
> since the plot labels were in portuguese and function names were in
> portuguese I changed the text to English in the most important
places.
> http://mega.ist.utl.pt/~pocm/collatz-en.scm
>
> Hope I can be of help! :)
At least I now know a couple of words of Portugese.
>
> Have fun,
I have a gift for finding faults and of course I managed to crash
your program within 10 minutes. I knew better than to try to plot
or print the sequence for my favorite number, the 1st 6th Generation
Type [1,2] Mersenne Hailstone, so I tried the following just to see
how long it would take
(collatz (- (expt 2 177525) 1))
which should have printed just Tamanho and Pico. But it crashed,
out of virtual memory. I guess you're storing the entire sequence
even though that's not required to find the stopping time and
excursion. Scheme should be able to handle the 80000+ digit
excursion value, but probably not if it has to store all
2.5 million large integers.
I'm working on a library of Collatz related functions for Python
and have run into the same situation, so I either have two seperate
functions (with and without sequence storage) or pass a parameter
to skip the memory usage so I can test ridiculous numbers if
desired.
Anyway, thanks again. I find the Collatz Conjecture endlessly
fascinating and there's never enough discussion on sci.math.
>
> Paulo Matos
pmatos
demonstrating my students some Scheme and to show them what Collatz
Conjecture was. I'll hopefully translate it all to english and improve
the data structures and algorithms to find some stuff faster.
:)
When that's ready, I'll post again. (maybe this weekend, who knows) ;)
Cheers,
Paulo Matos
Ernst
> fascinating and there's never enough discussion on sci.math.
Too funny. We should have a Web Site just for this. What happened to
the one that one dood started... He moved or something..
Hello everyone...
Was looking for inspiration to start working on a related application
and got a search hit on this thread. So I dropped in...
I'm not sure what "proper" things yall will be or are doing but I
thought to share that there are other systems for which single
attractor exists. Namely all powers of 3 for Y in A(x)+/-y,x/2
3(x)+3,x/2 and 3(x)+ 43046721, x/2 are both single attractor systems.
It's interesting imo.
There are multi-attractor
Y = 5 has 6!
--
A snippit
------
Report for 3(x) + 5 Contains 6 Attractor(s)
1 19 5 23 187 347 < The ratio 347/187 ( 1.855614973 ) is seen in
3(x)+y where y is a multiple of 5.
------
A while back I was poking around and thought there are convergent
systems and divergent systems... I think 7(x) + y y > 3 if I remember
right is divergent.. and 7(x) - y has convergent ones... Not too much
was done to show proof on this aspect and my memory of those quick runs
is a bit dated.
Good luck to ya!
Anyway There are some well worn steps on this path.
But yes, perhaps someone will see the whole picture.
Seems to be the fabric of something bigger.
Now if I can just motivate. <grin>
Ernst
Ernst
robert j. kolker Feb 9, 5:44 am show options
Newsgroups: sci.math
From: "robert j. kolker" <nowh...@nowhere.net> - Find messages by this
author
Date: Wed, 09 Feb 2005 08:44:47 -0500
Local: Wed, Feb 9 2005 5:44 am
Subject: Re: Collatz conjecture
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Bob Kolker
-----
The parity language works well in encryption imo.
Ernst
mensa...@aol.com
> > Anyway, thanks again. I find the Collatz Conjecture endlessly
> > fascinating and there's never enough discussion on sci.math.
>
>
> Too funny. We should have a Web Site just for this. What happened
to
> the one that one dood started... He moved or something..
Unfortunately, he moved a couple weeks after setting up the
web site and didn't get it back on-line for a month by which
time everyone seemed to have forgotten about it. Bad planning
on his part in retrospect.
>
>
> Hello everyone...
>
> Was looking for inspiration to start working on a related
application
> and got a search hit on this thread. So I dropped in...
