+summary & history of Proof of Full Collatz 3N+-1 simultaneous 3N+-3; disproof of 3N+1 stand alone #177 Correcting Math 4th ed

[Archimedes Plutonium]

I think the best write up is to emit the essentials in as few of words as possible in a summary and then write about the details in a history format.

Summary and history of my Proof of Full Collatz, 3N+-1 simultaneous 3N+-3; and disproof of Old Math's half baked Collatz 3N+1

Summary, quick proof:

Math Corollary: take any four consecutive even numbers, such as say 2, 4, 6, 8 and at least one of those has a factor of 2 three times. In our case it is 8. In the Full Collatz conjecture the choices of 3N with a -3, -1, +1, +3 from 3N+-1 simultaneously 3N+-3 always lands the newly formed even number with one that can be divided by 2 for three times. So, in our multiplication of an odd number by 3 we generate an even number divisible by 8. So Full Collatz always converges to 1 as each even number generated by 3N is lowered in value by 3/8. So let us say we start with 61 in Full Collatz we have 61;; 61, 183 with a choice of 183-3, 183-1, 183+1, 183+3 and we chose the even number with at least three factors of 2. In this case, it is 184 which descends to 23. And 3/8 of 61 is approximately 23. So, no matter what odd number we multiply by 3, we end up dividing the new number by at least 8.

The Old Math Collatz of just 3N+1 can never be proven true or false since no statement can be made of the 3N landing on an even number that is divisible by more than one factor of 2.

Collatz 3n+1 starting with 319804831 reaches a maximum of 1414236446719942480 which in New Math with a infinity borderline at 1*10^604 and its Algebraic Closure at 1*10^1208 would be violated by Old Math's Collatz since the 319804831 is in the 10^9 tier and the run produces numbers beyond the closure of 10^18. This number 319804831 is just one example of a run where the odd number produces a even number that has mostly just one factor of 2, so you have a run that is not 3/8 descending but a run that is mostly 3/2 ascending. The divergence of 5N+5 Collatz is mostly an ascending 5/2 runs.

Details of the history of my proof:

FULL COLLATZ PROOF:

Let me first walk through some of my historical insights I gained, some history of how I arrived at a proof of Collatz, for which I proved in a span of two months here in sci.math. Often the history is more informative and clear than the pure math write up.

The first thing I recognized as I started to look for a proof, is that there is a problem of the original half-baked Collatz in that what makes Collatz work is a Slide, where as soon as one has a even number slide number of 2,4,8,16,32, etc etc, they slide down to 1. But the problem here is that half-baked Collatz ignores mathematic's second primal Slide sequence of 3, 6, 12, 24, 48, etc etc. So I quickly realized why no-one could prove in Old Math the half-baked Collatz for it ignores many of the numbers of math. No proof of 3N+1 is possible when it does not engage in all the numbers of Number theory. And thus, when Full Collatz is proven true we can go back and see why the half baked Collatz has no proof in Old Math.

Alright, so the first thing I saw was two primal Slides in Collatz.

2, 4, 8, 16, 32, 64, 128, etc etc

3, 6, 12, 24, 48, 96, etc etc

Then there are derivative slides as compared to primal slides.

20, 40, 80, etc etc

600, 1200, 2400, etc etc

And, early on, I noticed there was a need for a sequence that has the opposite behaviour of Collatz-- internal trap of spinning around or divergence to infinity. Funny how no mathematician tackling Collatz ever seemed to look for a counterexample sequence like 5N+5 to compare to 3N+1.

Seeing the structure of 5N+5 and wondering how that can have internal spinning around in a trap and how it can diverge to infinity, and wondering why 3N+3 avoids all of that.

So, seeing there has to be two primal slides involved, I proclaimed the real Collatz conjecture involved 3N-3, 3N-1, 3N+1, 3N+3 written as 3N+-1 simultaneously 3N+-3. And that a conjecture on 3N+1 stand alone is not a viable conjecture. It is like saying 5N+5 never converges in any run to 1.

Alright, I should discuss this type of Collatz of 5N+5, for it shows us how a Collatz can have a "spinning around inside trapped in a spin". Such as 9 with 5N+5

9;;9, 50, 25, 130, 65, 330, 165, 830, 415, 2080, 1040, 520, 260, 130, 65, and spinning around endlessly.

Now let us check out 99 with 5N+5 and see if it diverges to infinity.

99;;99, 500, 250, 125, 630, 315, 1580, 790, 395, 1980, 990, 495, and as can be seen it keeps diverging to infinity, never descending.

And 999 is no different

999;;999, 5000, 2500, 1250, 625, 3130, 1565, 7830, 3915, and diverge to infinity.

So in the 5N+5 Collatz we have sequences that as soon as we hit a even number, it quickly ends up with ending digit of "5" taking us to infinity.

Now an interesting question for 5N+5 is whether 9 is some sort of marker where all the odd below 9 of 3, 5, 7 are internal spinning and whether all odd beyond 9, say 11 are divergent? So let me try 11.

11;; 11, 60, 30, 15, 80, 40, 20, 10, 5, 30 and here again we have internal spinning around.

Is there ever a case where 5N+5 converges to 1 like a normal Collatz? I doubt it, because no Slide Numbers of either 2, 4, 8, 16, 32, etc or 3, 6, 12, 24, etc etc or their derivatives, have the form 5N+5.

So in search of the proof of Collatz, I needed the mechanisms of how the runs always converge to 1. What makes the runs converge, is that the Slide Numbers have both the algebraic form of 3N+-1 simultaneously 3N+-3, and is that the only mechanism? For example, 32 is 3N-1 when N is 11, or, 16 is 3N+1 when N=5, or 24 is 3N+3 when N=7. But there is another, bigger mechanism at play in Collatz.

Then a insight hit me from doing the Staircase Conjecture proof.

On Sunday, May 22, 2016 at 2:27:43 AM UTC-5, Archimedes Plutonium wrote:
> Alright, I made great progress on Staircase and is proven true using Goldbach as theorem.
>
> I have not made that advance on Full Collatz. So what I am doing now is trying to capture what makes a proof of Staircase  and apply it to Full Collatz.
>
> What proves Staircase is a substitution of consecutive odd numbers to fill in the largest prime gap in a tier.

In other words, looking for the largest prime gap in each tier of 10, 100, 1000, etc etc. So is there a largest of something in each tier for Collatz, to make more progress?

> And afterwards, replace the odd-composite with a prime number that we are assured of by the Goldbach proof. We prove Staircase true, but we lack the details of how large the toolbox kit is for larger tiers.
>
> So, can I apply any of that to the Full Collatz?
>
> About the only details I can muster from Full Collatz, 3N+-1 simultaneously 3N+-3, or the half-baked Collatz with its 3N+1 alone, is the extent of the largest runs in each tier.
>
> So in Staircase proof I had to focus on the largest Prime Gap. In the Full Collatz, I likewise need something like that, and it appears to be the **largest run**.
>

So I need something solid as the largest run in each tier to prove Collatz. If the largest run is 9 then 99 then 999 then 9999 coinciding with 10, 100, 1000, 10^4 tier etc etc, then Collatz must have a proof. The 9s are not the largest run in each tier for the half-baked Collatz as Wikipedia shows 27.


> Now Wikipedia shows for the half-baked Collatz the large run of 27 which takes 111 steps, but when using the Full Collatz
>

27:: 27, (3*27)-1 =80, 40, 20, 10, 5, (3*5)+1=16, 8, 4, 2, 1 we end up with just 11steps

>
> So, we can ask what is the largest run in each tier of
>
> 100
> 1000
> 10^4
> 10^5
> etc
> etc
>
> If we find the largest run comes from the largest odd number in each tier-- 99 for 100, 999 for 1000.

Alright, looking real good so far. It looks like the largest odd number in a tier produces the largest run in that tier. And if so, then a proof of Full Collatz as true is achievable.

So in tier 10, is 9 making the largest run? Let me see by using just 3N+1 we have 9;;9, 28, 14, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1

Now it would be a waste of time to see if 7, 5 make larger runs since they are included in the 9 run.

Now let me try 99 by using 3N+-3 simultaneously 3N+-1
99;;99, 296, 148, 74, 37, 112, 56, 28, 14, 7, 24 to slide to 3 then 1.

27:: 27, (3*27)-1 =80, 40, 20, 10, 5, (3*5)+1=16, 8, 4, 2, 1

Using Full Collatz on 999
999;; 999, 3000, 1500, 750, 375, 1128, 564, 282, 141, 424, 212, 106, 53, 160, 80, 40, 20, 10, 5 slide to 1

And so I had to discuss the Algebraic Form of Full Collatz is a covering of all the Natural Numbers, excluding 0.

This is what I mean by *fully covering all the Natural Numbers*

Even Schemata

3N+-1       3N+-3

1-> 2,4       1->0,6

3-> 8, 10     3-> 6, 12

5 -> 14, 16   5 -> 12, 18

7 -> 20, 22    7 -> 18, 24

producing all the even numbers, and missing no even numbers for that, 3N+1 alone, misses 6, 12, 24,. .

Odd Schemata

3N+-1              3N+-3

=2 then n = 1        =0 then n=1

=4 then n =1       =6 then n=1

=8 then n=3       =6 then n=3
=10 then n=3     =12 then n=3

=14 then n=5     =12 then n=5
=16 then n=5     =18 then n=5

Showing where all even numbers produce all the odd numbers but if you had just 3N+1, many odd numbers would be missing.

So, half baked Collatz is not about all the Natural Numbers for it misses many even numbers, and as such, cannot be proven in math.

Now, what proved the Staircase conjecture is the tie-in with the Goldbach proof. Is there a tie-in in Collatz that proves it true? Is there a second and larger mechanism than the two Primal Slides discussed?

I found such a tie-in. What I found was a strange little known corollary that 4 consecutive even numbers must always have one of those even numbers with at least three factors of 2.

Now, that fact easily proves Collatz, the full Collatz because it gives a choice among four consecutive even numbers, and we will always pick the even number with the most factors of 2.

I am confident the history of mathematics has more examples than Collatz where a half baked conjecture was formulated which could never be proven true, since it had essential elements missing or misunderstood, or corrupted or contorted essential elements. The four color mapping corrupts borderlines and pretends borders do not exist in maps and thus are not colored; the Kepler packing ignores borders with its borderless notion of infinity; the no odd perfect conjecture ignores the fact that 9 has two 3s as factors not just one 3, so that 1+3 does not represent 9, but rather 3+3+1 represents 9, and missing being odd perfect by 2,  are conjectures missing essential elements and thus never provable until those contorted, corrupted definitions are made precise.

So, here, we need a Full Collatz, 3N+-1 simultaneously 3N+-3, not a half baked Collatz of 3N+1, for a proof of Collatz as I spoke before, is that the odd number multiply by 3 and forming a new even number has at least three divisions by 2. So a descent of at least 3/8 of starting value. If we start with 29 in Full Collatz, the next odd number that shows up is 3/8 x 29 is approx 11, so as seen 29;;29, 88, 44, 22, 11, etc.

Finally a proof emerges by realizing that in the choice of 4 consecutive even numbers that one of those even numbers must have a factor of 2 for three times.

Proof at last Re: 3/8 increments of fall in Collatz Re: Mechanism shown for 75, as to why Collatz works; prelim2inary Full Collatz 3N+-1 simultaneously 3N+-1 proof; disproof of half-baked Collatz 3N+1
Mechanism shown for 75, as to why Collatz works; prelim2inary Full Collatz 3N+-1 simultaneously 3N+-1 proof; disproof of half-baked Collatz 3N+1

Alright, what that mechanism means is given any odd number, say we are given 75, then the even numbers surrounding 75 from 3N-3, 3N-1, 3N+1, 3N+3 will always have a even number that has at minimum three divisions by 2. So for 75 the 3N is 225. And thus we have a choice 3N-3=222, or 3N-1=224 or 3N+1=226, or 3N+3 =228. One of those choices has at least three divisions by 2 and perhaps one has more than two. So, let us see. 222=2*111, 224=2*2*2*2*2*7, 226=2*113, 228=2*2*57.

So here we have the even number 224 which has a 2 as factor for 5 times.

So, it has been proven that in any given odd number larger than 1, that we always have a three factor of 2 even number produced. So that Collatz mechanism is that we multiply by 3 any odd number, but we end up by dividing by at least 8 any even number produced, because one of those even numbers is always divisible by 8.

So the Collatz run can be nicknamed the 3/8 run where the starting number is at least reduced to 3/8 its value upon division by 2 for three times. So that each arbitrary odd number that starts Collatz, except 1, converges to 1 and because of that continual convergence, the descent cannot have a spinning around internally, such as 9 does for 5N+5.

Research
Disclaimer: Due to the unconventional and speculative nature of this posting and thread, it would be inadvisable for students to apply any of the contents to their school course work.