>
> I'm not sure what "proper" things yall will be or are doing but I
> thought to share that there are other systems for which single
> attractor exists. Namely all powers of 3 for Y in A(x)+/-y,x/2
>
> 3(x)+3,x/2 and 3(x)+ 43046721, x/2 are both single attractor systems.
>
> It's interesting imo.
I don't remember where I was when you dropped out of the
newsgroup, but I've made some progress since you first
brought up the notion of attractors (previously, I had
confined my research to 3x+1, now I generalize to 3n+C
where C is an odd integer).
I now use a structure I call the Sequence Vector which is
defined to be: starting from an odd number, the list of
contiguous n/2 operations. For example, the sequence from
27 to 31 is
27 -> 82 -> 41 -> 124 -> 62 -> 31
so the sequence vector is [1,2] since there is one 3n+1
operation, one n/2 operation, one 3n+1 operation and
two n/2 operations. The structure [1,2] has an implied
3n+1 step preceeding each number in the vector. This comes
in handy in the calculations that follow.
Next, I define a Hailstone Function as the equation that
takes you from 31 back to 27. In this case, if 'a' is the
starting number (the hailstone), the seed (g) that generates
it would be
g = (((((((a*2)*2)-1)/3)*2)-1)/3)
That works out to be
g = (8a - 5)/9
In general, all sequence vectors have an associated Hailstone
Function of the form
g = (Xa - Z)/Y
Generalizing to 3n+C makes the Hailstone Function
g = (Xa - ZC)/Y
This is the equation of a straight line with slope X/Y and
intercept -ZC/Y. Now, since X is always a power of 2 and Y
is always a power of 3, the slope cannot ever be 1 which
means it cannot be parallel to the identity line f(x) = x.
It must, therefore, intercept the identity line.
I call the intercept the Crossover Point and it can be
calculated from the Hailstone Function parameters:
Crossover Point = ZC/(X-Y)
Now here's the cool part. If the Crossover Point is an
integer, then the seed equals the hailstone and you have
an attractor!
For the Crossover Point to be an integer, you need
the denominator X-Y to be 1 or -1
or
the factors of ZC must cancel all the factors of X-Y
Plugging the Hailstone Function parameters for [1,2]
into the Crossover Point equation for 3n+1, we get
5*1/(8-9) = 5/-1 = -5
so the sequence vector [1,2] has an attractor at -5.
And any 3n+C system will have attractors at +C, -C, -5C
and -17C.
Now the reason there is only the single know positive
attractor in 3n+1 is because C=1 and 1 can't cancel any
factors of the denominator of the Crossover Point.
Apparently, the truth of the Collatz Conjecture implies
that Z alone will never cancel _all_ the factors of X-Y.
Actually, the attractor at -17 is the only known case
where it does, the others having the trivial condition
of X-Y being 1 or -1. But the original Collatz Conjecture
was about positive integers otherwise we can just write
the whole conjecture off as being false.
Now, getting back to your observation that when C is a
power of 3, there is a single positive attractor, note
that since X is a power of 2 and Y is a power of 3, the
difference _cannot_ have 3 as a factor. In terms of the
Crossover Point, C being a power of 3 cannot cancel any
more factors in the denominator than 1 can and we are left
with just Z to establish attractors. Another way to say
this is that all C that are powers of 3 (which includes
1 since 3^0=1) have the same attractors. And if the
Collatz Conjecture is true, those attractors are +C, -C,
-5C and -17C.
>
> There are multi-attractor
>
> Y = 5 has 6!
Whenever C is _not_ a power of 3, we have the factors
of Z and C working together to cancel the factors of X-Y
and you'll get multiple positive attractors whenever C
supplies the factors that Z can't come up with. Try running
3n+85085. You'll get lots of attractors because 85085 is
5*7*11*13*17 and a lot more sequence vectors become loops.