--
Recently I re-opened the old newsgroup PAU of 1990s and there one can read my recent posts without the hassle of spammers, off-topic-misfits, front-page-hogs, stalking mockers, suppression-bullies, and demonizers.      

https://groups.google.com/forum/?hl=en#!forum/plutonium-atom-universe        
Archimedes Plutonium

[Archimedes Plutonium]

On Friday, May 27, 2016 at 12:26:43 AM UTC-5, Archimedes Plutonium wrote:
> I think the best write up is to emit the essentials in as few of words as possible in a summary and then write about the details in a history format.
>

(snipped)

Speaking of history:

Now someone should write a history of the Collatz conjecture to point out how dumb in logic most mathematicians are. To point out that we had a conjecture 70 years old with the math community feverishly trying to prove it, and yet not a single mathematician able to recognize you need the two Slides and you need a mechanism for why the convergence to 1.

What Collatz shows the world is how pitiful dumb are most mathematicians. That they feverishly slave to prove Collatz and yet not a one of them aware of missing the slide 3, 6, 12, 24, . . which would reveal that Collatz needed 3N+-1 along with 3N+-3 and that a Collatz of 3N+1 is a half baked conjecture.

Why in 70 years of scrambling to prove Collatz, why was no dumb clutz math professor able to see a mechanism? A mechanism as simple as Every four consecutive even numbers has a factor of 3 three times?

What were the chasing for in a proof of Collatz, if not that any 4 consecutive even numbers has a 2 factor three times?

So, someone needs to write a history of the attempts by mathematicians to prove Collatz to show the world how dumb and absent of logical reasoning the math community is.

AP

[abu.ku...@gmail.com]

the rule of slides, or just a trinary slide-rule?... anyway,
you should like to prove a)
neccesity and b)
sufficiency

[Archimedes Plutonium]

On Friday, May 27, 2016 at 12:26:43 AM UTC-5, Archimedes Plutonium wrote:
> I think the best write up is to emit the essentials in as few of words as possible in a summary and then write about the details in a history format.
>

Speaking of history:

Now someone should write a history of the Collatz conjecture to point out how dumb in logic most mathematicians are. To point out that we had a conjecture 70 years old with the math community feverishly trying to prove it, and yet not a single mathematician able to recognize you need the two Slides and you need a mechanism for why the convergence to 1.

What Collatz shows the world is how pitiful dumb are most mathematicians. That they feverishly slave to prove Collatz and yet not a one of them aware of missing the slide 3, 6, 12, 24, . . which would reveal that Collatz needed 3N+-1 along with 3N+-3 and that a Collatz of 3N+1 is a half baked conjecture.

Why in 70 years of scrambling to prove Collatz, why was no dumb clutz math professor able to see a mechanism? A mechanism as simple as Every four consecutive even numbers has at least one even number with a factor of 2 three times?

What were they chasing for in a proof of Collatz, if not that any 4 consecutive even numbers has at least one even number with a 2 factor three times?

So, someone needs to write a history of the attempts by mathematicians to prove Collatz to show the world how dumb and absent of logical reasoning the math community often is.

[abu.ku...@gmail.com]

slide, is three doublings?

every sequence of 2n-2, 2n, 2n+2, 2n+4 has a term with a factor of 8?

[abu.ku...@gmail.com]

so, it works for 10, 12, 14, 16, but, so, what

[Archimedes Plutonium]

Now, let me check something out as to how many factors of 2 are within all even numbers.

To prove the Full Collatz of 3N+-1 simultaneously 3N+-3 we observe that within all four consecutive even numbers, at least one of those even numbers has a factor of 2 for three times. So that if we chose 2,4, 6, 8 we have 8 with three factors of 2. If we chose 10, 12, 14, 16 we have 16 as the heaviest factor of 2.

This factor of 2 for 3 times is embedded in the sequence of 8, 16, 24, 32, 40, 48, 56, 64, 72, etc etc

Now the sequence embedding a factor of 2 for 2 times is 4, 8, 12, 16, 20, 24, 28, etc etc

Now the sequence embedding a factor of 2 for 1 times, is obviously every even number, and is uninteresting.

So the Old Math Collatz of solo 3N+1 has to land on an even number of a arbitrary 4 consecutive even numbers, say it lands on 10 which has only 1 factor of 2 and in every 4 consecutive even numbers, at least two of them have only one factor of 2. Is that true? Let me check that out. For of course, 10 and 14 have only one factor of 2 for 10, 12, 14, 16.

How about 996, 998, 1000, 1002, do two of those numbers have just 1 factor of 2?
For 996 is 2*2*249, 998 is 2*499, 1000=2*2*2*125, and 1002 is 2*501. So here we have two of the 4 consecutive even numbers as having just one factor of 2.

Now, is that true for all Consecutive Four Even numbers that two of them have just one factor of 2?

I believe so, and the proof is noting this pattern: This factor of 2 for 3 times is embedded in the sequence of 8, 16, 24, 32, 40, 48, 56, 64, 72, etc etc And the sequence embedding a factor of 2 for 2 times is 4, 8, 12, 16, 20, 24, 28, etc etc So what is remaining or left over are the even numbers 2, 6, 10, 14, 18, 22, 26, 30, etc etc

And, as we can see, that every four consecutive even numbers, must have Two Even numbers with only 1 factor of 2.

So in the Old Math Collatz, the 3N+1 has no guarantee that it can converge to 1 because the odd number can always generate a even number with only one factor of 2.

In the Full Collatz of 3N+-1 simultaneously 3N+-3, every odd number generates a even number that is divisible by at least 8.

Now a Collatz of 3N+-1 is far better because one of those lands on a even number with at least a factor of 2 twice. So for example 11 in 3N is 33 and -1 is 32 while +1 is 34.

So, is a conjecture of a Collatz of 3N+-1 proveable? Here we know that every even number generated has at least two factors of 2 and thus our 3 times is mitigated by division of 4. Thus a continual convergence to 1 and so this Collatz of 3N+-1 is proveable.

AP

[abu.ku...@gmail.com]

yeah, multiples of eight, but 2^4 = 4^2

[Archimedes Plutonium]

So the sequence of even numbers with only one factor of 2 is this sequence:

2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42, 46, 50, 54, 58, 62, 66, 70, 74, 78, 82, 86, 90, 94, etc etc

So, whenever a Collatz odd number forms one of these even numbers, it diverges to infinity since the multiplication of 3 is met by only one division by 2. And so, there is no way in proving that 3N+1 stand alone is going to have all runs converge to 1, or, in fact whether 3N-1 alone or 3N-3 alone or 3N+3 alone will have all runs converge to 1.

If you had 3N+1 simultaneous with 3N-1 you can prove that all runs converge to 1.
If you had 3N+3 simultaneous with 3N-3, you can prove that all runs converge to 1.
If you had 3N+-1 simultaneous with 3N+-3 you can prove that all runs converge to 1.