>
> --
>
> A snippit
> ------
> Report for 3(x) + 5 Contains 6 Attractor(s)
> 1 19 5 23 187 347 < The ratio 347/187 ( 1.855614973 ) is seen
in
> 3(x)+y where y is a multiple of 5.
The same logic applies here as for powers of 3. Any
attractor when C was 5 will also be an attractor when C
is a multiple of 5. Try C=35, you'll get the same sequence
vectors forming loops as the ones that formed in C=5 and
C=7. And remember that sequence vectors are not value
vectors. [2] is the +C attractor in every 3n+C system.
Note that each has a different sequence of values, but the
same sequence of 3n+C, n/2 operations.
3n+1: 1 -> 4 -> 2 -> 1
3n+3: 3 -> 12 -> 6 -> 3
3n+5: 5 -> 20 -> 10 -> 5
3n+7: 7 -> 28 -> 14 -> 7
3n+9: 9 -> 36 -> 18 -> 9
>
> ------
>
> A while back I was poking around and thought there are convergent
> systems and divergent systems... I think 7(x) + y y > 3 if I
remember
> right is divergent.. and 7(x) - y has convergent ones... Not too
much
> was done to show proof on this aspect and my memory of those quick
runs
> is a bit dated.
No comment here. I have not looked at Dn+C systems.
>
>
>
> Good luck to ya!
Glad to see you're back.
Ernst
Collatz is a series of some kind that is for sure... on the one side
it's the A(x)+/-y geometric and then the x/2 approaching 0 and the
transition back and forth.
I have no new Collatz data. There is a slope in the attractors data I
have not calculated but see. It may well be a constant of some type.
Lets see; been playing with encoding numbers a bit and I worked on
compressing data last year.
I have had success with encoding that I hope to share in the future.
That's all I got. The attractors program isn't public but it could
be made so.
Your presentation is looking good.
Man this thing sure is multidimensional.
Ernst
mensa...@aol.com
Ernst wrote:
> You always did write well.
Thanks. Sometimes writing out your thoughts helps clarify them.
Prior to yesterday's post, I don't think I had appreciated the
significance of the -17 attractor. The existance of this attractor
is a counter-example to the notion that there are no non-trivial
attractors when C=1. Sure, it's negative, but the Hailstone Function
and Crossover Point bridge the domains of positive and negative
integers. If there can be one non-trivial attractor, why can't there
be more?
A lot of proofs are based on proving that a certain thing cannot
exist. That relieves you of having to check every possibility.
But -17 shows that there can be no such proof. So where does that
leave us?
There is still the possiblity that you can prove the non-existance
of a _positive_ non-trivial attractor, but the sign isn't relevant
to the cancelling of factors in the Crossover Point fraction. There
is the possibility that certain factors only appear in X-Y when
it's negative, but I don't have any data on that.
>
> Collatz is a series of some kind that is for sure... on the one side
> it's the A(x)+/-y geometric and then the x/2 approaching 0 and the
> transition back and forth.
>
> I have no new Collatz data. There is a slope in the attractors data
I
> have not calculated but see. It may well be a constant of some type.
>
> Lets see; been playing with encoding numbers a bit and I worked on
> compressing data last year.
>
> I have had success with encoding that I hope to share in the future.
>
>
> That's all I got. The attractors program isn't public but it could
> be made so.
>
> Your presentation is looking good.
I got a lot more, if you want to see it.
Ernst
To tell you the truth, I worked with positive numbers.
So wasn't there more than one attractor for negative numbers in
3x+1,x/2?
I figure we will crack it or it will crack us :)
Ernst
mensa...@aol.com
Let's be realistic. The problem could well be undecidable (in
which case it would be true since undecidable means there are no
counter-examples). We are not going to crack it. But that doesn't
mean we have to abandon it. There is plenty to learn and if you
adopt a eudaemonic attitude, it won't matter whether you succeed
or not. Get your satisfaction from the journey and don't be
concerned about the destination. Take a lesson from James Harris
about the folly of being obsessed with great discoveries.