1) Now, how about 3N-1 simultaneous with 3N+3?
2) Or, how about 3N-3 simultaneous with 3N+1?

Well if we landed on 5 we have 3*5 = 15 and (1) yields 14 and 18.
Well if we landed on 11 we have 3*11 = 33 and (2) yields then 30 and 34.

So, if a Collatz formula is proveable or not proveable depends on whether it avoids even numbers with only one factor of 2.

4Research
4Disclaimer: Due to the unconventional and speculative nature of this posting and thread, it would be inadvisable for students to apply any of the contents to their school course work.
4
4AP

[abu.ku...@gmail.com]

sounda like a program; so,
just do it

[Archimedes Plutonium]

Now something I do not have time to do, is to see how the solo 3N+1 compares to the solo 3N+3. In many of my examples, the 3N+3 seemed to converge to 1 faster than the 3N+1.

Now of course, any of these take solo-- 3N+1, 3N-1, 3N+3 or 3N-3 is unproveable in math because a solo formula has a 50% chance on landing on a One Factor of 2 even number, so that half the time it is ascending and half the time descending. Whereas all four is proveable and a guaranteed 3/8 or better descent in every odd number transformed. And a 3N+1 simultaneously 3N-1 or 3N+3 simultaneously 3N-3 is also guaranteed proveable and a descent of at least 3/4. Not sure about a mixed bad of 3N+3 with 3N-1 or other variants, since they may land on the two even numbers with only One Factor of 2 available.

The entire Collatz conjecture is conquered, vanquished once we realize that given any four consecutive Even numbers say 18, 20, 22, 24, that one of them has at least two or more factors of 2 and one has 2 factors of 2 and the two remaining have only one factor of 2. So, the solo 3N+1 is not proveable whether it is true or false. But the Collatz of 3N+-1 simultaneous 3N+-3, or 3N+-1 or 3N+-3 are proveable.

On Sunday, October 5, 2014 at 2:25:34 AM UTC-5, David Bernier wrote:
(snipped)
>
> In August 2011, Terry Tao wrote a Blog Post about the Collatz
> Conjecture and related questions.
>
> He sketches a proof of
>
> Weak Collatz ==>
> exponential separation between powers of 2 and powers of 3 ,
>
> in the sense of Proposition 7 here:
> <
> https://terrytao.wordpress.com/2011/08/25/the-collatz-conjecture-littlewood-offord-theory-and-powers-of-2-and-3/#qak
> >
>
> I don't understand the proof. It involves additive combinatorics
> and something called Bohr sets.

So, then, we have Terry Tao, stuck on technical nonsense that is totally unrelated to proving Collatz.

When all he needed to do was focus on Four Consecutive Even Numbers. You see, Terry is a computation nerd, while AP is a math genius whose has the Logical mind, not the frivolous computation mind on Collatz.

What is funny about Terry Tao, is that the math community crowds around him for insights into Collatz for which Terry can never handle or prove, while AP proves it and yet no-one in the math community giving AP the credit I deserve. But that is okay with me, because people hate a genius for they are reminded of how dumb they are in logic and math. So, give Terry more honors and awards for doing worthless math, while AP proves another outstanding math conjecture.

What additive combinatorics and Bohr set is Bernier and Tao fascinated with? Bernier and Tao, is it any easier than four consecutive even numbers? Or is that too, too Strong proof of Collatz?

[David Bernier]

Well, I disagree. "Weak Collatz" is weaker than "Full Collatz". Yet, as
Terry Tao showed, "Weak Collatz" implies that, in a sense made precise
on Tao's Blog, powers of two and powers of three are "well-seprated",

cf.:

https://terrytao.wordpress.com/2011/08/25/the-collatz-conjecture-littlewood-offord-theory-and-powers-of-2-and-3/

N.B. Proposition 7:

"Weak Collatz" implies:

" there exists an epsilon > 0 and a constant C >0 such that
for any a, k >= 1,
if 2^a > 3^k, then : | 2^a - 3^k | > C* (1+epsilon)^k . "

It can be useful to take the log of the RHS:

log( C* (1+epsilon)^k ) = log(C) + k*log(1+epsilon) ,
which grows approximately linearly with k.

Tao cites/mention Baker's bound (due to Fields medallist Alan Baker)

( https://en.wikipedia.org/wiki/Alan_Baker_%28mathematician%29 )

in (7)

for a, k >= 1 let q(a, k) := | 2^a - 3^k |.

Then, there exists an absolute constant D >0 such that
| 2^a - 3^k | >= D* (2^a)/(a^C') .

If one C' works, so does C' +1, so w.l.o.g. ,
C' > 0, and 'a' can be imagined as a rather large natural number.
Cases of "small" 'a' can be taken care of by setting D = 10^100,
and the remaining cases are all in the Realm of "very large"
powers of 2, 3, respectively 2^a and 3^k ...

Now for the natural log of Baker's Bound:

log( D* (2^a)/(a^C') ) = log(D) + a*log(2) - C'*log(a).

Asymptotically, log(D) + a*log(2) - C'*log(a)
grows like a*log(2), i.e.
lim_{a -> oo} (log(D) + a*log(2) - C'*log(a))/(a*log(2)) = 1.

So, asymptotically, log( D* (2^a)/(a^C') ) grows linearly with a .

Baker's bound is very strong, it seems to me.

In particular, unless I'm mistaken, Baker's bound shows that:

"There exists an epsilon > 0 and a constant C >0 such that
for any a, k >= 1,
if 2^a > 3^k, then : | 2^a - 3^k | > C* (1+epsilon)^k . "

(comments welcome on Baker's Bound ==> The Above).



In that case, a " Big Theorem " which is known, is enough

to show:

"There exists an epsilon > 0 and a constant C >0 such that
for any a, k >= 1,
if 2^a > 3^k, then : | 2^a - 3^k | > C* (1+epsilon)^k . " (***)


However, I have no idea of my own on how to show (***).

(***) is a consequence of Weak Collatz, which itself is a consequence
of Full Collatz.

Weak Collatz is Conjecture 2 in Tao's Blog.
Weak Collatz says that any cycle in the Collatz iterations is of the
form: (4 -> 2 -> 1 -> 4 -> 2 -> 1 -> 4 -> 2 -> 1 ... )
and that no other finite period cycles exist.

Even if Weak Collatz is "granted",
supposing T(.) denotes the Collatz function:

T(n) = n/2 if n == 0 (mod 2) and
T(n) = 3n +1 if n == 1 (mod 2),

the existence of an unbounded sequence
(a_1, a_2, a_3, ... a_k, a_{k+1} ..... ) in N\{0}
such that T(a_1) = a_2,
T(a_2) = a_3,
...
in general:
T(a_k) = a_{k+1} for all k in N\{0}

*is not* ruled out.

Such an unbounded sequence of Collatz iterates starting
from some number a_1 and onwards with iterates a_2, a_3, ...
isn't proven impossible, with ease, just from assuming
the Weak Collatz Conjecture .


IOW, "Weak Collatz" is much weaker than "Full Collatz".
Secondly, "Weak Collatz" is still an open problem.

David Bernier

--
pub 4096R/41C769D6 2014-11-04
uid David Bernier (doubledeckerpot5)
<davi...@videotron.ca>

[David Bernier]

Rather, D = 1/(10^100).

Then,
| 2^a - 3^k | >= (1/(10^100))* (2^a)/(a^C')

with a >= 1, C'>0 fixed, is trivially true
for the 'a' where (2^a) < 10^100:

If (2^a) < 10^100, then

(1/(10^100))* (2^a)/(a^C') < 1* 1/(a^C').

a >= 1, C' >=0, so a^C' >= 1, whence
(1/(10^100))* (2^a)/(a^C') < 1*1 = 1, so
(1/(10^100))* (2^a)/(a^C') < 1, and
a fortiori, by the Uniqueness of Factorization into primes,

| 2^a - 3^k | >= 1 and
1 > (1/(10^100))* (2^a)/(a^C'),
so | 2^a - 3^k | > (1/(10^100))* (2^a)/(a^C') or
| 2^a - 3^k | > D* (2^a)/(a^C') ,
just as long as: 2^a < 10^100.


It's true that I don't understand the proof of Proposition 7.
I take it on faith and trust, given Tao's reputation, that
he gives references, and that that Blog post had 101 comments.

After the square sign that marks the QED for Proposition 7,
Tao writes in part:

" Thus we see that any proposed proof of the Collatz conjecture must
either use transcendence theory, or introduce new techniques that are
powerful enough to create exponential separation between powers of 2 and
powers of 3. "

In the following paragraph, he continues:

" Unfortunately, once one uses the transcendence theory bound (7), the
size [...] "

Formula (7) is the very powerful Baker Bound,

For a, k >= 1 there exist absolute constants D >0 and C' >0

such that: | 2^a - 3^k | >= D* (2^a)/(a^C') .

It seems to be the case, or I take it that, even using
the powerful Baker Bound from transcendence theory, the
Weak Collatz Conjecture isn't within Tao's reach.

Plus, the comments on Tao's blog posts are usually worth
reading or browsing.

I don't know why you think T. Tao does pitiful, frivolous or irrelevant

math; however since you also wrote:

"You see, Terry is a computation nerd, while AP is a math genius whose

has the Logical mind, not the frivolous [...] "

I take it your world view is that you are one of the top
mathematical minds in the world, and I think that says a lot.

[abu.ku...@gmail.com]

just needed that simple statement about T(n);
2 = 2 mod two

[Archimedes Plutonium]

Look how David purposely messes up by using "Full" rather than the correct use of Strong versus Weak. David does this because he wants to attack me by thinking he can throw me off-course in having to straighten out Full from Strong.

What David should be talking about in his first sentence is the fact that every 4 consecutive even numbers, pick any 4 at random that are consecutive, and two of them have only one factor of 2 and will destroy any attempted proof of 3N+1. But David is not smart in math, nor is Terry Tao as they continually glum onto irrelevant facts and ideas on solving Collatz.

Every four consecutive even numbers has two of them that are **only one factor of 2** while the other two have more than one factor of 2.

2, 4, 6, 8 where 2 and 6 have one factor of 2
then
4, 6, 8, 10 where 6 and 10 have one factor of 2
then
6, 8, 10, 12 where 6 and 10 have one factor of 2
then
8, 10, 12, 14 where 10 and 14 have one factor of 2
then
10, 12, 14, 16 where 10 and 14 again with only one factor of 2
then
12, 14, 16, 18 where 14 and 18 with only one factor of 2

Any Four Consecutive Even Numbers are always having two of them with just one factor of 2.

So, when you make Collatz be 3N+1 simultaneous 3N-1 you can prove it for one falls on a even with more than one factor of 2

When you make Collatz be 3N+3 simultaneous with 3N-3 you can prove it since one of those even will be of more than one factor of 2

When you make Collatz be 3N+-1 simultaneous with 3N+-3 you can prove it

But when you make Collatz be 3N+1 no proof is ever possible, because a solo formula can always have an odd number land on a even number with just one factor of 2.

> cf.:
>
> https://terrytao.wordpress.com/2011/08/25/the-collatz-conjecture-littlewood-offord-theory-and-powers-of-2-and-3/
>
> N.B. Proposition 7:
>
> "Weak Collatz" implies:
>
> " there exists an epsilon > 0 and a constant C >0 such that
> for any a, k >= 1,
> if 2^a > 3^k, then : | 2^a - 3^k | > C* (1+epsilon)^k . "
>
> It can be useful to take the log of the RHS:
>

Only if you are stupid in math, chasing after the Collatz, do you do things like this. To hide the fact that you have no clue as to the mechanism of Collatz, that a odd number has to land on either one of 4 consecutive even numbers for which two of those have only one factor of 2.

> log( C* (1+epsilon)^k ) = log(C) + k*log(1+epsilon) ,
> which grows approximately linearly with k.
>
> Tao cites/mention Baker's bound (due to Fields medallist Alan Baker)
>
> ( https://en.wikipedia.org/wiki/Alan_Baker_%28mathematician%29 )
>

So David wants to elevate everything that Baker and Tao do and say, even when Baker and Tao are too stupid to know that every four consecutive even numbers has two numbers that have more than one factor of 2, and is the key to proving any type of Collatz conjecture. David will read a 50 page silly article by Tao that never solves Collatz, and yet David will not read one sentence by AP about 4 consecutive even numbers.

> in (7)
>
> for a, k >= 1 let q(a, k) := | 2^a - 3^k |.
>
> Then, there exists an absolute constant D >0 such that
> | 2^a - 3^k | >= D* (2^a)/(a^C') .
>
> If one C' works, so does C' +1, so w.l.o.g. ,
> C' > 0, and 'a' can be imagined as a rather large natural number.
> Cases of "small" 'a' can be taken care of by setting D = 10^100,
> and the remaining cases are all in the Realm of "very large"
> powers of 2, 3, respectively 2^a and 3^k ...
>
> Now for the natural log of Baker's Bound:
>

When does Baker talk about 4 consecutive even numbers?