Yes, there are, in fact, three negative attractors in 3n+1:
-1, -5 and -17. From my work with Hailstone Functions, I've learned
that the positive and negative numbers should be considered as a
single system. They are not independant of each other. A sequence
that starts in the positive domain can never cross over into the
negative domain and vice versa, but that is not unusual in
mathematics, just look at the function f(x) = 1/x.
One of the lessons you should learn from Collatz is to abandon
value-centric thinking. Collatz is all about patterns and structures.
When you construct a Collatz tree, you start with the trivial loop
+C and building upwards from there. In 3n+1, the trunk of the tree
happens to be the powers of 2. In 3n+3, the powers of 2 are simply
an unimportant side branch. In 3n+5, the powers of 2 don't even
connect to the primary tree whose root is 5, they form a seperate
tree. But what's important is that the primary tree of 3n+5 (root 5)
has the same structure as the primary tree of 3n+1 (root 1).
So you should really refer to attractors by their structure and not
the values that are associated with them. As I pointed out in a
previous post, the same attractor structure takes on different values
in different systems. And Sequence Vectors are the perfect way to
describe attractor structures that is independent of the system.
And once you do that, you realize that the positive and negative
domains are two pieces of the same puzzle.
Therefore, we should say every 3n+C system has attractors whose
structure is [2], [1], [1,2] and [1,1,1,2,1,1,4]. These are the
sequence vectors that give you +1, -1, -5 and -17 in 3n+1. The same
sequence vectors give you +3, -3, -15 and -51 in 3n+3 and +C, -C,
-5C and -17C in any 3n+C system. Systems like 3n+5 and 3n+7 will have
additional attractors in both the positive and negative domains.
I've been studying 3n+C systems in order to learn how attractors work.
Unfortunately, 3n+1 doesn't have any non-trivial positive attractors,
leaving me with nothing to study. But 3n+85085 has lots of positive
attractors and much can be learned from it.
7 is an attractor in 3n+85085:
7 85106 42553 212744 106372 53186 26593 164864 82432 41216 20608
10304 5152 2576 1288 644 322 161 85568 42784 21392 10696 5348 2674
1337 89096 44548 22274 11137 118496 59248 29624 14812 7406 3703
96194 48097 229376 114688 57344 28672 14336 7168 3584 1792 896 448
224 112 56 28 14 7
The sequence vector is
sv = [1, 3, 10, 6, 3, 5, 1, 15]
The Hailstone Function parameters are
X = 2**sum(sv)
= 2**44
= 17592186044416
Y = 3**len(sv)
= 3**8
= 6561
Z = 3**0*2**29 + 3**1*2**28 + 3**2*2**23 + 3**3*2**20 +
3**4*2**14 + 3**5*2**4 + 3**6*2**1 + 3**7*2**0
= 1447320941
The Crossover Point is
C * Z
CP = ---------
X - Y
85085 * 1447320941
= ---------------------
17592186044416 - 6561
85085 * 1447320941
= --------------------
17592186037855
= 7
It's instructive to look at CP in terms of factors
5*7*11*13*17 * 11*173*373*2039
CP = ------------------------------
5*11*11*13*17*173*373*2039
and if we re-arrange the factors, we see that every factor in the
denominator gets cancelled by a factor in the numerator. We have
a single numerator factor left over and that becomes the integer
Crossover Point.
5 * 7 * 11 * 11 * 13 * 17 * 173 * 373 * 2039
CP = --------------------------------------------
5 * 11 * 11 * 13 * 17 * 173 * 373 * 2039
Or perhaps I should say _a_ Crossover Point. Because this is an
attractor, I could use any odd interger in the loop sequence as
the Crossover Point, by convention, we use the smallest odd element
in the sequence. Given a sequence vector (sv) that is an attractor,
all cyclic permutations of that sequence vector are the same
attractor. But the Hailstone Function is different for each
permutation, so how can they all be the same attractor?