> log( D* (2^a)/(a^C') ) = log(D) + a*log(2) - C'*log(a).
>
> Asymptotically, log(D) + a*log(2) - C'*log(a)
> grows like a*log(2), i.e.
> lim_{a -> oo} (log(D) + a*log(2) - C'*log(a))/(a*log(2)) = 1.
>
> So, asymptotically, log( D* (2^a)/(a^C') ) grows linearly with a .
>
> Baker's bound is very strong, it seems to me.
>

For David, strong is being blind to 4 consecutive even numbers.

> In particular, unless I'm mistaken, Baker's bound shows that:
>
> "There exists an epsilon > 0 and a constant C >0 such that
> for any a, k >= 1,
> if 2^a > 3^k, then : | 2^a - 3^k | > C* (1+epsilon)^k . "
>
> (comments welcome on Baker's Bound ==> The Above).
>
>

What is the bound on 4 consecutive even numbers, where two of those even have just one factor of 2.

>
> In that case, a " Big Theorem " which is known, is enough
>
> to show:
>
> "There exists an epsilon > 0 and a constant C >0 such that
> for any a, k >= 1,
> if 2^a > 3^k, then : | 2^a - 3^k | > C* (1+epsilon)^k . " (***)
>
>
> However, I have no idea of my own on how to show (***).
>

Why would anyone want an idea that has nothing to do with proving Collatz.

David, you are not a mathematician, for you cannot even read what another says about Collatz and comprehend what he says. You seem to be able to only parrot what you perceive are mathematicians. You have no mind of yourself.

You still do not know, nor accept the fact that 4 consecutive even numbers has two of them with just one factor of 2.

AP

[Archimedes Plutonium]

It is true that both David and Terry do not know nor acknowledge that 4 consecutive even numbers always has two of them with only one factor of 2 and thus allowing only proofs of Collatz be made on formulas that guarantee landing on even numbers with more than one factor of 2 such as 3N+-1 or 3N+-3 or Full Collatz of 3N+-1 simultaneously 3N+-3.

> After the square sign that marks the QED for Proposition 7,
> Tao writes in part:
>
> " Thus we see that any proposed proof of the Collatz conjecture must
> either use transcendence theory, or introduce new techniques that are
> powerful enough to create exponential separation between powers of 2 and
> powers of 3. "
>
> In the following paragraph, he continues:
>
> " Unfortunately, once one uses the transcendence theory bound (7), the
> size [...] "
>

David, why does not Terry Tao know that every 4 consecutive even numbers has two even numbers with only one factor of 2? Is that math too advanced for Terry? Does Princeton Univ not teach its students about consecutive even numbers?

> Formula (7) is the very powerful Baker Bound,
>
> For a, k >= 1 there exist absolute constants D >0 and C' >0
> such that: | 2^a - 3^k | >= D* (2^a)/(a^C') .
>
> It seems to be the case, or I take it that, even using
> the powerful Baker Bound from transcendence theory, the
> Weak Collatz Conjecture isn't within Tao's reach.
>

Terry Tao and Baker do what many modern academicians do when they have no idea of a proof. They hide their work into arcane, bizarre contraptions such as transcendence theory which is a catch word for garbage math.

> Plus, the comments on Tao's blog posts are usually worth
> reading or browsing.
>

Terry has never proven one single outstanding math conjecture, so that explains why the nonmath of David Bernier praises Terry.

> I don't know why you think T. Tao does pitiful, frivolous or irrelevant
> math; however since you also wrote:
>

David never seemed to go to a baseball game where you keep score on a scorecard. So let me see, what proof did Terry Tao of great conjectures ever prove? Did he prove No Odd Perfect? Did he prove twin primes infinite? Did he prove Collatz? Did he prove Goldbach? Did he prove Kepler Packing? Did he prove the true FLT, not the phony Wiles? No, AP proved these, and Terry never proved anything famous of conjectures, and this is why David Bernier follows him.

The only thing Terry has shown us in math, is how to become awarded in math for doing nothing but lead astray and producing more phony math results with a phony infinity, a infinity without a borderline.

> "You see, Terry is a computation nerd, while AP is a math genius whose
> has the Logical mind, not the frivolous [...] "
>
> I take it your world view is that you are one of the top
> mathematical minds in the world, and I think that says a lot.
>

What says alot is that David Bernier reads the AP Collatz proof and never once recognizes the fact that 4 consecutive even numbers has two of them with only one factor of 2. Never once tries to understand Collatz in that light of 4 consecutive even numbers. Is David that pitifully stupid in math that he cannot even force himself to write the fact of 4 consecutive even numbers, and so full of hatred of AP that he will never acknowledge 4 consecutive even numbers.

> David Bernier
>
> --
> pub 4096R/41C769D6 2014-11-04
> uid David Bernier (doubledeckerpot5)
> <davi...@videotron.ca>

I do not know which of these two are more blind at math-- David Bernier or Terry Tao.

Will neither of the two say-- in any 4 consecutive even numbers, two of them will have just one factor of 2. Come on Terry, come on David-- you can do it.

Say-- Any arbitrary four

consecutive, (that is right) any four consecutive even numbers, (come on, you can do it)

AP

[konyberg]

Why do you list several examples of your discovery? Why not state that every other natural number is odd?

KON

[Archimedes Plutonium]

On Wednesday, June 1, 2016 at 11:11:53 AM UTC-5, David Bernier wrote:
> On 06/01/2016 09:48 AM, David Bernier wrote:

Snipped Bernier's low opinion of AP, and his refusal to examine a proof of Collatz by AP.

Bernier is the typical person in the math community, judging from my 23 year experience with mathematicians in sci.math, where they refuse to ever examine a good proof of Collatz that is true yet whose author they hate. Refuse to look at the proof because they hate the author. Yet for an author they admire, such as Bernier admiring Tao, Bernier bends over backward to post and hang on every word of that author.

It is a sad reflection that mathematicians need reason and logic to do math, but where most mathematicians run on the fuel of hatred, rather than reason and logic. I do not really know why they ever bother to go into science or math, because they are crippled in the mind, from the start.

AP

[abu.ku...@gmail.com]

just dOO it

[Archimedes Plutonium]

On Wednesday, June 1, 2016 at 2:05:31 PM UTC-5, Archimedes Plutonium wrote:
(snipped)

>
> 2, 4, 6, 8 where 2 and 6 have one factor of 2
> then
> 4, 6, 8, 10 where 6 and 10 have one factor of 2
> then
> 6, 8, 10, 12 where 6 and 10 have one factor of 2
> then
> 8, 10, 12, 14 where 10 and 14 have one factor of 2
> then
> 10, 12, 14, 16 where 10 and 14 again with only one factor of 2
> then
> 12, 14, 16, 18 where 14 and 18 with only one factor of 2
>
> Any Four Consecutive Even Numbers are always having two of them with just one factor of 2.
>

Now here is a question for Don of red...@siu.edu to weigh in and answer, since both David Bernier and Terry Tao are too dumb to comprehend and answer.

The question is whether we can have a formula for an odd number that lands, always, only on a even number with just one factor of 2.

Don, we know that every arbitrary 4 consecutive even numbers has two of them as one factor of 2. Suppose we pick at random these 4 even numbers. 1832, 1834, 1836, 1838. Now just looking at them we would be hard pressed to tell which of those two are the one factor of 2, which of those is the 2 or more factors of 2 and which is the richest in factors of 2 with at least 3 factors of 2 or more. So let us break that set down into components and we have 1832= 2*2*2*229, then 1834=2*917. And here we stop momentarily for we know that 1838 is going to be the guaranteed other even number with just one factor of 2. So for 1836= 2*2*459 and finally, 1838=2*919.

So, the question I want to ask Don, since in the past he has been alert on algebraic form. The question is that 3N+1 or 3N+3 or 3N-1 or 3N-3, when each of those is taken solo, that they have a 50 to 50 chance of landing on a one factor 2 or landing on two or more factors of 2 of any given block of 4 consecutive even numbers.

So the question is Don, is there a formula that guarantees landing always on one of the one factors of 2? Now the formula 5N+5 often lands on a one factor of 2 such as 5*5+5 = 30 but sometimes on a two factor of 2 such as 5*3+5 = 20 which is a two factor of 2.

Don, is there a formula that always, and only lands on an even number with one factor of 2.

The sequence I am striving to reach for one factor of 2 is this: 2, 6, 10, 14, 18, 22, etc etc. It is the sequence of all even numbers with just one factor of 2.

So is there a Collatz type formula that always lands on a even number of that sequence. The 5N+5 comes close but not close enough.

What do you think Don?

AP

[Earle Jones27]

On 2016-05-27 05:26:35 +0000, Archimedes Plutonium said:

Archie: When setting out to prove a conjecture, it would be a good
idea to state exactly what the conjecture is: (Wikipedia is your
friend.)

Take any positive integer n. If n is even, divide it by 2 to get n / 2.
If n is odd, multiply it by 3 and add 1 to obtain 3n + 1. Repeat the
process indefinitely. The conjecture is that no matter what number you
start with, you will always eventually reach 1.

Now, proceed with your proof:

earle
*

> Now let me try 99 by using 3N+-3 simultaneously 3N+-199;;99, 296, 148,

> 74, 37, 112, 56, 28, 14, 7, 24 to slide to 3 then 1.
> 27:: 27, (3*27)-1 =80, 40, 20, 10, 5, (3*5)+1=16, 8, 4, 2, 1

> Using Full Collatz on 999999;; 999, 3000, 1500, 750, 375, 1128, 564,

> 282, 141, 424, 212, 106, 53, 160, 80, 40, 20, 10, 5 slide to 1
> And so I had to discuss the Algebraic Form of Full Collatz is a
> covering of all the Natural Numbers, excluding 0.
> This is what I mean by *fully covering all the Natural Numbers*
> Even Schemata
> 3N+-1       3N+-3
> 1-> 2,4       1->0,6
> 3-> 8, 10     3-> 6, 12
> 5 -> 14, 16   5 -> 12, 18
> 7 -> 20, 22    7 -> 18, 24
> producing all the even numbers, and missing no even numbers for that,
> 3N+1 alone, misses 6, 12, 24,. .
> Odd Schemata
> 3N+-1              3N+-3
> =2 then n = 1        =0 then n=1
> =4 then n =1       =6 then n=1

> =8 then n=3       =6 then n=3=10 then n=3     =12 then n=3
> =14 then n=5     =12 then n=5=16 then n=5     =18 then n=5

> 3N+1Mechanism shown for 75, as to why Collatz works; prelim2inary Full

> Collatz 3N+-1 simultaneously 3N+-1 proof; disproof of half-baked
> Collatz 3N+1
> Alright, what that mechanism means is given any odd number, say we are
> given 75, then the even numbers surrounding 75 from 3N-3, 3N-1, 3N+1,
> 3N+3 will always have a even number that has at minimum three divisions
> by 2. So for 75 the 3N is 225. And thus we have a choice 3N-3=222, or
> 3N-1=224 or 3N+1=226, or 3N+3 =228. One of those choices has at least
> three divisions by 2 and perhaps one has more than two. So, let us see.
> 222=2*111, 224=2*2*2*2*2*7, 226=2*113, 228=2*2*57.
> So here we have the even number 224 which has a 2 as factor for 5 times.
> So, it has been proven that in any given odd number larger than 1, that
> we always have a three factor of 2 even number produced. So that
> Collatz mechanism is that we multiply by 3 any odd number, but we end
> up by dividing by at least 8 any even number produced, because one of
> those even numbers is always divisible by 8.
> So the Collatz run can be nicknamed the 3/8 run where the starting
> number is at least reduced to 3/8 its value upon division by 2 for
> three times. So that each arbitrary odd number that starts Collatz,
> except 1, converges to 1 and because of that continual convergence, the
> descent cannot have a spinning around internally, such as 9 does for
> 5N+5.

> ResearchDisclaimer: Due to the unconventional and speculative nature of

[konyberg]

Try 2(2n - 1) for n = 1, 2, 3, ...

KON

[Archimedes Plutonium]

On Thursday, June 2, 2016 at 12:00:33 AM UTC-5, Earle Jones27 wrote:
> On 2016-05-27 05:26:35 +0000, Archimedes Plutonium said:
>
> Archie: When setting out to prove a conjecture, it would be a good
> idea to state exactly what the conjecture is: (Wikipedia is your
> friend.)
>

Sad that the standards of math in New Zealand are so pitifully low that this birdbrain gets a degree in math.

Where you have to teach the birdbrain English and Math in every thing you do:

I wrote:
Summary and history of my Proof of Full Collatz, 3N+-1 simultaneous 3N+-3; and disproof of Old Math's half baked Collatz 3N+1

A person with average intelligence can easily see Full Collatz and half baked Collatz, what they are.

I have found over the 23 years in sci.math, that people like Jones, a hate monger, attacks those he hates by always nit-picking and never able to serious consider what the person is doing.

Full Collatz: if you land on an odd number, you take four calculations-- 3N-1, 3N+1, 3N-3, 3N+3. You find the even number with the largest number of factors of 2 and proceed further, deleting the other three.

Half Baked Collatz: is the solo formula 3N+1 and is what Old Math calls Collatz, but because it is unprovable, it should be called the Half Baked or Half Arse Collatz. The Full Collatz shows why 3N+1 is unprovable.