First of all, the X and Y Hailstone Function parameters are invariant
under cyclic permutation, only the Z parameter chages. To see how
this aspect works, we do a factor analysis on each of the 8 cyclic
permutations.
sv cyclic permutation X Y Z CP
[1, 3, 10, 6, 3, 5, 1, 15] 17592186044416 6561 1447320941 7
[3, 10, 6, 3, 5, 1, 15, 1] 17592186044416 6561 8798264000339 42553
[10, 6, 3, 5, 1, 15, 1, 3] 17592186044416 6561 5498372254859 26593
[6, 3, 5, 1, 15, 1, 3, 10] 17592186044416 6561 33288381643 161
[3, 5, 1, 15, 1, 3, 10, 6] 17592186044416 6561 276438299731 1337
[5, 1, 15, 1, 3, 10, 6, 3] 17592186044416 6561 2302687617131 11137
[1, 15, 1, 3, 10, 6, 3, 5] 17592186044416 6561 765632777789 3703
[15, 1, 3, 10, 6, 3, 5, 1] 17592186044416 6561 9944542185611 48097
C factors Z factors
5 7 11 13 17 11 173 373 2039
5 7 11 13 17 11 173 373 2039 6079
5 7 11 13 17 11 29 131 173 373 2039
5 7 11 13 17 11 23 173 373 2039
5 7 11 13 17 11 173 191 373 2039
5 7 11 13 17 11 37 43 173 373 2039
5 7 11 13 17 11 23 23 173 373 2039
5 7 11 13 17 11 173 373 2039 6871
------------------------------------------------------------------
X-Y factors
5 11 13 17 11 173 373 2039
Amazin', ain't it? Note how every value of Z has a different
factorization and yet every one has the same factors in common - the
very factors needed to cancel all the X-Y factors. And the uncancelled
factors remaining become the the 8 Crossover Points which are the
8 odd numbers in the original sequence. And you can see how C and Z
work together to cancel the factors. If C were 1, some factors in
the denominator would go uncancelled and this sequence vector would
not be an attractor.
Space doesn't permit me to show you the factor analysis for attractor
42595. The sequence vector has 32 elements. There are 32 cyclic
permutations each of which has enough common factors to make every
one of the Crossover Points an integer.
And I haven't even done a factor analysis on the attractor that has
168 elements. It must truly be amazing.
It's been proved that a non-trivial positive 3n+1 loop must have at
least 275000 numbers. That's probably a sequence vector of length
91666. The attractor would have 91666 cyclic permutations. Imagine
what the factor analysis for that would look like.
So what does this all prove? Nothing. But now I have what I believe to
be a thorough understanding of how attractors work. All the analysis
above is courtesy of a suite of Python functions I developed for
Collatz research:
collatz_C(start,C,print_options)
returns R1,R2,attractor
collatz(start,print_options)
returns R1,R2
gclass(start,hailstone_function_parameters)
returns sequence_vector_generation
geni(generation,generation_member,hailstone_function_parameters)
returns ith_kth_generation_hailstone
build_sv(start,stop,iteration_limit,C)
returns sequence_vector
calc_XYZ(sequence_vector)
returns hailstone_function_parameters
crossover(hailstone_function_parameters,C)
returns CP_quotient, CP_remainder
zigzag(start,sequence_vector,C,print_options)
returns value_vector
Type12MH(generation,generation_member)
returns ith_kth_generation_Type[1,2]_Mersenne_Hailstone
As I said at the outset, the journey is it's own reward.
Ernst
Oh I don't doubt the Amazing aspect.
I see a few old topics in there.. Good job.
I'm not sure on the Political undertone; I've not read this group for
a while but a quick scan seems that it is very creative.
I doubt I can organize any better than you have although new
information is always a contribution and that's possible.
As I have done in the past ,
eudaemonically.
Ernst