But not this birdbrain in New Zealand is able to understand English and Math. He just wastes the time of everyone involved.

The math standards in New Zealand must be one of the lowest in the world judging from the posts of this birdbrain.

> Take any positive integer n. If n is even, divide it by 2 to get n / 2.
> If n is odd, multiply it by 3 and add 1 to obtain 3n + 1. Repeat the
> process indefinitely. The conjecture is that no matter what number you
> start with, you will always eventually reach 1.
>
> Now, proceed with your proof:
>
> earle

Now, do your students a favor and proceed to resign from teaching math which you know so little of.

Not worth saving, since it is people talk, and over half of sci.math posters are just people talkers

[Archimedes Plutonium]

Come to think of it, I can now appreciate why David Bernier would worship and hang on every word of Terry Tao about Collatz, Weak or whatever Collatz, when you read just one post of Jones on Collatz, that Bernier has an economy of math even though Tao's understanding of Collatz is muddled. If I were Bernier and read a post of Jones, I too would find some relief in Tao. Even though, none will get anywhere in math.

Not worth saving since just some more people talk.

And, I do not think New Zealand ever produced any great scientist of note. Even Rutherford had to get out of the country to revitalize himself in science.

[Archimedes Plutonium]

Alright Don, I did your work for you. You have to be faster in the future.

I think that 4N+2 for a Collatz will always land on a even number that has just one factor of 2 and hence, immediately diverging to infinity.

For 1 we have 4*1 +2 = 6

For 3 we have 4*3 +2 = 14

For 5 we have 4*5 +2 = 22

and our sequence of Evens with only one factor of 2 is 2, 6, 10, 14, 18, 22, 26, 30, 34, etc etc

So, unless, I made some mistake, we have a Collatz that always diverges.

Now, can we in a solo formula capture a Collatz that always converges?

Don, here I would need a formula for this sequence

4, 8, 12, 16, 20, 24, 28, etc etc

or a formula for this sequence which converges faster

8, 16, 24, 32, 40, etc etc

let me see, the formula 8N +8

For 1 we have 16
For 3 we have 32
For 5 we have 48

Looking good.

So, for a Collatz of 8N+8 we have maths most extreme convergence to 1.

Would you agree Don?

AP







> AP

[Richard Tobin]

In article <5421c523-bf18-4281...@googlegroups.com>,

Archimedes Plutonium <plutonium....@gmail.com> wrote:

>Don, we know that every arbitrary 4 consecutive even numbers has two of
>them as one factor of 2. Suppose we pick at random these 4 even numbers.
>1832, 1834, 1836, 1838. Now just looking at them we would be hard
>pressed to tell which of those two are the one factor of 2, which of
>those is the 2 or more factors of 2 and which is the richest in factors
>of 2 with at least 3 factors of 2 or more.

Since 100 is divisible by 4, obviously 1832 and 1836 are divisible
by 4 and the other two aren't.

200 is divisible by 8, so 1832 is divisible by 8 and 1836 isn't.

-- Richard

[konyberg]

I think you have totally misunderstood Collatz' conjucture!
Please read this: https://en.wikipedia.org/wiki/Collatz_conjecture

KON

[abu.ku...@gmail.com]

n.b, yhr collatz fractal is populated by the bugfeature (or
blackholes?

[konyberg]

torsdag 2. juni 2016 09.13.57 UTC+2 skrev Archimedes Plutonium følgende:

I don't think you knew this simple formula. 2, 6, 10, ... = 2(2n-1), n=1,2,3,....
Why would you otherwise keep on giving examples?

>
> I think that 4N+2 for a Collatz will always land on a even number that has just one factor of 2 and hence, immediately diverging to infinity.
>
> For 1 we have 4*1 +2 = 6
>
> For 3 we have 4*3 +2 = 14
>
> For 5 we have 4*5 +2 = 22
>
> and our sequence of Evens with only one factor of 2 is 2, 6, 10, 14, 18, 22, 26, 30, 34, etc etc
>
> So, unless, I made some mistake, we have a Collatz that always diverges.

Is 2, 6, 10, ... a Collatz sequence? I don't think so!
If you start with 2, then the sequence is 2, 1. No more, no less!

>
> Now, can we in a solo formula capture a Collatz that always converges?
>
> Don, here I would need a formula for this sequence
>
> 4, 8, 12, 16, 20, 24, 28, etc etc

Isn't that 4n, n=1, 2, 3, ...?

>
> or a formula for this sequence which converges faster
>
> 8, 16, 24, 32, 40, etc etc

An isn't that 8n, for n=1,2,3,...

>
> let me see, the formula 8N +8

No good!

>
> For 1 we have 16
> For 3 we have 32
> For 5 we have 48
>
> Looking good.
>
> So, for a Collatz of 8N+8 we have maths most extreme convergence to 1.
>
> Would you agree Don?

I don't know if Don agrees. But I don't!
If you mean a starting number that converges quickly, then use 2^n, n=1,2,3,...

So 48 isn't good.
>
> AP
>
>
>
>
>
>
>
> > AP

Why do you want number of 2, 6, 10, 14, .. as a starting number for Collatz sequences? They just add an element to the sequences!
Start with the odd numbers: 1, 3, 5, 7, 9, ..., 2n -1, n = 1, 2, 3, ...

An example C(14):
14 gives 14/2=7
3*7+1=22
22/2=11
3*11+1=34
34/2=17
3*17+1=52
52/2=26
26/2=13
3*13+1=40
40/2=20
20/2=10
10/2=5
3*5+1=16
16/2=8
8/2=4
4/2=2
2/2=1

So the Collatz sequence for the starting number 14 is:
C(14) = 14, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1

Can you figure it out?

KON

[abu.ku...@gmail.com]

polling all Donalds, except for one who'se on television

[konyberg]

Turn Collatz around and you have a fractal! Good one :)

KON

[abu.ku...@gmail.com]

the black areas in the fractal are artifacts of floatingpointies

[konyberg]

Hm, I would turn it around. The black (in Mandelbrot) is the set. The black area converges quickly, before the floating point errors. If you enlarge the set (going deeper in) then errors will creep in (in the colored parts). However, this can be avoided (not for ever).

KON

[Archimedes Plutonium]

Where in college, in mathematics, is it taught that 4 consecutive even numbers has two of those numbers with only one factor of 2, one with two factors of 2 and one with three or more factors of 2? Or is that fact no longer taught in schools?

AP

[abu.ku...@gmail.com]

this is like a.p and factors of two;
it doesn't really work in base_one, except maybe
by having two lines of the digit